Maths Study Material - Three Dimentional Geometry

THREE DIMENSIONAL GEOMETRY Coordinate of a point in space

There are infinite number of points in space. We want to identify each and every point of space with the help of three mutually perpendicular coordinate axes OX, OY and OZ. Three mutually perpendicular lines OX, OY, OZ are considered as the three axes. The plane formed with the help of x and y axes is called x-y plane, similarly y & z axes form y-z plane and z and x axes form z - x plane. Consider any point P in the space, Drop a perpendicular from that point to x -y plane, then the algebraic length of this perpendicular is considered as z-coordinate and from foot of the perpendicular drop perpendiculars to x and y axes. These algebraic lengths of perpendiculars are considered as y and x coordinates respectively.

1.

Vector representation of a point in space If coordinate of a point P in space is (x, y, z) then the position vector of the point P with respect to the same origin is x î + y ˆj + z kˆ .

2.

Distance formula

Distance between any two points (x 1, y1, z 1) and (x 2, y2, z2) is given as

( x 1 − x 2 ) 2 + ( y 1 − y 2 ) 2 + ( z1 − z 2 ) 2

Vector method We know that if position vector of two points A and B are given as OA and OB then

AB = | OB – OA | ⇒

3.

AB = |(x 2i + y2 j + z2k) – (x 1i + y1j + z1k)| ⇒AB =

( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 + ( z 2 − z1 ) 2

Distance of a point P from coordinate axes Let PA, PB and PC are distances of the point P(x, y, z) from the coordinate axes OX, OY and OZ respectively then

PA = y 2 + z 2 , PB = z 2 + x 2 , PC = x 2 + y 2 Example : Show that the points (0, 7, 10), (– 1, 6, 6) and (– 4, 9, 6) form a right angled isosceles triangle. Solution: Let A ≡ (0, 7, 10), B ≡ (–1, 6, 6), C ≡ (– 4, 9, 6) AB2 = (0 + 1)2 + (7 – 6)2 + (10 – 6) 2 = 18 ∴ AB = 3 2 Similarly ∴ BC = 3 2 , & AC = 6 Clearly AB2 +BC2 = AC2 ∴ ∠ ABC = 90° Also AB = BC Hence ∆ ABC is right angled isosceles. Example : Show by using distance formula that the points (4, 5, –5), (0, –11, 3) and (2, –3, –1) are collinear. Solution Let A ≡ (4, 5, –5), B ≡ (0, –11, 3), C ≡ (2, –3, –1).

AB =

( 4 − 0 )2 + (5 + 11)2 + ( −5 − 3)2 = 336 = 4 × 84 = 2 84

AC = ( 4 − 2)2 + (5 + 3)2 + ( −5 + 1)2 = 84 (0 − 2)2 + ( −11 + 3)2 + (3 + 1)2 = 84 BC + AC = AB Hence points A, B, C are collinear and C lies between A and B. Example : Find the locus of a point which moves such that the sum of its distances from points A(0, 0, – α) and B(0, 0, α) is constant. Solution. Let the variable point whose locus is required be P(x, y, z) Given PA + PB = constant = 2a (say)

BC =

∴

( x − 0)2 + ( y − 0)2 + ( z + α )2 + 1 2

⇒

= 2a –

( x − 0 )2 + ( y − 0 )2 + ( z − α )2 = 2a

x 2 + y 2 + ( z − α )2

⇒

x 2 + y2 + z2 + α2 + 2zα = 4a2 + x 2 + y2 + z2 + α2 – 2zα – 4a

⇒

4zα – 4a2 = – 4a

⇒

z 2α 2 a2

x 2 + y 2 + ( z − α )2

+ a2 – 2zα = x 2 + y2 + z2 + α2 – 2zα

x2 + y2 + ( z − α )2

 α2  x 2 + y 2 z2   + = 1 This is the required locus. x 2 + y2 + z2 1 − 2  = a2 – α2 or, 2 a  a − α 2 a2  Self practice problems : 1. One of the vertices of a cuboid is (1, 2, 3) and the edges from this vertex are along the +ve x-axis, +ve y-axis and +z axis respectively and are of length 2, 3, 2 respectively find out the vertices. Ans. (1, 2, 5), (3, 2, 5), (3, 2, 3), (1, – 1, 3), (1, – 1, 5), (3, – 1, 5), (3, – 1, 3). 2. Show that the points (0, 4, 1), (2, 3, –1), (4, 5, 0) and (2, 6, 2) are the vertices of a square. 3. Find the locus of point P if AP2 – BP2 = 18, where A ≡ (1, 2, – 3) and B ≡ (3, – 2, 1) Ans. 2x – 4y + 4z – 9 = 0

or,

4.

Section Formula

If point P divides the distance between the points A (x 1, y1, z1) and B (x 2, y2, z2) in the ratio of m : n, then coordinates of P are given as  mx 2 + nx 1 my 2 + ny 1 mz 2 + nz1  , ,   m+n m+n   m+n Note :- Mid point  x 1 + x 2 y 1 + y 2 z1 + z 2  , ,   2 2   2

5.

Centroid of a triangle  x 1 + x 2 + x 3 y 1 + y 2 + y 3 z1 + z 2 + z 3  , ,  G ≡ 3 3 3  

6. 7.

Incentre of triangle ABC:

 ax 1 + bx 2 + cx 3 ay 1 + by 2 + cy 3 az1 + bz 2 + cz 3  , ,   a+b+c a+b+c a+b+c  

Where AB = c, BC = a, CA = b

Centroid of a tetrahedron A (x 1, y1, z1) B (x2, y2, z2) C (x3, y3, z3) and D (x4, y4, z4) are the

vertices of a tetrahedron then coordinate of its centroid (G) is given as  xi yi z i   , ,  4 4 4    Example : Show that the points A(2, 3, 4), B(–1, 2, –3) and C(–4, 1, –10) are collinear. Also find the ratio in which C divides AB. Solution: Given A ≡ (2, 3, 4), B ≡ (–1, 2, –3), C ≡ (– 4, 1, –10).

∑ ∑ ∑

A (2, 3, 4) B (–1, 2, –3) Let C divide AB internally in the ratio k : 1, then  − k + 2 2k + 3 − 3k + 4  , , C≡   k +1   k +1 k +1

−k + 2 =–4 ⇒ 3k = – 6 ⇒ k = –2 k +1 2k + 3 −3k + 4 For this value of k, = 1, and = –10 k +1 k +1 Since k < 0, therefore k divides AB externally in the ratio 2 : 1 and pointsA(5, A, B, 4, C 6)are collinear. Example : The vertices of a triangle are A(5, 4, 6), B(1, –1, 3) and C(4, 3, 2). The internal bisector of ∠ BAC meets BC in D. Find AD.

∴

Solution

AC =

AB =

42 + 52 + 32 = 5 2

12 + 12 + 4 2 = 3 2

Since AD is the internal bisector of ∠ BAC BD AB 5 = = ∴ DC AC 3 ∴ D divides BC internally in the ratio 5 : 3 ∴

 5 × 4 + 3 × 1 5 × 3 + 3( −1) 5 × 2 + 3 × 3  , ,  D≡  5+3 5+3 5+3  

∴

AD =

2

2

23  12  19     5 −  + 4 −  + 6 −  8 8 8      

2

B (1, –1, 3)

or,

 23 12 19  D=  , ,   8 8 8 

D

C (4, 3, 2)

=

1530 unit 8

Example : If the points P, Q, R, S are (4, 7, 8), (– 1, – 2, 1), (2, 3, 4) and (1, 2, 5) respectively, show that PQ and RS intersect. Also find the point of intersection. Solution Let the lines PQ and RS intersect at point A. Let A divide PQ in the ratio λ : 1, then P(4, 7, 8) S(1, 2, 5) − λ + 4 − 2 λ + 7 λ + 8   λ l A≡  .... (1) , , . λ +1 λ +1  λ +1 1 Let A divide RS in the ratio k : 1, then A K  k + 2 2k + 3 5k + 4  , ,  A≡  ..... (2) R(2, 3, 4) Q(–1, –2, 1)  k +1 k +1 k +1  From (1) and (2), we have, −λ + 4 k + 2 = ..... (3) λ +1 k +1 −2λ + 7 2k + 3 = ..... (4) λ +1 k +1 λ + 8 5k + 4 = ..... (5) λ +1 k +1 From (3), – λk – λ + 4k + 4 = λk + 2λ + k + 2 or 2λk + 3λ – 3k – 2 = 0 ..... (6) From (4), –2λk – 2λ + 7k + 7 = 2λk + 3λ + 2k + 3 or 4λk + 5λ – 5k – 4 = 0 ..... (7) Multiplying equation (6) by 2, and subtracting from equation (7), we get –λ+k=0 or , λ=k Putting λ = k in equation (6), we get 2λ 2 + 3λ – 3λ – 2 = 0 or, λ = ± 1. But λ ≠ –1, as the co-ordinates of P would then be underfined and in this case PQ || RS, which is not true. ∴ λ = 1 = k. Clearly λ k = 1 satisfies eqn. (5). Hence our assumption is correct  − 1+ 4 − 2 + 7 1+ 8  3 5 9 , ,  ∴ A≡  or, A≡  , , . 2 2   2 2 2 2 Self practice problems: 1. Find the ratio in which xy plane divides the line joining the points A (1, 2, 3) and B (2, 3, 6). Ans. – 1 : 2 2. Find the co-ordinates of the foot of perpendicular drawn from the point A(1, 2, 1) to the line joining the point B(1, 4, 6) and C(5, 4, 4). Ans. (3, 4, 5) 3. 4.

5.

8.

8  Two vertices of a triangle are (4, –6, 3) and (2, –2, 1) and its centroid is  , − 1, 2  . Find the third vertex. 3  Ans. (2, 5, 2)

If centroid of the tetrahedron OABC, where co-ordinates of A, B, C are (a, 2, 3), (1, b, 2) and (2, 1, c) respectively be (1, 2, 3), then find the distance of point (a, b, c) from the origin. Ans. 107  1  Show that  − , 2, 0  is the circumcentre of the triangle whose vertices are A (1, 1, 0), B (1, 2, 1) and  2  C (– 2, 2, –1) and hence find its orthocentre. Ans. (1, 11, 0)

Direction Cosines And Direction Ratios

(i)

(ii) (iii)

Direction cosines: Let α, β, γ be the angles which a directed line makes with the positive directions of the axes of x, y and z respectively, then cos α, cosβ, cos γ are called the direction cosines of the line. The direction cosines are usually denoted by (, m, n). Thus = cos α, m = cos β, n = cos γ. If , m, n be the direction cosines of a line, then 2 + m 2 + n2 = 1 Direction ratios: Let a, b, c be proportional to the direction cosines , m, n then a, b, c are called the direction ratios. If a, b, c, are the direction ratios of any line L then a î + bˆj + ckˆ will be a vector parallel to the line L.

If , m, n are direction cosines of line L then î + m ˆj + n kˆ is a unit vector parallel to the line L. If , m, n be the direction cosines and a, b, c be the direction ratios of a v ector, then   a b c  =  ,m = ,n =  2 2 2 2 2 2 2 2 2  a +b +c a +b +c a +b +c   −c −b −a , n = , m = or = a2 + b 2 + c 2 a2 + b2 + c 2 a2 + b2 + c 2 (v) If OP = r, when O is the origin and the direction cosines of OP are , m, n then the coordinates of P are (r, mr, nr). , m,n, |AB| = r, and the coordinates of A is (x 1, y1, z1) then the coordinates of B is given as (x 1 + r , y1 + rm, z1 + rn) (vi) If the coordinates P and Q are (x 1, y1, z1) and (x 2, y2, z2) then the direction ratios of line PQ are, x 2 − x1 a = x 2 − x 1, b = y2 − y1 & c = z2 − z1 and the direction cosines of line PQ are = , | PQ | z 2 − z1 y 2 − y1 m= and n = | PQ | | PQ | (vii) Direction cosines of axes: Since the positive x −axis makes angles 0º, 90º, 90º with axes of x, y and z respectively. Therefore Direction cosines of x −axis are (1, 0, 0) Direction cosines of y−axis are (0, 1, 0) Direction cosines of z−axis are (0, 0, 1) Example : If a line makes angles α, β , γ with the co-ordinate axes, prove that sin2α + sin2β + sinγ2 = 2. Solution Since a line makes angles α, β , γ with the co-ordinate axes, hence cosα, cosβ , cosγ are its direction cosines ∴ cos2α + cos2β + cos2γ = 1 ⇒ (1 – sin2α) + (1 – sin2β ) + (1 – sin2γ) = 1 ⇒ sin2α + sin2β + sin2γ = 2. Example : Find the direction cosines , m, n of a line which are connected by the relations + m + n = 0, 2mn + 2m – n = 0 Solution Given, + m + n = 0 ..... (1) 2mn + 2m – n = 0 ..... (2) From (1), n = – ( + m). Putting n = – ( + m) in equation (2), we get, – 2m( + m) + 2m + ( + m) = 0 or, – 2m – 2m 2 + 2m + 2 + m = 0 or, 2 + m – 2m 2 = 0 (iv)

I f

d i r e c t i o n

c o s i n e s

o f

t h e

l i n e

A

B

a r e

2

or,

    +  – 2 = 0 m m

[dividing by m 2]

− 1± 1+ 8 − 1± 3 = = = 1, –2 m 2 2 Case I. when = 1 : In this case m = m From (1), 2 + n = 0 ⇒ n = – 2 ∴ :m:n=1:1:–2 ∴ Direction ratios of the line are 1, 1, – 2 ∴ Direction cosines are 1 1 ± ,± ,± 2 2 2 1 + 1 + ( −2 ) 12 + 12 + ( −2)2

or

or,

1 6

,

1 6

,

−2 6

or –

1 6

,–

1 6

,

−2 2

1 + 12 + ( −2)2

2 6

= – 2 : In this case = – 2m m From (1), – 2m + m + n = 0 ⇒ n=m ∴ : m : n = – 2m : m : m =–2:1:1 ∴ Direction ratios of the line are – 2, 1, 1. ∴ Direction cosines are −1 −2 −1 , , 2 2 2 2 2 2 2 ( −2 ) + 1 + 1 ( −2 ) + 1 + 1 ( −2) + 12 + 12

Case II. When

or,

−2 6

,

1 , 6

1 . 6

Self practice problems: 1. Find the direction cosine of a line lying in the xy plane and making angle 30° with x-axis.

1 3 ,n=0 ,= 2 2 A line makes an angle of 60° with each of x and y axes, find the angle which this line makes with z-axis. Ans. 45° A plane intersects the co-ordinates axes at point A(a, 0, 0), B(0, b, 0), C(0, 0, c) O is origin. Find the direction ratio of the line joining the vertex B to the centroid of face AOC. a c Ans. , – b, 3 3 A line makes angles α, β, γ, δ with the four diagonals of a cube, prove that 4 cos2α + cos2β + cos2γ + cos2δ = . 3

Ans.

2. 3.

4.

9.

m=±

Angle Between Two Line Segments: If two lines have direction ratios a1, b1, c 1 and a2, b2, c2 respectively then we can consider two vectors parallel to the lines as a1i + b1j + c 1k and a2i + b2j + c 2k and angle between them can be given as. a 1a 2 + b1b 2 + c1c 2 cos θ = . 2 a 1 + b12 + c12 a 22 + b 22 + c 22

(i)The line will be perpendicular if a1a2 + b1b2 + c1c2 = 0 (ii)The lines will be parallel if (iii)

c1 b a1 = 1 = b2 a2 c2

Two parallel lines have same direction cosines i.e. 1 = 2, m 1 = m 2, n1 = n2

Example :

What is the angle between the lines whose direction cosines are −

Solution cosθ

Let θ be the required angle, then = 1 2 + m 1 m 2 + n 1 n 2  3   3   1   1   3   3   = − 4  − 4  +  4   4  + − 2  .  2            3 1 3 1 + − =− = ⇒ θ = 120°, 16 16 4 2 Find the angle between any two diagonals of a cube. The cube has four diagonals Y

Example : Solution

G(0, a, 0)

E

D (0, a, a)

O (0, 0, 0)

F(a, a, 0)

(a, a, a)

A X (a, 0, 0) B(a, 0, a)

C(0, 0, a)

OE, AD, CF and GB The direction ratios of OE are a, a, a or, 1, 1, 1

3 1 3 3 1 3 , ,− ;− , , ? 4 4 2 4 4 2

Z’

its direction cosines are

1

,

1

,

1

. 3 3 3 Direction ratios of AD are – a, a, a. or, – 1, 1, 1. −1 1 1 ∴ its direction cosines are . , , 3 3 3 Similarly, direction cosines of CF and GB respectively are 1 1 −1 1 −1 1 , , , , and . 3 3 3 3 3 3 We take any two diagonals, say OE and AD Let θ be the acute angle between them, then

∴

 1   −1  1   1   1   1  1  + .  + .  = cosθ =            3  3 3  3  3  3  3

or,

 1 θ = cos–1   . 3

Example : If two pairs of opposite edges of a tetrahedron are mutually perpendicular, show that the third pair will also be mutually perpendicular. Solution: Let OABC be the tetrahedron where O is the origin and co-ordinates of A, B, C be (x 1, y1, z 1), (x 2, y2, z 2), (x 3, y3, x 3) respectively. A (x1, y1, z1)

O (0, 0, 0)

B C (x3, y3, z3) (x2, y2, z2) OA ⊥ BC and OB ⊥ CA . Let We have to prove that OC ⊥ BA . Now, direction ratios of OA are x 1 – 0, y1 – 0, z1 – 0 or, x 1, y1, z1 direction ratios of BC are (x 3 – x 2), (y 3 – y2), (z3 – z2). OA ⊥ BC . ∵ ∴ x 1(x 3 – x 2) + y1(y3 – y2) + z1(z3 – z2) = 0 ..... (1) Similarly, ∵ OB ⊥ CA ∴ x 2(x 1 – x 3) + y2(y1 – y3) + z2(z1 – z3) = 0 ..... (2) Adding equations (1) and (2), we get x 3(x 1 – x 2) + y3(y1 – y2) + z3(z1 – z2) = 0 ∴ OC ⊥ BA [∵ direction ratios of OC are x 3, y3, z3 and that of BA are (x 1 – x 2), (y1 – y2), (z 1 – z 2)] Self practice problems: 1. Find the angle between the lines whose direction cosines are given by + m + n = 0 and 2 + m 2 – n2 = 0 Ans. 60° 2. P (6, 3, 2) Q (5, 1, 4) R (3, 3, 5) are vertices of a ∆ find ∠Q. Ans. 90° 3. Show that the direction cosines of a line which is perpendicular to the lines having directions cosines 1 m 1 n1 and 2 m 2 n2 respectively are proportional to m 1n2 – m2n1 , n12 – n21, 1m 2 – 2m1

1 0 . Projection of a line segment on a line (i)

If the coordinates P and Q are (x 1, y1, z1) and (x 2, y 2, z2) then the projection of the line segments

, m, n is (x 2 − x1 ) + m(y 2 − y1 ) + n(z 2 − z1 ) a ˆ = . b Vector form: projection of a vector a on another vector b is a . b |b| → In the above case we can consider PQ as (x 2 – x 1) î + (y2 – y 1) ˆj + (z2 – z 1) kˆ in place of a and î + m ˆj + n kˆ in place of b . (iii) | r |, m | r | & n | r | are the projection of r in OX, OY & r = | r | ( î + m ˆj + n kˆ ) OZ axes. (iv) PQ on a line having direction cosines

(ii)

Solved Example : Find the projection of the line joining (1, 2, 3) and (–1, 4, 2) on the line having direction ratios 2, 3, – 6. Solution Let A ≡ (1, 2, 3), B ≡ (–1, 4, 2) B A 90°

90°

P L Direction ratios of the given line PQ are 2, 3, – 6 2 2 + 3 2 + ( −6 )2 = 7

∴

M

Q

direction cosines of PQ are

2 3 6 , ,– 7 7 7 Projection of AB on PQ = (x 2 – x 1) + m(y2 – y 1) + n(z2 – z1) 2 3 6 (–1 – 1) + (4 – 2) – (2 – 3) = 7 7 7 Self practice problems: 1. A (6, 3, 2), B (5, 1, 1,), C(3, –1, 3) D (0, 2, 5) Find the projection of line segment AB on CD line.

=

−4 + 6 + 6 8 = 7 7

Ans.

5/7

2.

3.

The projections of a directed line segment on co-ordinate axes are – 2, 3, – 6. Find its length and 12 4 3 direction cosines. Ans. 13 ; , , 13 13 13 Find the projection of the line segment joining (2, – 1, 3) and (4, 2, 5) on a line which makes equal 7 acute angles with co-ordinate axes. Ans. 3

A PLANE

If line joining any two points on a surface lies completely on it then the surface is a plane. OR If line joining any two points on a surface is perpendicular to some fixed straight line. Then this surface is called a plane. This fixed line is called the normal to the plane.

1 1 . Equation Of A Plane (i) (ii) (iii) (iv)

Normal form of the equation of a plane is x + my + nz = p, where, ,m n are the direction cosines of the normal to the plane and p is the distance of the plane from the origin. General form: ax + by + cz + d = 0 is the equation of a plane, where a, b, c are the direction ratios of the normal to the plane. The equation of a plane passing through the point (x 1, y1, z 1) is given by a (x − x 1) + b( y − y1) + c (z − z1) = 0 where a, b, c are the direction ratios of the normal to the plane. Plane through three points: The equation of the plane through three non−collinear points x

(x 1, y1, z 1), (x 2, y2, z2), (x 3, y3, z 3) is

y

z 1

x1 y1 z1 1 =0 x 2 y2 z2 1 x 3 y3 z3 1

(v)

(vi) Note:

(a) (b) (c)

(d) (e)

x y z + + =1 a b c Vector form: The equation of a plane passing through a point having position vector a & normal to vector n is ( r − a ). n = 0 or r . n = a . n

Intercept Form: The equation of a plane cutting intercept a, b, c on the axes is

Vector equation of a plane normal to unitvector nˆ and at a distance d from the origin is . =d r n Coordinate planes (i) Equation of yz −plane is x = 0 (ii) Equation of xz−plane is y = 0 (iii) Equation of xy−plane is z = 0 Planes parallel to the axes: If a = 0, the plane is parallel to x −axis i.e. equation of the plane parallel to the x −axis is by + cz + d = 0. Similarly, equation of planes parallel to y−axis and parallel to z −axis are ax + cz +d = 0 and ax + by + d = 0 respectively. Plane through origin: Equation of plane passing through origin is ax + by + cz = 0. Transformation of the equation of a plane to the normal form: To reduce any equation ax + by + cz − d = 0 to the normal form, first write the constant term on the right hand side and make it positive, then divide each term by a 2 + b 2 + c2 , where a, b, c are coefficients of x, y and z respectively e.g.

ax

(g)

by

+

cz

=

d

± a 2 + b2 + c2 ± a 2 + b 2 + c2 ± a 2 + b 2 + c2 Where (+) sign is to be taken if d > 0 and (−) sign is to be taken if d < 0. Any plane parallel to the given plane ax + by + cz + d = 0 is ax + by + cz + λ = 0. ± a 2 + b2 + c2

(f)

+

Distance between two parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is | d1 − d2 | given as a2 + b2 + c 2 Equation of a plane passing through a given point & parallel to the given vectors: The equation of a plane passing through a point having position vector a and parallel to

b & c is r = a + λ b + µ c (parametric form) where λ & µ are scalars. or r . ( b × c) = a . ( b × c) (non parametric form) (h)

A plane ax + by + cz + d = 0 divides the line segment joining (x 1, y1, z1) and (x 2, y2, z2). in the ratio

 ax1 + by1 + cz1 + d  −   ax2 + by2 + cz2 + d 

(i)

The xy−plane divides the line segment joining the points (x 1, y 1, z 1) and (x 2, y2, z2) in the ratio

y1 x1 z1 − z . Similarly yz−plane in − x and zx −plane in − y 2 2 2 (j)

Coplanarity of four points The points A(x 1 y1 z1), B(x 2 y 2 z2) C(x 3 y3 z 3) and D(x 4 y 4 z4) are coplaner then x 2 − x1 y 2 − y1 z 2 − z1 x 3 − x1 y 3 − y1 z 3 − z1 =0 x 4 − x 1 y 4 − y 1 z 4 − z1

very similar in vector method the points A ( r1 ), B( r2 ), C( r3 ) and D( r4 ) are coplanar if [ r4 – r1 , r4 – r2 , r4 – r3 ] = 0 Example : Find the equation of the plane upon which the length of normal from origin is 10 and direction ratios of this normal are 3, 2, 6. Solution If p be the length of perpendicular from origin to the plane and , m, n be the direction cosines of this normal, then its equation is x + my + nz = p ..... (1) Here p = 10; Direction ratios of normal to the plane are 3, 2, 6

∴ Direction cosines of normal to the required plane are 32 + 22 + 62 = 7 3 2 6 = ,m= ,n= 7 7 7 Putting the values of , m, n, p in (1), equation of required plane is 3 2 6 x + y + z = 10 or, 3x + 2y + 6z = 70 7 7 7 Example : Show that the points (0, – 1, 0), (2, 1, – 1), (1, 1, 1), (3, 3, 0) are coplanar. Solution Let A ≡ (0, – 1, 0), B ≡ (2, 1, – 1), C ≡ (1, 1, 1) and D ≡ (3, 3, 0) Equation of a plane through A (0, – 1, 0) is a (x – 0) + b (y + 1) + c (z – 0) = 0 or, ax + by + cz + b = 0 ..... (1) If plane (1) passes through B (2, 1, – 1) and C (1, 1, 1) Then 2a + 2b – c = 0 ..... (2) and a + 2b + c = 0 ..... (3) From (2) and (3), we have a b c = = 2 + 2 − 1− 2 4 − 2 a b c = = or, = k (say) 4 −3 2 Putting the value of a, b, c, in (1), equation of required plane is 4kx – 3k(y + 1) + 2kz = 0 or, 4x – 3y + 2z – 3 = 0 ..... (2) Clearly point D (3, 3, 0) lies on plane (2) Thus points D lies on the plane passing through A, B, C and hence points A, B, C and D are coplanar. Example : If P be any point on the plane x + my + nz = p and Q be a point on the line OP such that OP . OQ = p2, show that the locus of the point Q is p(x + my + nz) = x 2 + y 2 + z2. Solution Let P ≡ (α, β, γ), Q ≡ (x 1, y1, z1) Direction ratios of OP are α, β, γ and direction ratios of OQ are x 1, y1, z1. Since O, Q, P are collinear, we have α β γ = = ..... (1) x1 y1 z1 = k (say) As P (α, β, γ) lies on the plane x + my + nz = p, α + mβ + nγ = p or k(x 1 + my 1 + nz 1) = p ..... (2) Given, OP . OQ = p2

∴

α 2 + β2 + γ 2

x12 + y12 + z12 = p2

or,

k 2 ( x12 + y12 + z12 )

x12 + y12 + z12 = p2

or, k ( x12 + y12 + z12 ) = p2 On dividing (2) by (3), we get, x1 + my 1 + nz1 1 = p x12 + y12 + z12

..... (3)

or, p (x 1 + my1 + nz1) = x12 + y12 + z12 Hence the locus of point Q is p (x + my + nz) = x 2 + y2 + z 2. x y z Example : A point P moves on a plane + + = 1. A plane through P and perpendicular to OP meets the a b c

co-ordinate axes in A, B and C. If the planes through A, B and C parallel to the planes x = 0, y = 0, z = 0 intersect in Q, find the locus of Q. Solution Given plane is x y z + + =1 ..... (1) a b c Let P ≡ (h, k, ) h k ..... (2) Then + + = 1 a b c OP =

h2 + k 2 + 2

h

Direction cosines of OP are

∴

or,

∴

2

2

2

,

k 2

2

2

,

2

h +k + h +k + h + k2 + 2 Equation of the plane through P and normal to OP is h k x+ y+ = h2 + k 2 + 2 2 2 2 2 2 2 2 2 2 h +k + h +k + h +k + hx + ky + z = (h2 + k2 + 2)  h2 + k 2 + 2   h2 + k 2 + 2  , 0 , , 0, 0  , B ≡  0, A ≡    k h    

 h2 + k 2 + 2  C ≡  0, 0,    Let Q ≡ (α, β , γ), then

α= Now

1

α

2

h2 + k 2 + 2 h2 + k 2 + 2 h2 + k 2 + 2 ,γ= ,β= k h

+

From (3), h =

∴

1

β

2

+

1

γ

2

=

h2 + k 2 + 2 (h2 + k 2 + 2 )2

=

1 (h2 + k 2 + 2 )

..... (3) ..... (4)

h2 + k 2 + 2 α

h h2 + k 2 + 2 = a aα

Similarly

k h2 + k 2 + 2 h2 + k 2 + 2 = and = b bβ c cγ

h 2 + k 2 + 2 h2 + k 2 + 2 h2 + k 2 + 2 h k + + = + + = 1 [from (2)] aα bβ cγ a b c 1 1 1 1 1 1 1 + + = = + + [from (4)] or, aα bβ cγ h 2 + k 2 + 2 α 2 β 2 γ 2 ∴ Required locus of Q (α, β , γ) is 1 1 1 1 1 1 . + + = + + ax by cz x 2 y 2 z 2 Self practice problems : 1. Check wether this point are coplanar if yes find the equation of plane containing them A ≡ (1, 1, 1) B ≡ (0, – 1, 0) C ≡ (2, 1, –1) D ≡ (3, 3, 0) Ans. yes, 4x – 3y + 2z = 3 2. Find the plane passing through point (– 3, – 3, 1) and perpendicular to the line joining the points (2, 6, 1) and (1, 3, 0). Ans. x + 3y + z + 11 = 0 3. Find the equation of plane parallel to x + 5y – 4z + 5 = 0 and cutting intercepts on the axes whose rum 3000 is 150. Ans. x + 5y – 4z = 19 4. Find the equation of plane passing through (2, 2, 1) and (9, 3, 6) and perpendicular to the plane x + 3y + 3z = 8. Ans. 3x + 4y – 5z = 9 5. Find the equation of the plane | | to î + ˆj + kˆ and î − ˆj and passing through (1, 1, 2). Ans. x + y – 2z + 2 = 0 6. Find the equation of the plane passing through the point (1, 1, – 1) and perpendicular to the planes x + 2y + 3z – 7 = 0 and 2x – 3y + 4z = 0. Ans. 17x + 2y – 7z = 26

∴

1 2 . Sides of a plane:

A plane divides the three dimensional space in two equal parts. Two points A (x 1 y 1 z 1 ) and B (x 2 y 2 z 2) are on the same side of the plane ax + by + cz + d = 0 if ax 1 + by1 + cz1 + d and

ax 2 + by2 + cz2 + d are both positive or both negative and are opposite side of plane if both of these values are in opposite sign. Example : Show that the points (1, 2, 3) and (2, – 1, 4) lie on opposite sides of the plane x + 4y + z – 3 = 0. Solution Since the numbers 1+ 4 × 2 + 3 – 3 = 9 and 2 – 4 + 4 – 3 = – 1 are of opposite sign., the points are on opposite sides of the plane.

1 3 . A Plane & A Point

a 2 + b 2 + c2 The length of the perpendicular from a point having position vector a to plane r . n = d is | a .n − d | given by p = . |n|

.

The coordinates of the foot of perpendicular from the point (x, y, z) to the plane x ' − x 1 y ' − y 1 z' − z 1 (ax1 + by1 + cz1 + d) = = ax + by + cz + d = 0 aregain by =– a b c a2 + b 2 + c 2 (iv) To find image of a point w.r.t. a plane. Let P (x 1, y 1, z 1) is a given point and ax + by + cz + d = 0 is given plane Let (x ′, y ′, z′) is the image point. then (a) x ′ – x 1 = λa, y ′ – y1 = λb, z′ – z 1 = λc ⇒ x ′ = λ a + x 1, y ′ = λ b + y 1, z ′ = λ c + z1  x′ + x 1   y′ + y 1   z′ + z 1   + b  + c  =0 (b) a  2   2   2  from (i) put the values of x ′, y′, z′ in (ii) and get the values of λ and resubtitute in (i) to get (x ′ y′ z ′). The coordinate of the image of point (x 1 , y 1 , z1) w.r.t the plane ax + by + cz + d = 0 are given x ' − x 1 y ' − y 1 z' − z1 (ax 1 + by1 + cz1 + d) = = by =–2 a b c a2 + b 2 + c 2 (v) The distance between two parallel planes ax + by + cx + d = 0 and ax + by + cx + d’ = 0 is | d − d' | given by 2 a + b2 + c 2 Example : Find the image of the point P (3, 5, 7) in the plane 2x + y + z = 16. Solution Given plane is 2x + y + z = 16 ..... (1) P ≡ (3, 5, 7) Direction ratios of normal to plane (1) are 2, 1, 1 Let Q be the image of point P in plane (1). Let PQ meet plane (1) in R then PQ ⊥ plane (1) Let R ≡ (2r + 3, r + 5, r + 7) Since R lies on plane (1) ∴ 2(2r + 3) + r + 5 + r + 7 = 0 or, 6r + 18 = 0 ∴ r=–3 ∴ R ≡ (– 3, 2, 4) Let Q ≡ ( α, β , γ ) Since R is the middle point of PQ α+3 ⇒ α=–9 ∴ –3= 2 β+5 2= ⇒ β=–1 2 γ+7 ⇒ γ=1 ∴ Q = (– 9, – 1, 1). 4= 2 Example : Find the distance between the planes 2x – y + 2z = 4 and 6x – 3y + 6z = 2. Solution Given planes are 2x – y + 2z – 4 = 0 ..... (1) and 6x – 3y + 6z – 2 = 0 ..... (2) a1 b1 c 1 We find that a = b = c Hence planes (1) and (2) are parallel. 2 2 2 (iii)

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ax'+ by'+ cz'+ d

2 =0 3 Required distance between the planes 2 4− 10 10 3 = = = 2 2 2 3 .3 9 2 + ( −1) + 2

Plane (2) may be written as 2x – y + 2z –

∴

..... (3)

Example : A plane passes through a fixed point (a, b, c). Show that the locus of the foot of perpendicular to it from the origin is the sphere x 2 + y 2 + z2 – ax – by – cz = 0

0 98930 58881 , BHOPAL, (M.P.)

(ii)

Distance of the point (x ′ , y′ , z′ ) from the plane ax + by + cz+ d = 0 is given by

TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000,

(i)


P(α, β, γ)

x + my + nz + d = 0 ..... (1) Plane passes through the fixed point (a, b, c) ∴ a + mb + nc + d = 0 ..... (2) Let P ( α, β , γ) be the foot of perpendicular from origin to plane (1). Direction ratios of OP are α – 0, β – 0, γ – 0 i.e. α, β , γ From equation (1), it is clear that the direction ratios of normal to the plane i.e. OP are , m, n ; α, β , γ and , m, n are the direction ratios of the same line OP α β γ 1 ∴ = = = (say) m n k ∴ = k α, m = k β , n = k γ ..... (3) Putting the values of , m, n in equation (2), we get ..... (4) kaα + kbβ + kcγ + d = 0 Since α, β , γ lies in plane (1) ∴ α + m β + nγ + d = 0 ..... (5) Putting the values of , m, n from (3) in (5), we get k α2 + k β 2 + k γ 2 + d = 0 ..... (6) or kα2 + k β 2 + kγ2 – kaα – kbβ – kc γ = 0 [putting the value of d from (4) in (6)] or α2 + β 2 + γ 2 – a α – b β – c γ = 0 Therefore, locus of foot of perpendicular P (α, β , γ) is x 2 + y 2 + z2 – ax – by – cz = 0 ..... (7) Self practice problems: Find the intercepts of the plane 3x + 4y – 7z = 84 on the axes. Also find the length of perpendicular 1. from origin to this line and direction cosines of this normal. 1 3 4 −7 ; , , Ans. a = 28, b = 21, c = – 12, p = 74 74 74 74 2. Find : (i) perpendicular distance (ii) foot of perpendicular (iii) image of (1, 0, 2) in the plane 2x + y + z = 5 1  4 1 13  5 1 7  Ans. (i) (ii)  , , (iii)  , ,  6 3 6 6  3 3 3

1 4 . Angle Between Two Planes: (i)

Consider two planes ax + by + cz + d = 0 and a′ x + b′ y + c ′ z + d′ = 0. Angle between these planes is the angle between their normals. Since direction ratios of their normals are (a, b, c) and (a′ , b′ , c′ ) respectively, hence θ, the angle between them, is given by

aa'+ bb'+ cc' cos θ =

a 2 + b2 + c2

a ' 2 + b ' 2 + c' 2

a b c = = a' b' c' n1 . n 2 The angle θ between the planes r . n = d1 and r . n 2 = d2 is given by, cos θ = | n1 | . | n 2 | Planes are perpendicular if n1 . n 2 = 0 & planes are parallel if n1 = λ n 2 . Planes are perpendicular if aa′ + bb′ + cc′ = 0 and planes are parallel if

(ii)

1 5 . Angle Bisectors (i)

The equations of the planes bisecting the angle between two given planes a1x + b1y + c 1z + d1 = 0 and a2x + b2y + c 2z + d2 = 0 are

a1x + b1y + c1z + d1

=±

a12 + b12 + c12 (ii)

a 2 x + b 2 y + c2 z + d 2 a 22 + b22 + c22

Equation of bisector of the angle containing origin: First make both the constant terms positive. Then the positive sign in

a1x + b1y + c1z + d1 a12

(iii)

+

b12

+ c12

=±

a 2 x + b2 y + c2 z + d 2 gives the bisector of a 22 + b22 + c22

the angle which contains the origin. Bisector of acute/obtuse angle: First make both the constant terms positive. Then a1a2 + b1 b2 + c 1c 2 > 0 ⇒ origin lies on obtuse angle

0 98930 58881 , BHOPAL, (M.P.) Vec&3D/Page : 12 of 77

Let the equation of the variable plane be O(0, 0, 0)


Solution

origin lies in acute angle

1 6 . Family of Planes (i)

The equation of plane passing through the intersection of the planes r . n1 = d 1 & r . n 2 = d2 is r . (n1 + λ n 2 ) = d1 + λd2 where λ is arbitrary scalar Example : The plane x – y – z = 4 is rotated through 90° about its line of intersection with the plane x + y + 2z = 4. Find its equation in the new position. Solution Given planes are x–y–z=4 ..... (1) and x + y + 2z = 4 ..... (2) Since the required plane passes through the line of intersection of planes (1) and (2) ∴ its equation may be taken as x + y + 2z – 4 + k (x – y – z – 4) = 0 or (1 + k)x + (1 – k)y + (2 – k)z – 4 – 4k = 0 ..... (3) Since planes (1) and (3) are mutually perpendicular, ∴ (1 + k) – (1 – k) – (2 – k) = 0 2 or, 1+k–1+k–2+k=0 or, k= 3 2 Putting k = in equation (3), we get, 3 5x + y + 4z = 20 This is the equation of the required plane. Example : Find the equation of the plane through the point (1, 1, 1) which passes through the line of intersection of the planes x + y + z = 6 and 2x + 3y + 4z + 5 = 0. Solution Given planes are x+y+z–6=0 ..... (1) and 2x + 3y + 4z + 5 = 0 ..... (2) Given point is P (1, 1, 1). Equation of any plane through the line of intersection of planes (1) and (2) is x + y + z – 6 + k (2x + 3y + 4z + 5) = 0 ..... (3) If plane (3) passes through point P, then 3 1 + 1 + 1 – 6 + k (2 + 3 + 4 + 5) = 0 or, k= 14 From (1) required plane is 20x + 23y + 26z – 69 = 0 Example : Find the planes bisecting the angles between planes 2x + y + 2z = 9 and 3x – 4y + 12z + 13 = 0. Which of these bisector planes bisects the acute angle between the given planes. Does origin lie in the acute angle or obtuse angle between the given planes ? Solution Given planes are – 2x – y – 2z + 9 = 0 ..... (1) and 3x – 4y + 12z + 13 = 0 ..... (2) Equations of bisecting planes are −2x − y − 2z + 9 3x − 4y + 12z + 13 =± 2 2 2 ( −2) + ( −1) + ( −2) 3 2 + ( −4)2 + (12)2 or, 13 [– 2x – y – 2z + 9] = ± 3 (3x – 4y + 12z + 13) or, 35x + y + 62z = 78, ..... (3) [Taking +ve sign] and 17x + 25y – 10z = 156 ..... (4) [Taking – ve sign] Now a1a2 + b1b2 + c1c 2 = (– 2) (3) + (– 1) (– 4) + (– 2) (12) = – 6 + 4 – 24 = – 26 < 0 ∴ Bisector of acute angle is given by 35x + y + 62z = 78 ∵ a1a2 + b1b2 + c1c 2 < 0, origin lies in the acute angle between the planes. Example : If the planes x – cy – bz = 0, cx – y + az = 0 and bx + ay – z = 0 pass through a straight line, then find the value of a2 + b2 + c 2 + 2abc. Solution Given planes are x – cy – bz = 0 ..... (1) cx – y + az = 0 ..... (2) bx + ay – z = 0 ..... (3) Equation of any plane passing through the line of intersection of planes (1) and (2) may be taken as x – cy – bz + λ (cx – y + az) = 0 or, x (1 + λc) – y (c + λ) + z (– b + aλ) = 0 ..... (4) If planes (3) and (4) are the same, then equations (3) and (4) will be identical. 1 + cλ −(c + λ ) −b + aλ = = ∴ b a −1 (i) (ii) (iii) From (i) and (ii), a + ac λ = – bc – bλ (ii)


Any plane passing through the line of intersection of non−parallel planes or equation of the plane through the given line in serval form. a1x + b 1y + c 1z + d1 = 0 & a2x + b2y + c2z + d2 = 0 is a1x + b1y + c 1z + d1 + λ (a2x + b2y + c2z + d2) = 0


⇒


a1a2 + b1 b2 + c 1c 2 < 0

..... (5)


c + λ = – ab + a2λ or

λ=

(ab + c ) 1 − a2

..... (6)

From (5) and (6), we have, −(a + bc ) −(ab + c ) = . ac + b (1 − a 2 ) 3 2 or, a – a + bc – a bc = a2bc + ac2 + ab2 + bc or, a2bc + ac 2 + ab2 + a3 + a2bc – a = 0 or, a2 + b2 + c2 + 2abc = 1. Self practice problems: 1. A tetrahedron has vertices at O(0, 0, 0), A(1, 2, 1), B(2, 1, 3) and C(–1, 1, 2). Prove that the angle  19  between the faces OAB and ABC will be cos–1   .  35  2. Find the equation of plane passing through the line of intersection of the planes 4x – 5y – 4z = 1 and 10 2x + y + 2z = 8 and the point (2, 1, 3). Ans. 32x – 5y + 8z – 83 = 0, λ = 3 3. Find the equations of the planes bisecting the angles between the planes x + 2y + 2z – 3 = 0, 3x + 4y + 12z + 1 = 0 and sepecify the plane which bisects the acute angle Ans. 2x + 7y – 5z = 21, 11x + 19y + 31z = 18; 2x + 7y – 5z = 21 between them. 4. Show that the origin lies in the acute angle between the planes x + 2y + 2z – 9 = 0 and 4x – 3y + 12z + 13 = 0 5. Prove that the planes 12x – 15y + 16z – 28 = 0, 6x + 6y – 7z – 8 = 0 and 2x + 35y – 39z + 12 = 0 have a common line of intersection.

17. Area of a triangle:

Let A (x 1, y1, z 1), B (x 2, y2, z 2), C (x 3, y 3, z 3) be the vertices of a triangle, then ∆ =

where ∆x

1 = 2

y1 y2 y3

z1 1 z1 1 z2 1 , ∆y = z 2 2 z3 1 z3

x1 1 x1 x 2 1 and ∆z = x 2 x3 1 x3

(∆2x + ∆2y + ∆2z )

y1 1 y2 1 y3 1

→

→

Vector Method − From two vector AB and AC . Then area is given by → 1 1 → | AB x AC | = 2 2

i

j

k

x 2 − x1 x 3 − x1

y 2 − y1 y 3 − y1

z2 − z1 z3 − z1

Example : Through a point P (h, k, ) a plane is drawn at right angles to OP to meet the co-ordinate axes in A,

B and C. If OP = p, show that the area of ∆ ABC is

p5 . 2hk

Solution OP = h 2 + k 2 + 2 = p Direction cosines of OP are h k , , 2 2 2 2 2 2 2 h +k + h +k + h + k 2 + 2 Since OP is normal to the plane, therefore, equation of the plane will be, h k x+ y+ z = h2 + k 2 + 2 2 2 2 2 2 2 2 2 2 h +k + h +k + h +k + or, hx + ky + z = h2 + k2 + 2 = p2 ..... (1) 2   p2   p2    , 0, 0   0,   0, 0, p  , 0 ∴ A≡  h , B ≡  k  ,C≡        2 2 2 Now area of ∆ ABC, ∆ = A xy + A yz + A zx Now Axy = area of projection of ∆ ABC on xy-plane = area of ∆ AOB

= Mod of

1 2

p2 h

0

0

p k

1 =

0

0

1

1

2

1 p4 2 | hk |


λ=–


(a + bc ) (ac + b) From (ii) and (iii),

or,

=

1 p8 1 p8 1 p8 + + 4 h 2 k 2 4 k 2 2 4 2h 2 p8 2 2 2

4h k

p10

( 2 + k 2 + h 2 ) =

or,

2 2 2

4h k

∆=

1 8 . Volume Of A Tetrahedron: Tetrahedron

p5 . 2hk

Volume of a tetrahedron with vertices A (x 1, y 1, z 1), B( x 2, y2, z2), C (x 3, y 3, z 3) and

x1 x2


D (x 4, y 4, z 4) is given by V =

1 6 x3 x4

y1 y2

z1 1 z2 1

y3

z3 1

y4

z4 1

A LINE 1 9 . Equation Of A Line (i)

A straight line in space is characterised by the intersection of two planes which are not parallel and therefore, the equation of a straight line is a solution of the system constituted by the equations of the two planes, a1x + b1y + c1z + d1 = 0 and a2x + b2y + +c2z + d2 =0. This form is also known as non− symmetrical form.

(ii)

The equation of a line passing through the point (x 1, y1, z1) and having direction ratios a, b, c is =

(iii) (iv)

x − x1 a

y − y1 z − z1 = = r. This form is called symmetric form. A general point on the line is given by (x 1 + b c

ar, y1 + br, z 1 + cr). Vector equation: Vector equation of a straight line passing through a fixed point with position vector a and parallel to a given vector b is r = a + λ b where λ is a scalar.. The equation of the line passing through the points (x 1, y1, z1) and (x 2, y2, z2) is

z − z1 x − x1 y − y1 = = z x 2 − x1 y 2 − y1 2 − z1

Vector equation of a straight line passing through two points with position vectors a & b is r = a + λ ( b − a ). (vi) Reduction of cartesion form of equation of a line to vector form & vice versa x − x1 y − y1 z − z1 = (x ˆ + y ˆ + z ˆ ) + λ (a ˆ + b ˆ + c ˆ ). = = ⇔ j r i k 1i 1 j 1k a b c Note: Straight lines parallel to co-ordinate axes: Straight lines Equation Straight lines Equation (i) Through origin y = mx, z = nx (v) Parallel to x −axis y = p, z = q (ii) x −axis y = 0, z = 0 (vi) Parallel to y −axis x = h, z = q (iii) y−axis x = 0, z = 0 (vii) Parallel to z −axis x = h, y = p (iv) z−axis x = 0, y = 0 Example : Find the equation of the line through the points (3, 4, –7) and (1, – 1, 6) in vector form as well as in cartesian form. Solution Let A ≡ (3, 4, – 7), B ≡ (1, – 1, 6) (v)

Now

→

→

→

→

→

a = OA = 3 i + 4 j – 7 k , →

→

→

→

→

= b = OB = i – j + 6 k →

→

→

Equation of the line through A( a ) and B( b ) →

→

→

→

→

is →

→

→

→

r = a +t (b – a ) →

or ..... (1) r = 3 i + 4 j – 7 k + t (–2 i – 5 j + 13 k ) Equation in cartesian form : x −3 y −4 z+7 x−3 y−4 z+7 = = = = Equation of AB is or, 3 −1 4 +1 − 7 − 6 2 5 − 13 x −1 y + 2 z − 3 = = Example : Find the co-ordinates of those points on the line which is at a distance of 2 3 6 3 units from point (1, –2, 3).


∆2 =

∴


1 p4 1 p4 and Azx = 2 | k | 2 | h |

Similarly, Ayz =

3 6 2   r + 1, r − 2, r + 3  7 7 7  3 6 2  Q ≡  r − 1, r − 2, r + 3  7 7 7   Distance of Q from P = | r | According to question | r | = 3 ∴ r=±3 Putting the value of r, we have 5 39  23 3   1  13 ,   Q ≡ − , − , or Q ≡ − , − 7 7 7 7 7 7    Example : Find the equation of the line drawn through point (1, 0, 2) to meet at right angles the line x +1 y − 2 z +1 = = 3 −2 −1 Solution Given line is x +1 y − 2 z +1 = = ..... (1) 3 −2 −1 Let P ≡ (1, 0, 2) Co-ordinates of any point on line (1) may be taken as Q ≡ (3r – 1, – 2r + 2, – r – 1) Direction ratios of PQ are 3r – 2, – 2r + 2, – r – 3 Direction ratios of line AB are 3, – 2, – 1 Since PQ ⊥ AB ∴ 3 (3r – 2) – 2 (– 2r + 2) – 1 (– r – 3) = 0 1 ⇒ 9r – 6 + 4r – 4 + r + 3 = 0 ⇒ 14r = 7 ⇒ r= 2 Therefore, direction ratios of PQ are 1 7 – , 1, – or, – 1, 2, – 7 2 2 Equation of line PQ is x −1 y = 0 z − 2 x −1 y z−2 = = = = or, −1 2 −7 1 −2 7


Let

x −1 y − 2 z − 3 x − 4 y −1 = = = and = z intersect. Find also the 2 3 4 5 2 point of intersection of these lines. x −1 y − 2 z − 3 = = Solution Given lines are ..... (1) 2 3 4 x − 4 y −1 z − 0 = = and ..... (2) 5 2 1 Any point on line (1) is P (2r + 1, 3r + 2, 4r +3) and any point on line (2) is Q (5λ + 4, 2λ + 1, λ) Lines (1) and (2) will intersect if P and Q coincide for some value of λ and r. ⇒ 2r – 5λ = 3 ..... (1) ∴ 2r + 1 = 5λ + 4 3r + 2 + 2λ + 1 ⇒ 3r – 2λ = – 1 ..... (2) 4r + 3 = λ ⇒ 4r – λ = – 3 ..... (3) Solving (1) and (2), we get r = – 1, λ = – 1 Clearly these values of r and λ satisfy eqn. (3) Now P ≡ (– 1, – 1, – 1) Hence lines (1) and (2) intersect at (– 1, – 1, – 1). Self practice problems: 1. Find the equation of the line passing through point (1, 0, 2) having direction ratio 3, – 1, 5. Prove that x −1 y z−2 − = this line passes through (4, – 1, 7). Ans. 3 −1 5 x − 2 y +1 z −7 = = 2. Find the equation of the line parallel to line and passing through the point (3, 0, 5). 3 1 9

Example :

Show that the two lines


Given line is


x −1 y + 2 z − 3 = = ..... (1) 2 3 6 Let P ≡ (1, –2, 3) Direction ratios of line (1) are 2, 3, 6 2 3 6 ∴ Direction cosines of line (1) are , , 7 7 7 Equation of line (1) may be written as x −1 y + 2 z − 3 = = ..... (2) 2 3 6 7 7 7 Co-ordinates of any point on line (2) may be taken as

Solution

Ans.

 13 23  , 0  ,  5 5 

2 0 . Reduction Of Non-Symmetrical Form To Symmetrical Form: Let equation of the line in non−symmetrical form be a1x + b1y + c1z + d1 = 0, a2x + b2y + c2z + d2 = 0. To find the equation of the line in symmetrical form, we must know (i) its direction ratios (ii) coordinate of any point on it. (i) Direction ratios: Let , m, n be the direction ratios of the line. Since the line lies in both the planes, it must be perpendicular to normals of both planes. So a1 + b1m + c1n = 0, a2 + b2m + c2n = 0. From these equations, proportional values of , m, n can be found by


cross−multiplication as

m n = = b1c2 − b2 c1 c1a 2 − c2a1 a1b2 − a 2 b1

i j k The vector a1 b1 c 1 = i (b1c2 – b2c1) + j (c1a2 – c2a1) + k (a1b2 – a2b1) will be parallel a2 b 2 c 2 to the line of intersection of the two given planes. hence : m: n = (b1c2 – b2c1): (c1a2 – c2a1): (a1b2 – a2b1) (ii) Point on the line − Note that as , m, n cannot be zero simultaneously, so at least one must be non−zero. Let a1b2 − a2b1 ≠ 0, then the line cannot be parallel to xy plane, so it intersect it. Let it intersect xy−plane in (x 1, y1, 0). Then a1x 1 + b1y1 + d1 = 0 and a2x 1 + b2y1 + d2 = 0.Solving these,we get a point on the line. Then its equation becomes.

Alternative method

b1d 2 − b 2d1 d a − d 2a 1 y− 1 2 z−0 a1b 2 − a 2 b1 a1b2 − a 2 b1 = = a1b2 − a 2 b1 b1c2 − b2 c1 c1a 2 − c2a1

x−

x − x1 y − y1 z−0 = = or b1c2 − b2 c1 c1a 2 − c2a1 a1b2 − a 2 b1 Note: If ≠ 0, take a point on yz−plane as (0, y1, z1) and if m ≠ 0, take a point on xz−plane as (x 1, 0, z1). Alternative method a1 b1 If a ≠ b Put z = 0 in both the equations and solve the equations a1x + b1y + d1 = 0, a2x + b2y + d2 =0 2 2 otherwise Put y = 0 and solve the equations a1x + c1z + d1 = 0 and a2x + c2z + d2 = 0 Example : Find the equation of the line of intersection of planes 4x + 4y – 5z = 12, 8x + 12y – 13z = 32 in the symmetric form. Solution Given planes are 4x + 4y – 5z – 12 = 0 ..... (1) and 8x + 12y – 13z – 32 = 0 ..... (2) Let , m, n be the direction ratios of the line of intersection : then 4 + 4m – 5n = 0 ..... (3) and 8 + 12m – 13n = 0 m m n m n n = = = = = = ∴ or, or, − 52 + 60 − 40 + 52 48 − 32 8 12 16 2 3 4 Hence direction ratios of line of intersection are 2, 3, 4. Here 4 ≠ 0, therefore line of intersection is not parallel to xy-plane. Let the line of intersection meet the xy-plane at P (α, β, 0). Then P lies on planes (1) and (2) ∴ 4α + 4β + 12 = 0 or, α+β–3=0 ..... (5) and 8α + 12β – 32 = 0 or, 2α + 3β – 8 = 0 ..... (6) Solving (5) and (6), we get α β 1 α β 1 = = = = or, −8+9 −6+8 3−2 1 2 1 ∴ α = 1, β = 2 x −1 y − 2 z − 0 = = . Hence equation of line of intersection in symmetrical form is 2 3 4 Example : Find the angle between the lines x – 3y – 4 = 0, 4y – z + 5 = 0 and x + 3y – 11 =0, 2y – z + 6 = 0. Solution Given lines are x − 3 y − 4 = 0  ..... (1) 4y − z + 5 = 0 x + 3 y − 11 = 0  and ..... (2) 2y − z + 6 = 0  Let 1, m 1, n1 and 1, m 2, n2 be the direction cosines of lines (1) and (2) respectively


3.


x −3 y z−5 = = 3 1 3 Find the coordinates of the point when the line through (3, 4, 1) and (5, 1, 6) crosses the xy plane.

Ans.

line (1) is perpendicular to the normals of each of the planes x – 3y – 4 = 0 and 4y – z + 5 = 0 ∴ 1 – 3m 1 + 0.n1 = 0 ..... (3) and 01 + 4m 1 – n1 = 0 ..... (4) Solving equations (3) and (4), we get, m1 n 1 1 m1 n1 = 1 = = = = k (let). or, 0 − ( −1) 4 − 0 3−0 3 1 4 Since line (2) is perpendicular to the normals of each of the planes x + 3y – 11 = 0 and 2y – z + 6 = 0, ∴ 2 + 3m 2 = 0 ..... (5) and 2m 2 – n2 = 0 ..... (6) 2 ∴ 2 = – 3m 2 or, = m2 −3 n2 and n2 = 2m 2 or, = m 2. 2 2 m n = 3 = 2 = t (let). ∴ −3 1 2 If θ be the angle between lines (1) and (2), then cosθ = 12 + m 1m 2 + n1n2 = (3k) (– 3t) + (k) (t) + (4k) (2t) = – 9kt + kt + 8kt = 0 ∴ θ = 90°. Self practice problems: 1. Find the equation of the line of intersection of the plane 4x + 4y – 5z = 12 y−2 x −1 z−0 8x + 12y – 13z = 32 Ans. = = 3 2 4 2. Show that the angle between the two lines defined by the equations x = y and xy + yz + zx = 0 is  1 cos–1   3 3. Prove that the three planes 2x + y – 4z – 17 = 0, 3x + 2y – 2z – 25 = 0, 2x – 4y + 3z + 25 = 0 intersect at a point and find its co-ordinates. Ans. (3, 7, – 1)

2 1 . Foot, Length And Equation Of Perpendicular From A Point To A Line: (i)Cartesian form: Let equation of the line be

x −a y − b z− c = = = r (say) ..........(i) m n

and A (α, β, γ ) be the point. Any point on line (i) is P (r + a, mr + b, nr + c) ......... (ii) If it is the foot of the perpendicular from A on the line, then AP is perpendicular to the line. So (r + a − α) + m (mr + b − β) + n (nr + c − γ) = 0 i.e. r = (α − a) + (β − b) m + ( γ − c)n since 2 + m 2 + n2 = 1. Putting this value of r in (ii), we get the foot of perpendicular from point A on the given line. Since foot of perpendicular P is 2

2

known, then the length of perpendicular is given by AP = (r + a − α ) + ( mr + b − β) + ( nr + c − γ )

2

the

y −β z− γ x−α = = (ii) Vector Form: Equation of a line r + a − α mr + b − β nr + c − γ passing through a point having position vector α and perpendicular to the lines r = a1 + λ b1 and r = a 2 + λ b 2 is parallel to b1 x b 2 . So the vector equation of such a line is r = α + λ ( b1 x b 2 ). Position vector β of the  2 (a − α) . b   b − α . Position vector of image of a point α in a straight line r = a + λ b is given by β = 2 a −  2  |b|   (a −α) . b   b . The equation of the perpendicular is r = α + µ the foot of the perpendicular on line is f = a −  2 | b |   equation of perpendicular is given by

  (a − α ) .b   2  ( a − α ) −     |b| 

 b . 

22. To find image of a point w. r. t a line x − x2 y − y2 z − z2 = = is a given line a b c Let (x′, y′, z′) is the image of the point P (x 1, y1, z1) with respect to the line L. Then

Let L ≡

(i)

a (x 1 – x′) + b (y1 – y′) + c (z1 – z′) = 0

0 98930 58881 , BHOPAL, (M.P.) Vec&3D/Page : 18 of 77 TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000,


∵

x1 + x′ y1 + y′ z1 + z′ − x2 − y2 − z2 2 2 (ii) = = 2 =λ a b c from (ii) get the value of x′, y′, z′ in terms of λ as x′ = 2aλ + 2x 2 – x 1, y′ = 2bα – 2y2 – y1, z′ = 2cλ + 2z2 – z1 now put the values of x′, y′, z′ in (i) get λ and resubtitute the value of λ to to get (x′ y′ z′). x +1 y − 3 z + 2 = = Example : Find the length of the perpendicular from P (2, – 3, 1) to the line . 2 3 −1 x +1 y − 3 z + 2 = = Solution Given line is ..... (1) 2 3 −1 P ≡ (2, – 3, 1) Co-ordinates of any point on line (1) may be taken as Q ≡ (2r – 1, 3r + 3, – r – 2) Direction ratios of PQ are 2r – 3, 3r + 6, – r – 3 Direction ratios of AB are 2, 3, – 1 Since PQ ⊥ AB ∴ 2 (2r – 3) + 3 (3r + 6) – 1 (– r – 3) = 0 −15 or, 14r + 15 = 0 ∴ r= 14  − 22 − 3 − 13  531 , ,  ∴ Q≡  ∴ PQ = units. 14 14   7 14 Second method : Given line is x +1 y −3 z + 2 = = 2 3 −1 P (2, –3, 1) P ≡ (2, – 3, 1) 2 3 1 Direction ratios of line (1) are , ,– 14 14 14 RQ = length of projection of RP on AB A R Q B (–1, 3, –2) 2 3 1 15 (2 + 1) + ( −3 − 3) − (1 + 2) = = 14 14 4 14 PR2 = 32 + 62 + 32 = 54

PQ =

PR 2 − RQ 2 = Self practice problems:

∴

1.

2. 3.

54 −

225 = 14

531 14

x − 11 y + 2 z + 8 = = . 10 −4 − 11 Also find the image of the point in the line. Ans. 14 , N ≡ (1, 2, 3), Ι ≡ (0, 5, 1) x y −1 z − 2 = Find the image of the point (1, 6, 3) in the line = . Ans. (1, 0 , 7) 1 2 3 x − 15 y − 29 z − 5 = = Find the foot and hence the length of perpendicular from (5, 7, 3) to the line . 3 8 −5 x−5 y −7 z−3 = = Find also the equation of the perpendicular. Ans. (9, 13, 15) ; 14 ; 2 3 6

Find the length and foot of perpendicular drawn from point (2, –1, 5) to the line

2 3 . Angle Between A Plane And A Line: (i)

If θ is the angle between line

 sin θ =   

x − x1 y − y1 z − z1 = = and the plane ax + by + cz + d = 0, then m n

 . 2 + m 2 + n 2 

a + bm + cn

(a 2 + b2 + c2 )

(ii)

 b.n  Vector form: If θ is the angle between a line r = ( a + λ b ) and r . n = d then sin θ =   .

(iii)

Condition for perpendicularity

(iv)

Condition for parallel

| b | | n |

m n = = b xn = 0 c a b a + bm + cn = 0

b.n = 0

24. Condition For A Line To Lie In A Plane (i) (ii)

x − x1 y − y1 z − z1 = = would lie in a plane m n ax + by + cz + d = 0, if ax 1 + by1 + cz1 + d = 0 & a + bm + cn = 0. Vector form: Line r = a + λ b would lie in the plane r . n = d if b . n = 0 & a . n = d Cartesian form: Line

2 5 . Coplanar Lines: (i)

x − α' x − α y −β z − γ y − β' z − γ ' and = If the given lines are = = = , then condition n m ' m' n' α − α' β − β' γ − γ ' m n = 0 & plane containing the above for intersection/coplanarity is ' m' n' two lines

(ii)

x −α y −β z− γ m n =0 is m' n' '

Condition of coplanarity if both the lines are in general form Let the lines be ax + by + cz + d = 0 = a′ x + b′ y + c′ z + d′ & αx + βy + γ z + δ = 0 = α′x + β′y + γ′z + δ′

a b c a' b' c' They are coplanar if α β γ α' β' γ '

d d' =0 δ δ'

Alternative method: get vector along the line of shortest distance as

i j k m n ′ m′ n′

Now get unit vector along this vector uˆ = i + mj + nk Let v = (α – α′ ) î + (B – B′) ˆj + (y – y′) S. D. = u. v Example : Find the distance of the point (1, 0, – 3) from the plane x – y – z = 9 measured parallel to the x−2 y+2 z−6 = = line . 2 3 −6 Solution Given plane is x – y – z = 9 ..... (1) x−2 y+2 z−6 = = Given line AB is ..... (2) 2 3 −6 Equation of a line passing through the point Q(1, 0, – 3) and parallel to line (2) is x −1 y z + 3 = = = r.. ..... (3) 2 3 −6 Co-ordinates of point on line (3) may be taken as P (2r + 1, 3r, – 6r – 3) If P is the point of intersection of line (3) and plane (1), then P lies on plane (1), ∴ (2r + 1) – (3r) – (– 6r – 3) = 9 r=1 or, P ≡ (3, 3, – 9) Distance between points Q (1, 0, – 3) and P (3, 3, – 9) PQ =

(3 − 1)2 + (3 − 0 )2 + ( −9 − (3)) 2 =

4 + 9 + 36 = 7.

B Q (1, 0, – 3)

A P Example: Find the equation of the plane passing through (1, 2, 0) which contains the line

x + 3 y −1 z − 2 = = . 3 4 −2

Equation of any plane passing through (1, 2, 0) may be taken as a (x – 1) + b (y – 2) + c (z – 0) = 0 ..... (1) where a, b, c are the direction ratios of the normal to the plane. Given line is x + 3 y −1 z − 2 = = ..... (2) 3 4 −2 If plane (1) contains the given line, then 3a + 4b – 2c = 0 ..... (3) Also point (– 3, 1, 2) on line (2) lies in plane (1) ∴ a (– 3 – 1) + b (1 – 2) + c (2 – 0) = 0 or, – 4a – b + 2c = 0 ..... (4) Solving equations (3) and (4), we get, a b c = = 8 − 2 8 − 6 − 3 + 16 a b c = = or, = k (say). ..... (5) 6 2 13 Substituting the values of a, b and c in equation (1), we get, 6 (x – 1) + 2 (y – 2) + 13 (z – 0) = 0. or, 6x + 2y + 13z – 10 = 0. This is the required equation. x −1 y +1 z − 3 = = on the plane x + 2y + z = 9. Example : Find the equation of the projection of the line 2 −1 4 B Solution A Let the given line AB be x −1 y +1 z − 3 = = ..... (1) 2 −1 4 Given plane is x + 2y + z = 9 ..... (2) D C Let DC be the projection of AB on plane (2) Clearly plane ABCD is perpendicular to plane (2). Equation of any plane through AB may be taken as (this plane passes through the point (1, – 1, 3) on line AB) a (x – 1) + b (y + 1) + c (z – 3) = 0 ..... (3) where 2a – b + 4c = 0 ..... (4) [∵ normal to plane (3) is perpendicular to line (1)] Since plane (3) is perpendicular to plane (2), ∴ a + 2b + c = 0 ..... (5) Solving equations (4) & (5), we get, a b c = = . −9 2 5 Substituting these values of a, b and c in equation (3), we get 9 (x – 1) – 2 (y + 1) – 5 (z – 3) = 0 or, 9x – 2y – 5z + 4 = 0 ...... (6) Since projection DC of AB on plane (2) is the line of intersection of plane ABCD and plane (2), therefore equation of DC will be 9 x − 2y − 5z + 4 = 0 .....(i)   and ..... (7) x + 2y + z − 9 = 0 .....(ii) Let , m, n be the direction ratios of the line of intersection of planes (i) and (ii) ∴ 9 – 2m – 5n = 0 ..... (8) and + 2m + n = 0 ..... (9) m n = = ∴ − 2 + 10 − 5 − 9 18 + 2 x − 3 y +1 z + 2 x −7 y z+2 = = = = Example : Show that the lines and are coplanar. Also find the 2 −3 1 −3 1 2 equation of the plane containing them. Solution Given lines are x − 3 y +1 z + 2 = = = r (say) ..... (1) 2 −3 1 x −7 y z+7 = = and = R (say) ..... (2) −3 1 2 If possible, let lines (1) and (2) intersect at P. Any point on line (1) may be taken as (2r + 3, – 3r – 1, r – 2) = P (let). Any point on line (2) may be taken as (– 3R + 7, R, 2R – 7) = P (let). ∴ 2r + 3 = – 3R + 7 or, 2r + 3R = 4 ..... (3) Also – 3r – 1 = R or, – 3r – R = 1 ..... (4) Solution

and r – 2 = 2R – 7 or, r – 2R = – 5. ..... (5) Solving equations (3) and (4), we get, r = – 1, R = 2 Clearly r = – 1, R = 2 satisfies equation (5). Hence lines (1) and (2) intersect. ∴ lines (1) and (2) are coplanar. Equation of the plane containing lines (1) and (2) is x −3 y +1 z + 2 2 −3 1 =0 −3 1 2 or, (x – 3) (– 6 – 1) – (y + 1) (4 + 3) + (z + 2) (2 – 9) = 0 or, – 7 (x – 3) – 7 (y + 1) – 7 (z + 2) = 0 or, x–3+y+1+z+2=0 or, x + y + z = 0. Self practice problems: x−2 y+3 z−6 = = 1. Find the values of a and b for which the line is perpendicular to the plane a 4 −2 3x – 2y + bz + 10 = 0. Ans. a = 3, b = – 2 x −1 y − 2 z − 3 x−2 y −3 z−4 = = = = 2. Prove that the lines and are coplanar. Also find the equation 2 3 3 3 4 5 of the plane in which they lie. Ans. x – 2y + z = 0 3. Find the plane containing the line Ans. 4.

5.

13x + 3y – 72 – 7 = 0

y−3 z−4 x−2 x +1 y −1 −z + 1 = = and parallel to the line = = 3 5 2 1 −2 1

y−2 x−4 x −1 z−3 y −1 = & = z are intersecting each other. Find thire = = 3 5 2 2 4 intersection and the plane containing the line. Ans. (– 1, – 1, – 1) & 5x – 8y + 11z – 2 = 0 Show that the lines r = (– î – 3 ˆj – 5 kˆ ) + λ (–3 î – 5 ˆj – 7 kˆ ) & r (2 î + 4 ˆj + 6 kˆ ) + µ ( î +4 ˆj + 7 kˆ ) are

Show that the line

(

2 6 . Skew Lines: Lines (i)

)

r . î − 2ˆj + kˆ = 0

coplanar and find the plane containing the line. Ans.

The straight lines which are not parallel and non−coplanar i.e. non−intersecting are called α '−α β'−β γ '− γ m n ≠ 0, then lines are skew.. ' m' n' Shortest distance: Suppose the equation of the lines are

skew lines. If ∆ = (ii)

S.D. =

(iii)

(iv)

x −α y −β z− γ x − α' y − β' z − γ ' = = = = and m n n' ' m' (α − α' ) (mn'− m' n) + (β − β' ) (n − n' ) + ( γ − γ' ) (m'−' m)

∑ (mn'−m' n)

2

α'−α β'−β γ '− γ m n ÷ ∑ (mn′ − m′n)2 = m' n' ' Vector Form: For lines a1 + λ b1 & a 2 + λ b 2 to be skew ( b1 x b 2 ). ( a 2 − a1 ) ≠ 0 or [ b1 b 2 ( a 2 − a1 )] ≠ 0. Shortest distance between the two parallel lines r = a 1 + λ b & (a 2 − a1 ) x b r = a 2 + µ b is d = . |b|

Example : Find the shortest distance and the vector equation of the line of shortest distance between the lines given by → → →  → → → r = 3 r + 8 j + 3 k + λ  3 r − j + k    Given lines are →

Solution

→ → →  → → → r = 3 r + 8 j + 3 k + λ  3 i − j + k   

and

→ → → → → →  r = −3 r − 7 j + 6 k + µ  − 3 i + 2 j + 4 k   

→

→

..... (1)

→ → →  r = −3 r − 7 j + 6 k + µ  − 3 i + 2 j + 4 k  and   Equation of lines (1) and (2) in cartesian form is x −3 y−8 z−3 = = AB : =λ 3 −1 1 →

and

→



CD :  

→

1 3

A

→

  

..... (2)

L

B

90° 90° C

M

D

=µ

Let L ≡ (3λ + 3, – λ + 8, λ + 3) and M ≡ (– 3µ – 3, 2µ – 7, 4µ + 6) Direction ratios of LM are 3λ + 3µ + 6, – λ – 2µ + 15, λ – 4µ – 3. Since LM ⊥ AB ∴ 3 (3λ + 3µ + 6) – 1 (– λ – 2µ + 15) + 1 (λ – 4µ – 3) = 0 or, 11λ + 7µ = 0 ..... (5) Again LM ⊥ CD ∴ – 3 (3λ + 3µ + 6) + 2 (– λ – 2µ + 15) + 4 ( λ – 4µ – 3) = 0 or, – 7λ – 29µ – 0 ..... (6) Solving (5) and (6), we get λ = 0, µ = 0 ∴ L ≡ (3, 8, 3), M ≡ (– 3, – 7, 6) Hence shortest distance LM =

(3 + 3 ) 2 + (8 + 7 ) 2 + (3 − 6 ) 2

= 270 = 3 30 units Vector equation of LM is → → → → →  → r = 3 i + 8 j + 3 k + t  6 i + 15 j − 3 k    x−3 y−8 z−3 = = Note : Cartesian equation of LM is . 6 15 −3 Example : Prove that the shortest distance between any two opposite edges of a tetrahedron formed by →

the planes y + z = 0, x + z = 0, x + y = 0, x + y + z = Solution Given planes are y+z=0 ..... (i)

3 a is

2 a. x+z=0

x+y=0 ..... (iii) x+y+z= 3a Clearly planes (i), (ii) and (iii) meet at O(0, 0, 0) Let the tetrahedron be OABC Let the equation to one of the pair of opposite edges OA and BC be y + z = 0, x + z = 0 ..... (1)

..... (ii) ..... (iv)

, 0) P O (0, 0

x + y = 0, x + y + z = 3 a ..... (2) equation (1) and (2) can be expressed in symmetrical form as x −0 y−0 z−0 = = ..... (3) 1 1 −1 C Q (0, 0, x −0 y−0 z− 3a 3 a ) = = and, ..... (4) 1 −1 0 d. r. of OA and BC are (1, – 1) and (1, – 1, 0). Let PQ be the shortest distance between OA and BC having direction cosine (, m, n) ∴ PQ is perpendicular to both OA and BC. ∴ +m–n=0 and –m=0 Solving (5) and (6), we get, m n = = = k (say) 1 1 2 O also, 2 + m 2 + n2 = 1 A 1 ∴ k2 + k2 + 4k2 = 1 ⇒ k = 6 1 1 2 ∴ = ,m= ,n= C B 6 6 6 Shortest distance between OA and BC i.e. PQ = The length of projection of OC on PQ A P O = | (x 2 – x 1) + (y2 – y1) m + (z2 – z1) n | 90° 1 1 2 0 . + 0 . + 3 a . = = 2 a. 6 6 6 90° Self practice problems: C Q B

A

D

1.

Find the shortest distance between the lines its equation.

2.

27.

x −1 y − 2 z − 3 x−2 y −4 z−5 = = = = and . Find also 2 3 4 3 4 5

1

, 6x – y = 10 – 3y = 6z – 25 6 Prove that the shortest distance between the diagonals of a rectangular parallelopiped whose sides are bc ca ab , , a, b, c and the edges not meeting it are 2 2 2 2 2 b +c c +a a + b2 Ans.

Sphere: General equation of a sphere is given by x

2

+ y2 + z2 + 2ux + 2vy + 2wz + d = 0 (−u,

−v, −w) is the centre and u 2 + v 2 + w 2 − d is the radius of the sphere. Example : Find the equation of the sphere having centre at (1, 2, 3) and touching the plane x + 2y + 3z = 0. Solution : Given plane is x + 2y + 3z = 0 ..... (1) Let H be the centre of the required sphere. Given H ≡ (1, 2, 3) H Radius of the sphere, HP = length of perpendicular from H to plane (1) P | 1+ 2 × 2 + 3 × 3 | = = 14 14 Equation of the required sphere is (x – 1)2 + (y – 2)2 + (z – 3)2 = 14 or x 2 + y2 + z2 – 2x – 4y – 6z = 0 →

Example :

→

→

→

Find the equation of the sphere if it touches the plane r .( 2 i − 2 j − k ) = 0 and the position →

→

→

→

→

→

vector of its centre is 3 i + 6 j − 4 k

→

Solution Given plane is r .(2 i − 2 j − k ) = 0 Let H be the centre of the sphere, then →

→

→

→

..... (1)

→

OH = 3 i + 6 j − 4 k = c (say) Radius of the sphere = length of perpendicular from H ot plane (1) →

=

→

→

→

| c .(2 i − 2 j − k ) | →

→

→

|2 i −2 j− k | →

→

→

→

→

→

| (3 i + 6 j − 4 k ).(2 i − 2 j − k ) |

=

→

→

→

|2 i −2 j− k |

| 6 − 12 + 4 | 2 = = a (say) 3 3 Equation of the required sphere is

=

→

→

| r−c | = a →

or

→

→

→

→

→

| x i + y j + z k − (3 i + 6 j − 4 k ) | =

2 3

or

→ → 4 → | (x – 3) i + (y – 6) j + (z + 4) k | 2 = 9

4 or 9 (x 2 + y2 + z2 – 6x – 12y + 8z + 61) = 4 9 2 2 2 or 9x + 9y + 9z – 54x – 108y + 72z + 545 = 0 Example : Find the equation of the sphere passing through the points (3, 0, 0), (0, – 1, 0), (0, 0, – 2) and whose centre lies on the plane 3x + 2y + 4z = 1 Solution Let the equation of the sphere be x 2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 ..... (1) Let A ≡ (3, 0, 0), B ≡ (0, – 1, 0), C ≡ (0, 0, – 2) Since sphere (1) passes through A, B and C, ∴ 9 + 6u + d = 0 ..... (2) 1 – 2v + d = 0 ..... (3) 4 – 4w + d = 0 ..... (4) Since centre (– u, – v, – w) of the sphere lies on plane 3x + 2y + 4z = 1 ∴ – 3u – 2v – 4w = 1 ..... (5) (2) – (3) ⇒ 6u + 2v = – 8 ..... (6) (3) – (4) ⇒ – 2v + 4w = 3 ..... (7) −2 v − 8 From (6), u = ..... (8) 6 From (7), 4w = 3 + 2v ..... (9)

or

(x – 3)2 + (y – 6)2 + (z + 4)2 =

Self practice problems: 1.

Find the value of k for which the plane x + y + z = x 2 + y2 + z2 – 2x – 2y – 2z – 6 = 0. Ans. 3 ±3

2.

Find the equation to the sphere passing through (1, – 3, 4), (1, – 5, 2) and (1, – 3, 0) which has its centre in the plane x + y + z = 0 Ans. x 2 + y2 + z2 – 2x + 6y – 4z + 10 = 0

3.

Find the equation of the sphere having centre on the line 2x – 3y = 0, 5y + 2z = 0 and passing through the points (0, – 2, – 4) and (2, – 1, – 1). Ans. x 2 + y2 + z2 – 6x – 4y + 10 z + 12 = 0

4.

Find the centre and radius of the circle in which the plane 3x + 2y – z – 7 14 = 0 intersects the sphere x 2 + y2 + z2 = 81. Ans.

5.

3 k touches the sphere

4 2 units

A plane passes through a fixed point (a, b, c) and cuts the axes in A, B, C. Show that the locus of the centre of the sphere OABC is a b c + + x y z = 2.


Example : Find the equation of the sphere with the points (1, 2, 2) and (2, 3, 4) as the extremities of a diameter. Find the co-ordinates of its centre. Solution Let A ≡ (1, 2, 2), B ≡ (2, 3, 4) Equation of the sphere having (x 1, y1, z1) and (x 2, y2, z2) as the extremities of a diameter is (x – x 1) (x – x 2) + (y – y1) (y – y2) + (z – z1) (z – z2) = 0 Here x 1 = 1, x 2 = 2, y1 = 2, y2 = 3, z1 = 2, z2 = 4 ∴ required equation of the sphere is (x – 1) (x – 2) + (y – 2) (y – 3) + (z – 2) (z – 4) = 0 or x 2 + y2 + z2 – 3x – 5y – 6z + 16 = 0 Centre of the sphere is middle point of AB 3 5  ∴ Centre is  , , 3  2 2 



Putting the values of u, v and w in (5), we get 2v + 8 − 2v – 3 – 2 v = 1 2 ⇒ 2v + 8 – 4v – 6 – 4v = 2 ⇒ v = 0 0−8 4 =− From (8), u= 6 3 3 From (9), 4w = 3 ∴ w= 4 From (3), d = 2v – 1 = 0 – 1 = – 1 From (1), equation of required sphere is 0−8 8 3 − x 2 + y 2 + z2 – x+ z–1=0 6 3 2 or 6x 2 + 6y2 + 6z2 – 16x + 9z – 6 = 0

Vector 1.

Vectors & Their Representation:

Vector quantities are specified by definite magnitude and definite directions. A vector is generally represented by a directed line segment, say AB . A is called the initial point & B is called the terminal point. The magnitude of vector AB is expressed by  AB . Zero Vector: A vector of zero magnitude is a zero vector. i.e. which has the same initial & terminal Vector point, is called a Zero Vector. It is denoted by O. The direction of zero vector is indeterminate. Unit Vector: A vector of unit magnitude in the direction of a vector a is called unit vector along a Vector a and is denoted by aˆ symbolically, aˆ = . |a| Equal Vectors: Vectors Two vectors are said to be equal if they have the same magnitude, direction & represent the same physical quantity. Collinear Vectors: Two vectors are said to be collinear if their directed line segments are parallel irrespective of their directions. Collinear vectors are also called parallel vectors. If they have the same direction they are named as like vectors otherwise unlike vectors. Symbolically, two non zero vectors a and b are collinear if and only if, a = Kb , where K ∈ R a3 a1 a2 Vectors a = a1 î + a 2 ˆj + a 3kˆ and b = b1 î + b 2 ˆj + b 3kˆ are collinear if b = b = b 1 2 3 Coplanar Vectors: A given number of vectors are called coplanar if their line segments are all Vectors parallel to the same plane. Note that “T WO VECTORS ARE ALWAYS C OPLANAR ”. Solved Example Find unit vector of î − 2ˆj + 3kˆ if Solution a = a x î + a y ˆj + a zkˆ a = î − 2ˆj + 3kˆ 2 2 2 then ∴ | a | = ax + ay + az | a | = 14 3 1 a 2 ˆj + ˆ aˆ = | a | = 14 î – 14 k 14 Solved Example Find values of x & y for which the vectors a = (x + 2) î – (x – y) ˆj + kˆ b = (x – 1) î + (2x + y) ˆj + 2 kˆ are parallel. y−x x+2 1 a and b are parallel if x − 1 = 2 x + y = 2 x = – 5, y = – 20

Solution

2.

Angle Between two Vectors It is the smaller angle formed when the initial points or the terminal points of the two vectors are brought together. It should be noted that 0º ≤ θ ≤ 180º .

3.

Addition Of Vectors:

If two vectors a

→ → & b are represented by OA & OB , then their sum a + b is a vector represented

→

by OC , where OC is the diagonal of the parallelogram OACB.

a + b = b + a (commutative) (a + b) + c = a + ( b + c) (associativity) a+0 = a = 0+a |a+b|≤|a| +|b| a + ( −a ) = 0 = ( −a ) + a

| a − b | ≥||a| −|b|| a±b

=

a b A vector in the direction of the bisector of the angle between the two vectors a & b is + . Hence a b bisector of the angle between the two vectors a and b is λ a + b , where λ ∈ R+. Bisector of the exterior angle between a & b is λ a − b , λ ∈ R+.

(

4.

| a |2 + | b |2 ±2 | a || b | cos θ where θ is the angle between the vectors

)

(

Multiplication Of A Vector By A Scalar:

)

If a is a vector & m is a scalar, then m a is a vector parallel to a whose modulus is m times that of a . This multiplication is called SCALAR MULTIPLICATION. If a and b are vectors & m, n are scalars, then:

m ( a ) = (a ) m = m a m (na) = n(ma) = (mn )a (m + n ) a = m a + n a m (a + b ) = m a + m b Solved Example: If a = î + 2ˆj + 3kˆ and b = 2î + 4 ˆj − 5kˆ represent two adjacent sides of a parallelogram, find unit vectors parallel to the diagonals of the parallelogram. Solution. Let ABCD be a parallelogram such that AB = a and BC = b . Then, AB + BC = AC ⇒ AC = a + b = 3î + 6ˆj − 2kˆ

⇒ Now,

⇒ BD = AD − AB = b − a ⇒ AC = 3î + 6ˆj − 2kˆ

| AC | =

9 + 36 + 4 = 7

and,

ˆ ˆ ˆ BD = i + 2 j − 8k

⇒

| BD | =

1 + 4 + 64 =

and

AB + BD = AD

∴ and,

AD + AD = AB

(

AC 1 Unit vector along AC = = 3î + 6ˆj − 2kˆ 7 | AC |

BD Unit vector along BD = = | BD |

Solved Example

1 69

69

)

(î + 2ˆj − 8kˆ )

ABCDE is a pentagon. Prove that the resultant of the forces AB , AE , BC , DC , ED

and AC is 3 AC . Solution. Let R be the resultant force ∴ R = AB + AE + BC + DC + ED + AC

∴ R = ( AB + BC ) + ( AE + ED + DC ) + AC = AC + AC + AC = 3 AC . Hence proved. Self Practice Problems : 1. Express : (i) The vectors BC CA and AB in terms of the vectors OA , OB and OC (ii) The vectors OA , OB and in terms of the vectors OC , OB and OC . 2.

Ans. (i) BC = OC − OB , CA = OA − OC , AB = OB − OA Given a regular hexagon ABCDEF with centre O, show that

(i) OB – OA = OC – OD (ii) OD + OA = 2 OB + OF (iii) AD + EB + PC = 4 AB ˆ ˆ ˆ 3. The vector − i + j − k bisects the angle between the vectors c and 3î + 4ˆj . Determine the unit vector 1 2 ˆ 14 ˆ − î + j− k along c . Ans. 3 15 15 4. The sum of the two unit vectors is a unit vector. Show that the magnitude of the their difference is 3 .

5.

Position Vector Of A Point: Point

let O be a fixed origin, then the position vector of a point P is the vector OP . If a and b are position vectors of two points A and B, then, AB = b − a = pv of B − pv of A.

DISTANCE DISTANCE FORMULA F ORMULA

Distance between the two points A (a) and B (b) is AB = a − b SECTION FORMULA If a and b are the position vectors of two points A & B then the p.v. of na + mb a point which divides AB in the ratio m: n is given by: r = . m+n a+b Note p.v. of mid point of AB = . 2 ABCD is a parallelogram. If L, M be the middle point of BC and CD, express AL and 3 AC . AM in terms of AB and AD , also show that AL + AM = 2 Solution.Let the position vectors of points B and D be respectively b and a referred to A as origin of reference.

Solved Example:

Then AC = AD + DC = AD + AB

[∵

DC = AB ]

= d + b

i.e.

AB = b , AD = d position vector of C referred to A is d + b

∵

∴

AL = p.v. of L, the mid point of BC . 1 1 1 = [p.v. of D + p.v. of C] = b + d + b = AB + AD 2 2 2 1 1 AM = 2 d + d + b = AD + 2 AB 1 1 ∴ AL + AM = b + b d + d+ 2 2 3 3 3 3 b + d = = (b + d ) = AC . 2 2 2 2 Solved Example If ABCD is a parallelogram and E is the mid point of AB, show by vector method that DE trisects and is trisected by AC. Solution. Let AB = a and AD = b Then BC = AD = b and AC = AB + AD = a + b Also let K be a point on AC, such that AK : AC = 1 : 3

(

[

)

]

1 1 AK = .........(i) (a + b ) AC ⇒ 3 3 Again E being the mid point of AB, we have 1 a AE = 2 Let M be the point on DE such that DM : ME = 2 : 1 b+a AD + 2AE AM = ∴ = ..........(ii) 3 1+ 2 1 From (i) and (ii) we find that : AK = ( a + b ) = AM , and so we conclude that K and M coincide. i.e. DE 3 trisect AC and is trisected by AC. Hence proved. Self Practice Problems 1. If a, b are position vectors of the points (1, –1), (–2, m), find the value of m for which a and b are collinear. Ans. m = 2 2. The position vectors of the points A, B, C, D are î + ˆj + kˆ , 2î + 5ˆj , 3î + 2ˆj − 3kˆ , î − 6ˆj − kˆ respectively.. Show that the lines AB and CD are parallel and find the ratio of their lengths. Ans. 1 : 2 3. The vertices P, Q and S of a triangle PQS have position vectors p, q and s respectively.. (i) Find m , the position vector of M, the mid-point of PQ, in terms of p and q . (ii) Find t , the position vector of T on SM such that ST : TM = 2 : 1, in terms of p, q and s . (iii) If the parallelogram PQRS is now completed. Express r , the position vector of the point R in terms of p, q and s Prove that P, T and R are collinear. 1 1 1 Ans. t = 2 (p + q + s ) , m = 2 ( p + q) , r = 2 q+p−s 4. D, E, F are the mid-points of the sides BC, CA, AB respectively of a triangle. Show FE = 1/2 BC and

or,

5. 6.

AK =

that the sum of the vectors AD , BE , CF is zero. The median AD of a triangle ABC is bisected at E and BE is produced to meet the side AC in F; show that AF = 1/3 AC and EF = 1/4 BF. Point L, M, N divide the sides BC, CA, AB of ∆ABC in the ratios 1 : 4, 3 : 2, 3 : 7 respectively. Prove that AL + BM + CN is a vector parallel to CK , when K divides AB in the ratio 1 : 3.

6.

Scalar Product Of Two Vectors: Geometrical interpretation of Scalar Product Let a and b be vectors represented by OA and OB respectively. Let θ be the angle between OA and OB . Draw BL ⊥ OA and AM ⊥ OB. From ∆s OBL and OAM, we have OL = OB cos θ and OM = OA cos θ. Here OL and OM are known as

1. 2.

projections of b on a and a on b respectively.. Now, = | a | | b | cos θ a.b = | a | (OB cos θ ) = | a | (OL) = (Magnitude of a ) (Projection of b on a ) ........(i) Again a . b = | a | | b | cos θ = | b | (| a | cos θ ) = | b | (OA cos θ) = | b | (OM) = (magnitude of b ) (Projection of a on b ) ........(ii) Thus geometrically interpreted, the scalar product of two vectors is the product of modulus of either vector and the projection of the other in its direction. a.b i.i = j.j = k.k = 1; i.j = j.k = k.i = 0 projection of a on b = |b|

a = 3. 4.

if a = a1i + a2j + a3k & b = b1i + b2j + b3k then a . b = a1b1 + a2b2 + a3b3

a 12 + a 2 2 + a 3 2

b =

,

b12 + b 2 2 + b 32

a.b 0≤φ≤π the angle φ between a & b is given by cos φ = |a| |b| a . b = a b cos θ (0 ≤ θ ≤ π) , note that if θ is acute then a . b > 0 & if θ is obtuse then a . b < 0

6.

2 a . a = a = a 2 , a.b = b.a (commutative) a.b = 0 ⇔ a ⊥ b (a ≠ 0 b ≠ 0 )

7.

(m a ) . b = a . (m b) = m (a . b) (associative) where m is scalar..

5.

a . (b + c ) = a . b + a . c (distributive)

Maximum value of a . b is  a   b  (ii) Minimum value of a . b is –  a   b  a . i i + a . j j + a . k k. (iii) Any vector a can be written as, a = Solved Example Find the value of p for which the vectors a = 3î + 2ˆj + 9kˆ and b = î + pˆj + 3kˆ are (i) perpendicular (ii) parallel Solution. (i) ⇒ = 0 ⇒ a.b 3î + 2ˆj + 9kˆ . î + pˆj + 3kˆ = 0 a⊥b ⇒ 3 + 2p + 27 = 0 ⇒ p = – 15 ˆ ˆ ˆ ˆ ˆ (ii) We know that the vectors a = a1i + a 2 j + a 3k and b = b1i + b 2 j + b 3kˆ are parallel iff a = λb ⇔ a1î + a 2 ˆj + a 3kˆ = λ b1î + b 2 ˆj + b 3kˆ ⇔ a1 = λb1, a2 = λb2, a3 = λb3 a1 a2 a3 ⇔ b = b = b ( =λ) 1 2 3 So, vectors a = 3î + 2ˆj + 9kˆ and b = î + pˆj + 3kˆ are parallel iff Note : (i)

( ) ( ) ( ) (

(

)

2 9 3 = = p 3 1 Solved Example: If a + b + c = 0 , | a | Solution. We have, a + b + c = 0 ⇒ ⇒ a +b = –c 2 2 a+b = |c| ⇒ ⇒ 2 2 ⇒ +2 a a + b

) (

(

)

)

⇒

3=

2 ⇒ p

p=2/3 = 3, | b | = 5 and | c | = 7, find the angle between a and b .

(a + b ) . (a + b ) = (− c ) . (− c ) a

b

2

+ b

2

+ 2a . b = c

cos θ = c

2

2

π . 3 Solved Example Find the values of x for which the angle between the vectors a = 2x 2 î + 4x ˆj + kˆ and b = 7 î – 2 ˆj + x kˆ is obtuse. a.b Solution. The angle q between vectors a and b is given by cos θ = | a||b|

⇒

9 + 25 + 2 (3) (5) cos θ = 49

⇒

cos θ =

1 2

⇒

θ=

Now, θ is obtuse

a.b <0 | a||b|

⇒ ⇒

cos θ < 0 a.b < 0 14x 2 – 8x + x < 0

⇒ [∵, | a |, | b | > 0 ]

⇒

17x (2x – 1) < 0

⇒ x(2x – 1) < 0 ⇒ 0 < x <

⇒

1 2

Hence, the angle between the given vectors is obtuse if x ∈ (0, 1/2) Solved Example:D is the mid point of the side BC of a triangle ABC, show that AB2 + AC2 = 2 (AD2 + BD2) Solution. We have AB = AD + DB ⇒

AB2 = ( AD + DB )2 = AD2 + DB + 2 AD . DB

Also we have

.............(i)

2 ⇒ AC2 = ( AD + DC) AC = AD + DC = AD2 + DC2 + 2 AD . DC Adding (i) and (ii), we get

.............(ii)

AB2 + AC2 = 2AD2 + 2BD2 + 2 AD . (DB + DC ) = 2(DA2 + DB2), for DB + DC = 0

Solved Example If a = î + ˆj + kˆ and b = 2 î – ˆj + 3 kˆ , then find (i) Component of b along a . (ii) Component of b perpendicular to along a . Solution. (i) Component of b along a is  a.b    a  | a |2    Here a . b = 2 – 1 + 3 = 4 | a |2 = 3  a.b  4 4   Hence  2  a = a = ( î + ˆj + kˆ ) | | 3 3  a   a.b  1   (ii) Component of b perpendicular to along a is b –  2î − 7ˆj + 5kˆ 2  a.= | | 3  a  θ Self Practice Problems :1. If a and b are unit vectors and θ is angle between them, prove that tan = 2 |a−b| . 2. Find the values of x and y is the vectors a = 3î + xˆj − kˆ and b = 2î + ˆj + ykˆ are mutually |a+b| 31 41 perpendicular vectors of equal magnitude. Ans. x = – , y= 12 12 3. Let a = x 2 î + 2ˆj − 2kˆ , b = î − ˆj + kˆ and c = x 2 î + 5ˆj − 4kˆ be three vectors. Find the values of x for which the angle between a and b is acute and the angle between b and c is obtuse. Ans.(–3, –2) ∪ (2, 3) 4. The points O, A, B, C, D, are such that OA = a , OB = b , OC = 2a + 3b , OD = a + 2b . Give that the

(

5.

7. 1.

)

length of OA is three times the length of OB show that BD and AC are perpendicular.. ABCD is a tetrahedron and G is the centroid of the base BCD. Prove that AB2 + AC2 + AD2 = GB2 + GC2 + GD2 + 3GA2

Vector Product Of Two Vectors:

If a & b are two vectors & θ is the angle between them then a x b = a b sin θ n , where n is the unit

vector perpendicular to both a & b such that a , b & n forms a right handed screw system.

2.

Geometrically a x b = area of the parallelogram whose two adjacent sides are represented by a & b .

3.

î × î = ˆj × ˆj = kˆ × kˆ = 0 ; î × ˆj = kˆ, ˆj × kˆ = î, kˆ × î = ˆj

4.

If a = a1 î +a2 ˆj + a3 kˆ

î ˆj kˆ & b = b1 î + b2 ˆj + b3 kˆ then a × b = a1 a 2 a 3 b1 b 2

5. 6. 7. 8. 9.

( )

(

)

axb Unit vector perpendicular to the plane of a & b is n = ± axb r axb A vector of magnitude ‘r’ & perpendicular to the palne of a & b is ± axb axb If θ is the angle between a & b then sin θ = a b

( )

If a , b & c are the pv’s of 3 points A, B & C then the vector area of triangle ABC = 1 a x b + bx c + cx a . The points A, B & C are collinear if a x b + b x c + c x a = 0

2

[

]

1 d1 xd 2 2 2 2 2 2 a .a a . b Lagrange's Identity: for any two vectors a & b ;(a x b) = a b − (a . b) = a .b b.b

b3

a x b ≠ b x a (not commutative) (m a ) × b = a × m b = m a × b (associative) where m is a scalar.. a x ( b + c ) = (a x b ) + (a x c ) (distributive) a x b = 0 ⇔ a & b are parallel (collinear) (a ≠ 0 , b ≠ 0) i.e. a = K b , where K is a scalar..

Area of any quadrilateral whose diagonal vectors are d1 & d 2 is given by

Solved Example Find a vector of magnitude 9, which is perpendicular to both the vectors 4î + ˆj + 3kˆ and − 2î + ˆj − 2kˆ . Solution. Let a = 4î − ˆj + 3kˆ and b = − 2î + ˆj − 2kˆ . Then, î ˆj kˆ 4 − 1 3 = (2 – 3) ˆ – (–8 + 6) ˆ + (4 – 2) i j kˆ = − î + 2ˆj + 2kˆ −2 1 −2 ⇒ | a × b | = ( −1)2 + 2 2 + 2 2 = 3  a×b  9   = ∴ Required vector = 9  ( −î + 2ˆj + 2kˆ ) = − 3î + 6ˆj + 6kˆ  3 | a×b | Solved Example For any three vectors a, b, c . Show that a × (b + c ) + b × (c + a) + c × (a + b) = 0 . Solution. We have, a × (b + c ) + b × (c + a) + c × (a + b) = a × b + a × c + b × c + b × a + c × a + c × b [Using distributive law] = a×b + a×c +b×c −a×b −a×c −b×c [∵ a × b = −b × a etc] Solved Example: For any vector a , prove that | a × î |2 + | a × ˆj |2 + | a × kˆ |2 = 2 | a |2 Solution. Let a = a1î + a 2 ˆj + a 3 kˆ . Then a × î = (a1î + a 2 ˆj + a 3kˆ ) × î = a1 ( î × î ) + a2 ( ˆj × î ) + a3 (kˆ × î ) = –a2 kˆ + a 3 î ⇒ | a × î |2 = a22 + a32 a × ˆj = (a1î + a 2 ˆj + a 3kˆ ) × ˆj = a1kˆ − a 3 î ⇒ ⇒ | a × ˆj |2 = a21 + a32 | a × kˆ |2 = a12 + a22 ∴ | a × î |2 + | a × ˆj |2 + | a × kˆ |2 = a22 + a33 + a12 + a32 + a12 + a22 2 (d12 + a22 + a32) = 2 | a |2 a×b =

Solved Example: Let OA = a , OB = 10 a + 2b and OC = b where O is origin. Let p denote the area of the quadrilateral OABC and q denote the area of the parallelogram with OA and OC as adjacent sides. Prove that p = 6q. Solution. We have, p = Area of the quadrilateral OABC 1 = | OB × AC | 2 1 = | OB × (OC − OA ) | 2 1 = | (10a + 2b) × (b − a) | 2 1 = | 10(a × b − 10(a × a) + 2(b × b) − 2(b × a) | 2 1 = | 10(a × b) − 0 + 0 + 2(a × b) | 2 and, q = Area of the parallelogram with OA and OC as adjacent sides = | OA × OC | = | a × b | ........(ii) From (i) and (ii), we get p = 6q Self Practice Problems : 1. If p and q are unit vectors forming an angle of 30º; find the area of the parallelogram having a = p + 2q and b = 2p + q as its diagonals. Ans. 3/4 sq. units 2. Show that {( a + b + c ) × ( c – b )} . a = 2 [ a b c ]. 3. Prove that the normal to the plane containing the three points whose position vectors are a, b, c lies in the direction b × c + c × a + a × b 4. ABC is a triangle and EF is any straight line parallel to BC meeting AC, AB in E, F respectively. If BR and CQ be drawn parallel to AC, AB respectively to meet EF in R and Q respectively, prove that ∆ ARB = ∆ACQ.

8.

Scalar Triple Product:

The scalar triple product of three vectors a , b & c is defined as: a x b . c = a b c sin θ cos φ where

θ is the angle between a & b & φ is the angle between a x b & c . It is also written as [ a b c ] and spelled as box product.

Scalar triple product geometrically represents the volume of the parallelopiped whose three coterminous edges are

represented by a , b & c i. e. V = [ a b c ]

In a scalar triple product the position of dot & cross can be interchanged i.e.

a . (b x c) = − a .( cx b) i. e. [ a b c ] = − [ a c b ]

1 2 3 If a = a1i+a2j+a3k; b = b1i+b2j+b3k & c = c1i+c2j+c3k then [a b c] = b1 b 2 b 3 .

a . ( b x c ) = (a x b ). c

OR [ a b c ] = [ b c a ] = [ c a b ]

a a a

c1 c2 c3 In general, if a = a 1 l + a 2 m + a 3 n ; b = b1 l + b 2 m + b 3 n & c = c1 l + c2 m + c3 n a1 a 2 a 3 then a b c = b1 b 2 b 3 l m n ; where , m & n are non coplanar vectors. c1 c2 c3 If a , b , c are coplanar ⇔ [ a b c ] = 0 . Scalar product of three vectors, two of which are equal or parallel is 0 i.e. [ a b c ] = 0 , If a , b , c are non − coplanar then [ a b c ] > 0 for right handed system & [ a b c ] < 0 for left handed

[ ]

[

system.

]

[(a + b) c d ] = [ a c d ] + [ b c d ] The volume of the tetrahedron OABC with O as origin & the pv’s of A, B and C being a , b & c respectively 1 is given by V = [ a b c ] 6 [i j k] = 1

[ K a b c ] = K[ a b c ]

The positon vector of the centroid of a tetrahedron if the pv’s of its vertices are a , b , c & d are given by

1 [a + b + c + d] . 4

Note that this is also the point of concurrency of the lines joining the vertices to the centroids of the opposite faces and is also called the centre of the tetrahedron. In case the tetrahedron is regular it is equidistant from the vertices and the four faces of the tetrahedron.

[

]

[

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[ ]

Remember that: a − b b − c c − a = 0 & a + b b + c c + a = 2 a b c . Solved Example Find the volume of a parallelopiped whose sides are given by − 3î + 7ˆj + 5kˆ , − 5î + 7ˆj − 3kˆ and 7î − 5ˆj − 3kˆ Solution. Let a = −3î + 7ˆj + 5kˆ , b = −5î + 7ˆj + 3kˆ and c = 7î − 5ˆj − 3kˆ . We know that the volume of a parallelopiped whose three adjacent edges are a, b, c is [a, b, c ] . −3 7 5 −5 7 −3 = –3 (–21 – 15) – 7 (15 + 21) + 5 (25 – 49) Now, [a b c ] = 7 −5 −3 = 108 – 252 – 120 = –264 So, required volume of the parallelopiped = [a, b, c ] = | – 264 | = 264 cubic units Solved Example: Simplify [a − b b − c c − a] Solution. We have : [by def.] [a − b b − c c − a] = {(a − b) × (b − c )} . (c − a) = (a × b − a × c − b × b + b × c ) . ( c − a ) [by dist. law] = (a × b + c × a + b × c ) . ( c − a ) [ ∵ b×b = 0 ] = (a × b) . c – (a × b) . a + (c × a) . c – (c × a ) . a + (b × c ) . c – (b × c ) . a [by dist. law] = [a b c ] – [a b a] + [c a c ] – [c a a] + [b c c ] – [b c a] = [a b c ] – [b c a] [∵ scalar triple product when any two vectors are equal is zero ] = [a b c ] – [a b c ] =0 [∵ [b c a] = [a b c ] ] Solved Example: Find the volume of the tetrahedron whose four vertices have position vectors a b c and d . Solution. Let four vertices be A, B, C, D with p. v. a b c and d . respectively.. ∴ = (a – d) DA = (b – d ) DB = (c – d) DC 1 Hence volume = [a – d b – d c – d] 6 1 = ( a – d ) . [( b – d ) × ( c – d )] 6 1 = (a – d) . [b × c – b × d + c × d ] 6 1 = {[ a b c ] – [ a b d ] + [ a c d ] – [ d b c ]} 6 1 = {[ a b c ] – [ a b d ] + [ a c d ] – [ b c d ]} 6 Solved Example: Show that the vectors a = −2 i + 4 j − 2k , b = 4 i − 2 j − 2k and c = −2 i − 2 j + 4k are coplanar.. −2 4 −2 4 −2 −2 = 0 Solution: −2 −2 4 Self Practice Problems : 1. Show that a . (b + c ) × (a + b + c ) = 0 2. One vertex of a parallelopiped is at the point A (1, –1, –2) in the rectangular cartesian co-ordinate. If three adjacent vertices are at B(–1, 0, 2), C(2, –2, 3) and D(4, 2, 1), then find the volume of the parallelopiped. Ans. 72 The vectors are coplanar since [a b c ] =

3.

Find the value of m such that the vectors 2î − ˆj + kˆ , î + 2ˆj − 3kˆ and 3î + mˆj + 5kˆ are coplanar.. Ans. – 4

4.

9.

Show that the vector a, b, c , are coplanar if and only if b + c , c + a , a + b are coplanar..

Vector Triple Product:

Let a , b , c be any three vectors, then the expression a x ( b x c ) is a vector & is called a vector triple product. Geometrical Interpretation of a x ( b x c )

vectors a & ( b x c ) . Now a x ( b x c ) is a vector perpendicular to the plane containing a & ( b x c ) but b x c is a vector perpendicular to the plane containing b & c , therefore a x ( b x c ) is a vector which lies in the plane of b & c and perpendicular to a . Hence we can express a x ( b x c ) in terms of b & c i.e. a x ( b x c ) = xb + yc where x & y are scalars. a x ( b x c ) = (a . c) b − (a . b) c (a x b) x c = (a . c) b − (b . c) a (a x b) x c ≠ a x (b x c) , in general Consider the expression a x ( b x c ) which itself is a vector, since it is a cross product of two

For any vector a , prove that î × (a × î ) + ˆj × (a × ˆj ) + kˆ × (a × k ) = 2a Solution. Let a = a1î + a 2 ˆj + a 3kˆ . Then, î × (a × î ) + ˆj × (aˆ × ˆj ) + kˆ × (a × kˆ ) = {(î . î )a − ( î . a)î } + {( ˆj . ˆj )a − ( ˆj . a)ˆj} + {(kˆ . kˆ )a − (kˆ . a)kˆ } = {(a − ( î . a) î } + {a − ( ˆj . a)ˆj} + {a − (kˆ . a)kˆ } = 3a − {( î . a)î + ( ˆj . a)ˆj + (kˆ . a)kˆ = 3a − (a1î + a 2 ˆj + a 3kˆ ) = 3a − a = 2a Solved Example Prove that a × {b × (c × a)} = (b . d)(a × c ) – (b. c ) (a × d) Solution. We have, a × {b × (c × a)} = a × {(b . d) c − (b . c ) d} = a × {(b . d) c } − a × {(b . c ) d} [by dist. law] = (b . d) (a × c ) − (b . c ) − (b . c ) (a × d) Solved Example: Let a = α î + 2ˆj – 3kˆ , b = î + 2αˆj – 2kˆ and c = 2î – αˆj + kˆ . Find the v alue(s) of α, if any, such that × ( c × a ) = 0. Find the vector product when α = 0. a×b × b×c Solution.: × ( c×a ) a×b × b×c = a b c b × ( c×a ) = a b c a.b c − b.c a which vanishes if (i) a . b c = b . c a (ii) a b c = 0 (i) a . b c = b . c a leads to the equation 2 α3 + 10 α + 12 = 0, α2 + 6α = 0 and 6α – 6 = 0, which do not have a common solution. (ii) a b c = 0 α 2 −3 2 1 2α − 2 =0 ⇒ 3α = 2 ⇒ α= ⇒ 3 2 −α 1 when α = 0, a b c = – 10, a . b = 6, b . c = 0 and the vector product is – 60 2î + kˆ . b × a + a | a |2 −1 a×b + a Sol ExaIf A + B = a , A . a = 1 and A × B = b , then prove that A = and B = . | a |2 | a |2 Solution.: Given A + B = a .....(i) 2 ⇒ ⇒ ⇒ 1 + a.B = |a| a. A +B = a.a a.A+a . B = a.a 2 ⇒ Given A × B = b ⇒ a.B = |a| – 1 a× A ×B = a × b ⇒ a .B A – a . A B = a×b ⇒ [using equation (2)] | a |2 −1 A − B = a × b solving equation (1) and (5), simultaneously, we get b × a + a | a |2 −1 a×b + a and B = A = | a |2 | a |2

Solved Example

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Sol. Ex. Solve for r , the simultaneous equations r × b = c × b , r . a = 0 provided a is not perpnedicular to (r − c) × b = 0 Solution ⇒ r − c and b are collinear ∴ ⇒ r = c + kb ........(i) r − c = kb ⇒ r .a = 0 (c + k b ) . a = 0 a.c a.c r =c− b ⇒ k=– putting in (i) we get a.b a.b Solved Example :If x × a + kx = b , where k is a scalar and a, b are any two vectors, then determine terms of a, b and k. Solution: x × a + kx = b ..........(i) Premultiple the given equation vectorially by a a × ( x × a ) + k (a × x ) = a × b ⇒ (a . a) x − (a . x ) a + k(a × x ) = a × b ..........(ii) Premultiply (i) scalarly by a [a x a ] + k ( a . x ) = a . b k(a . x ) = a . b .......(iii) Substituting x × a from (i) and a . x from (iii) in (ii) we get  (a . b )  1 k b + ( a × b ) + a x = a2 + k 2  k   Self Practice Problems : 1. Prove that a × (b × c ) + b × (c × a) + c × (a × b) = 0 . 2.

3. 4. 5. 6.

b.

x in

Find the unit vector coplanar with î + ˆj + 2 kˆ and î + 2 ˆj + kˆ and perpendicular to î + ˆj + kˆ . 1 1 Ans. ( – ˆj + kˆ ) or, ( ˆj – kˆ ) 2 2 Prove that a × {a × (a × b)} = (a . a) (b × a) . 1 1 Given that x + 2 (p . x ) p = q , show that p . x = p . q and find x in terms of p and q . p 2 If x . a = 0, x . b = 0 and x . c = 0 for some non-zero vector x , then show that [a b c ] = 0 ( r . a ) (b × c ) ( r . b ) (c × a ) ( r . c ) (a × b ) Prove that r = + + where a, b, c are three non-coplanar vectors [abc ] [abc ] [abc ]

1 0 . Reciprocal System Of Vectors:

If a, b, c & a' , b' , c' are two sets of non coplanar

vectors such that a.a' = b. b' = c. c' = 1 then the two systems are called Reciprocal System of vectors. bxc cxa axb a= b= c= Note: abc abc abc

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]

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Solved Example If a b c and a′, b′, c ′ be the reciprocal system of vectors, prove that (i) a . a′ + b . b′ + c . c ′ = 3 (ii) a × a′ + b × b′ + c × c ′ = 0 Solution. (i) We have : a . a ′ = b . b′ = c . c ′ = 1 a . a ′ + b . b′ + c . c ′ = 1 + 1 + 1 = 3

(ii) We have :

and ∴

1 a′ = λ (b × c ) , b ′ = λ (c × a) and c ′ = λ ( (a × b) , where λ = [a b c ] a × a′ = a × λ(b × c ) = λ {a × (b × c )} = λ {(a . c ) b − (a . b) c } b × b′ = b × λ(c × a ) = λ {b × (c × a )} = λ {(b . a) c − (b . c ) a} c × c ′ = c × λ(a × b ) = λ {c × (a × b )} = λ {(c . b) a − (c . a) b} a × a′ + b × b ′ + c × c ′ = λ {(a . c ) b − (a . b) c } + λ {(b . a) c − (b . c ) a} + λ {(c . b) a − (c . a) b} = λ [(a . c ) b − (a . b) c + (b . a ) c − (b . c ) a + (c . b) a − (c . a) b] = λ [(a . c ) b − (a . b) c + (a . b ) c − (b . c ) a + (b . c ) a − (a . c ) b] = λ0 = 0

1 1 . Linear Combinations:

Given a finite set of vectors a , b , c ,...... then the vector r = xa + yb + zc + ........ is called a linear

combination of a , b , c ,...... for any x, y, z..... ∈ R. We have the following results:

(a)

If a , b are non zero, non−collinear vectors then xa + yb = x' a + y' b ⇒ x = x' ; y = y'

(b)

FundamentalTheorem: Let a , b be non zero, non collinear vectors. Then any vector r coplanar

with a , b can be expressed uniquely as a linear combination of a , b

i.e. There exist some uniquly x, y ∈ R such that xa + yb = r . (c)

(d)

(e)

(f)

If a , b , c are non−zero, non−coplanar vectors then:

xa + yb + zc = x' a + y' b + z' c ⇒ x = x' , y = y' , z = z' Fundamental Theorem In Space: Let a , b , c be non−zero, non−coplanar vectors in space. Then any vector r , can be uniquly expressed as a linear combination of a , b , c i.e. There exist some unique x,y ∈ R such that xa + yb + zc = r . I f x1 , x 2 ,...... x n are n non zero v ect ors, & k 1 , k 2 , . . . . . k n are n scal ars & i f t h e l i near c om bi n at i on k1x1 + k 2 x 2 +........ k n x n = 0 ⇒ k1 = 0, k 2 = 0..... k n = 0 t hen we say t hat vectors x1 , x 2 ,...... x n are L INEARLY INDEPENDENT VECTORS. If x1 , x 2 ,...... x n are not L INEARLY I NDEPENDENT then they are said to be L INEARLY DEPENDENT vectors. i.e. if k 1x1 + k 2 x 2 + ........ + k n x n = 0 & if there exists at least one kr ≠ 0 then x1 , x 2 ,...... x n are said to

be L I N E AR LY L Y D E PE N DE N T . Note 1: If kr ≠ 0; k1x1

+ k 2 x 2 + k 3x 3 + ....... + k r x r + ...... + k n x n = 0 − k r x r = k1x1 + k 2 x 2 + ....... + k r −1 . x r −1 + k r +1 . x r +1 +......+ k n x n 1 1 1 1 1 −kr x r = k1 x1 + k 2 x 2 + .....+ k r −1. x r −1 +.....+ k n xn kr kr kr kr kr x r = c1x1 + c 2 x 2 +......+ c r −1x r −1 + c r x r −1 +......+ c n x n i.e. x r is expressed as a linear combination of vectors. x1 , x 2 ,.......... x r −1 , x r +1 ,........... x n

Hence x r with x1 , x 2 ,.... x r −1 , x r +1 .... x n forms a linearly dependent set of vectors. Note 2: If a = 3 î + 2 ˆj + 5 kˆ then a is expressed as a L INEAR COM BINATION of vectors î , ˆj , kˆ Also, a , î , ˆj , kˆ form a linearly dependent set of vectors. In general, every set of four vectors is a linearly dependent system. î , ˆj , kˆ are Linearly Independent set of vectors. For K î + K ˆj + K kˆ = 0 ⇒ K = K = K = 0

1 2 3 1 2 3 Two vectors a & b are linearly dependent ⇒ a is parallel to b i.e. a x b = 0 ⇒ linear dependence of a & b . Conversely if a x b ≠ 0 then a & b are linearly independent. If three vectors a, b, c are linearly dependent, then they are coplanar i.e. [ a , b, c ] = 0 , conversely,,

if [ a , b, c ] ≠ 0 , then the vectors are linearly independent. Solved Example:

Given A that the points a – 2 b + 3 c , 2 a + 3 b – 4 c , – 7 b + 10 c , A, B, C have

position vector prove that vectors AB and AC are linearly dependent. Solution. Let A, B, C be the given points and O be the point of reference then and OC = – 7 b + 10 c OA = a – 2 b + 3 c , OB = 2 a + 3 b – 4 c

Now AB = p.v. of B – p.v. of A = OB – OA = ( a + 5 b – 7 c ) = – AB ∴ AC = λ AB where λ = – 1. Hence AB and AC are linearly dependent Solved Example: Prove that the vectors 5 a + 6 b + 7 c , 7 a – 8 b + 9 c and 3 a + 20 b + 5 c are linearly dependent a , b , c being linearly independent vectors. Solution. We know that if these vectors are linearly dependent , then we can express one of them as a linear combination of the other two.

Now let us assume that the given vector are coplanar, then we can write 5 a + 6 b + 7 c = ( 7 a – 8 b + 9 c ) + m (3 a + 20 b + 5 c ) where , m are scalars Comparing the coefficients of a , b and c on both sides of the equation 5 = 7 + 3 ..........(i) 6 = – 8 + 20 m ..........(ii) 7 = 9 + 5m ..........(iii) From (i) and (iii) we get 1 4 = 8 ⇒ = = m which evidently satisfies (ii) equation too. 2 Hence the given vectors are linearly dependent . Self Practice Problems : 1. Does there exist scalars u, v, w such that ue1 + ve 2 + we 3 = i where e1 = k , e 2 = j + k , e 3 = − j + 2k ? Ans. No 2. Consider a base a, b, c and a vector − 2a + 3b − c . Compute the co-ordinates of this vector relatively to the base p, q, r where p = 2a − 3b , q = a − 2b + c , r = −3a + b + 2c . Ans. (0, –7/5, 1/5) 3. If a and b are non-collinear vectors and A =(x + 4y) a + (2x + y + 1) b and B = (y – 2x + 2) a + (2x – 3y – 1) b , find x and y such that 3 A = 2B . Ans. x = 2, y = –1 4. If vectors a, b,c be linearly independent, then show that :(i) a − 2b + 3c , − 2a + 3b − 4c , − b + 2c are linearly dependent (ii) a − 3b + 2c , − 2a − 4b − c , 3a + 2b − c are linearly independent. ˆ ˆ ˆ ˆ 5. Given that i − j , i − 2 j are two vectors. Find a unit vector coplanar with these vectors and perpendicular

6.

7.

Note:

to the first vector î − ˆj . Find also the unit vector which is perpendicular to the plane of the two given 1 vectors. Do you thus obtain an orthonormal triad? Ans. es ( î + ˆj ) ; k; Yes 2 If with reference to a right handed system of mutually perpendicular unit vectors î, ˆj, kˆ α = 3 i − j , β = 2 i + j − 3k express β in the form β = β1 + β2 where β1 is parallel to α & β2 is perpendicular to α . 3 1 1 3 β1 = i − j , β 2 = i + j − 3k Ans. 2 2 2 2 Prove that a vector r in space can be expressed linearly in terms of three non-coplanar, non-null [ r b c ] a + [ r c a ] b + [ r a b] c vectors a, b, c in the form r = [a b c ]

Test Of Collinearity: Three points A,B,C with position vectors a, b, c respectively are collinear, if &

only if there exist scalars x , y, z not all zero simultaneously such that; xa + yb + zc = 0 , where x + y + z = 0.

Four points A, B, C, D with position vectors a, b, c, d respectively are coplanar

Note: Test Of Coplanarity:

if and only if there exist scalars x, y, z, w not all zero simultaneously such that xa + yb + zc + wd = 0 where, x + y + z + w = 0. Solved Example Show that the vectors 2a − b + 3c , a + b − 2c and a + b − 3c are non-coplanar vectors. Solution. Let, the given vectors be coplanar. Then one of the given vectors is expressible in terms of the other two. Let 2a − b + 3c = x a + b − 2c + y a + b − 3c , for some scalars x and y.. ⇒ 2a − b + 3c = (x + y) a (x + y) b + (–2x – 3y) c ⇒ 2 = x + y, –1 = x + y and 3 = 2x – 3y. Solving, first and third of these equations, we get x = 9 and y = –7. Clearly, these values do not satisfy the thrid equation. Hence, the given vectors are not coplanar. Solved Example: Prove that four points 2a + 3b − c , a − 2b + 3c , 3a + 4b − 2c and a − 6b + 6c are coplanar..

(

)

(

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Solution. Let the given four points be P, Q, R and S respectively. These points are coplanar if the vectors PQ , PR and PS are coplanar. These vectors are coplanar iff one of them can be expressed as a linear combination of other two. So, let PQ = x PR + y PS ⇒ − a − 5b + 4c = x a + b − c + y − a − 9b − 7c ⇒ − a − 5b + 4c = (x – y) a + (x – 9y) b + (–x + 7y) c ⇒x – y = –1, x – 9y = –5, –x + 7y = 4 [Equating coeff. of a, b, c on both sides]

(

)

(

)

Solving the first of these three equations, we get x = –

1 1 ,y= . 2 2

These values also satisfy the third equation. Hence, the given four points are coplanar. Self Practice Problems : 1. If, a, b, c, d are any four vectors in 3-dimensional space with the same initial point and such that 3a − 2b + c − 2d = 0 , show that the terminal A, B, C, D of these vectors are coplanar. Find the point at which AC and BD meet. Find the ratio in which P divides AC and BD. 2. Show that the vector a − b + c , b − c − a and 2a − 3b − 4c are non-coplanar, where a, b, c , are any noncoplanar vectors. 3. Find the value of λ for which the four points with position vectors − ˆj − kˆ , 4î + 5ˆj + λkˆ . 3î + 9ˆj + 4kˆ and − 4î + 4ˆj + 4kˆ are coplanar..

1 2 . Application Of Vectors: (a)

Ans.

λ=1

Work done against a constant force F over a displacement s

The tangential velocity V of a body moving in a circle is given by V = w x r where r is the pv of the point P..

is defined as W = F . s (b)

(c)

(d)

is along the normal to the plane OPN such that r , F & M form a right handed system. Moment of the couple = ( r1 − r2 ) x F where r1 & r2 are pv’s of the point of the application of the forces F & −F .

The moment of F about ’O’ is defined as M = r x F where r is the pv of P wrt ’O’. The direction of M

Solved Example: Forces of magnitudes 5 and 3 units acting in the directions 6î + 2ˆj + 3kˆ and 3î + 2ˆj + 6kˆ respectively act on a particle which is displaced from the point (2, 2, –1) to (4, 3, 1). Find the work done by the forces. Solution. Let F be the resultant force and d be the displacement vector. Then, (3î + 2ˆj + 6kˆ ) (6î + 2ˆj + 3kˆ ) 1 5 = +3 (39 î + 4ˆj + 33kˆ ) F = 7 9 + 4 + 36 36 + 4 + 9 and, d = ( 4î + 3ˆj + kˆ ) – (2î + 2ˆj − kˆ ) = 2î + ˆj + 2kˆ 1 ∴ Total work done = F . d = (39î + 4ˆj + 33kˆ ) . ( 2î + ˆj + 2kˆ ) 7 1 148 (78 + 4 + 66) = units. = 7 7 Self Practice Problems :1. A point describes a circle uniformly in the î , ˆj plane taking 12 seconds to

complete one revolution. If its initial position vector relative to the centre is î , and the rotation is from î to ˆj , find the position vector at the end of 7 seconds. Also find the velocity vector.. Ans. 1 / 2

(ˆj −

)

(

3 î , p/12 î − 3 ˆj

)

2.

The force represented by 3 î + 2kˆ is acting through the point 5î + 4ˆj − 3kˆ . Find its moment about the point î + 3ˆj + kˆ . Ans. 2î − 20 ˆj − 3kˆ

3.

Find the moment of the comple formed by the forces 5î + kˆ and − 5î − kˆ acting at the points î − ˆj − 5kˆ (9, –1, 2) and (3, –2, 1) respectively Ans.

Miscellaneous Solved Examples Solved Example: Show that the points A, B, C with position vectors 2î − ˆj + kˆ , î − 3ˆj − 5kˆ and 3î − 4ˆj − 4kˆ respectively, are the vertices of a right angled triangle. Also find the remaining angles of the triangle. Solution. We have, = Position vector of B – Position vector of A AB = ( î − 3ˆj − 5kˆ ) – (2î − ˆj + kˆ ) = − î − 2ˆj − 6kˆ

BC

= Position vector of C – Position vector of B

= (3î − 4ˆj − 4kˆ ) – ( î − 3ˆj − 5kˆ ) = 2î − ˆj + kˆ and,

CA

= Position vector of A – Position vector of C

= ( 2î − ˆj + kˆ ) – (3î − 4ˆj − 4kˆ ) = − î + 3ˆj + 5kˆ Since AB + BC + CA = ( −î − 2ˆj − 6kˆ ) + (2î − ˆj + kˆ ) + ( −î + 3ˆj + 5kˆ ) = 0 So, A, B and C are the vertices of a triangle. Now, BC . CA = (2î − ˆj + kˆ ) . ( − î + 3ˆj + 5kˆ ) = –2 – 3 + 5 = 0 ⇒

BC ⊥ CA

⇒

∠BCA =

π 2

Hence, ABC is a right angled triangle.

Since a is the angle between the vectors AB and AC . Therefore AB . AC ( − î − 2ˆj − 6kˆ ) . ( î − 3ˆj − 5kˆ ) cos A = = | AB | | AC | ( −1)2 + ( −2)2 + ( −6)2 12 + ( −3)2 + ( −5)2 −1 + 6 + 30

35

35 35 = A = cos–1 1 + 4 + 36 1 + 9 + 25 41 35 41 41 ˆ ˆ ˆ ˆ ˆ ˆ ( i + 2 j + 6k ) (2 i − j + k ) BA 2−2+6 6 6 cos B = = = ⇒ B = cos–1 2 2 2 2 2 2 ⇒ cos B = 41 6 1 + 2 + 6 2 + ( −1) + (1) | BA | | BC | 41 41 Solved Example: If a, b, c are three mutually perpendicular vectors of equal magnitude, prove that a + b + c is equally inclined with vectors a, b and c . Solution.: Let | a | = | b | = | c | = λ (say). Since a, b, c are mutually perpendicular vectors, therefore a . b = b . c = c . a = 0 ..............(i) 2 a+b+c Now, = a . a + b . b + c . c + 2a . b + 2b . c + 2c . a 2 = | a | | + | b |2 + | c |2 [Using (i) ] = 3λ2 [ ∵ | a | = | b | = | c | = λ] ..............(ii) ∴ | a + b + c | = 3λ Suppose a + b + c makes angles θ1, θ2, θ3 with a, b and c respectively. Then, a . (a + b + c ) a.a+a.b+a.c = cosθ1 = |a||a+b+c | | a || a +b + c | |a| | a |2 λ 1 = = = = [Using (ii)] |a+b+c | | a || a +b +c | 3λ 3

=

∴

=

 1   θ 1 cos–1    3 1   and θ = cos–1 3 3 

  1    Similarly, θ2 = cos–1  ∴ θ1 = θ2 = θ 3.  3    Hence, a + b + c is equally inclineded with a, b and c Solved Example: Prove using vectors : If two medians of a triangle are equal, then it is isosceles. Solution. : Let ABC be a triangle and let BE and CF be two equal medians. Taking A as the origin, let the position vectors of B and C be b and c respectively. Then, 1 1 1 P.V. of E = ∴ c and, P.V. of F = BE = 2 (c − 2b) 2 2 b 1 CF = 2 (b − 2c ) Now, BE = CF ⇒ | BE | = | CF | 2 2 1 1 (b − 2c ) = ⇒ | BE |2 = | CF |2 ⇒ 2 (c − 2b) 2 1 1 ⇒ | b − 2c |2 ⇒ | c − 2b |2 = | b − 2c |2 | c − 2b |2 = 4 4 ⇒ (c − 2b) . (c − 2b) = (b − 2c ) . (b − 2c )

c . c – 4b . c + 4b . b = b . b – 4b . c + 4c . c ⇒ | c |2 – 4b . c + 4 | b |2 = | b |2 – 4b . c + 4 | c |2 ⇒ 3 | b |2 = 3 | c |2 ⇒ | b |2 = | c |2 ⇒ AB = AC Hence, triangle ABC is an isosceles triangle. Solved Example: Using vectors : Prove that cos (A + B) = cos A cos B – sin A sin B Solution. Let OX and OY be the coordinate axes and let î and ˆj be unit vectors along OX and OY respectively. Let ∠XOP = A and ∠XOQ = B. Drawn PL ⊥ OX and QM ⊥ OX. Clearly angle between OP and OQ is A + B

⇒

( )

In ∆OLP, OL = OP cos A and LP = OP sin A. Therefore OL = (OP cos A) î and LP = (OP sin A) − ˆj Now. OL + LP = OP ⇒ OP = OP [(cos A ) î – (sin A) ˆj ] In ∆OMQ, OM = OQ cos B and MQ = OQ sin B. Therefore, OM = (OQ cos B) î , MQ = (OQ sin B) ˆj

Now, OM + MQ = OQ From (i) and (ii), we get OP . OQ = OP [(cos A) î – (sin A) ˆj ] . OQ [(cos B) î + (sin B) ˆj ] = OP . OQ [cos A cos B – sin A sin B] But, OP . OQ = | OP | | OQ | cos (A + B) = OP . OQ cos (A + B) ∴ OP . OQ cos (A + B) = OP . OQ [cos A cos B – sin A sin B] ⇒ cos (A + B) = cos A cos B – sin A sin B Solved Example: Prove that in any triangle ABC (i) c2 = a2 + b2 – 2ab cos C (ii) c = bcosA + acosB. Solution.

(i)

In ∆ABC, AB + BC + CA = 0

or, BC + CA = – AB Squaring both sides

......(i)

( BC )2 + ( CA ) 2 + ( BC ). CA + ( AB )2 ⇒ ⇒

a2 + b2 + 2 ( BC . CA ) = c2 c2 = a2 + b2 – 2ab cosC

⇒

c2 = a2 + b2 = 2 ab cos (π – C)

( BC + CA ). AB = – AB . AB BC . AB + CA . AB = – c2 – ac cosB – bc cos A = – c2 acosB + bcosA = c. Solved Ex.: If D, E, F are the mid-points of the sides of a triangle ABC, prove by vector method that area of 1 ∆DEF = (area of ∆ABC) 4 Solution. Taking A as the origin, let the position vectors of B and C be b and c respectively. Then, the 1 1 1 position vectors of D, E and F are (b + c ) , c and b respectively.. 2 2 2 1 1 −b Now, = (b + c ) = DE c – 2 2 2 1 1 −c and = – ( (b + c ) = DF 2 2 b 2 1 1  − b × − c  ∴ Vector area of ∆DEF (DE × DF) = 2  2 2  2 1 1  1 ( AB × AC )   = (b × c ) = 8  4 2 1 1 = (vector area of ∆ABC) Hence, area of ∆DEF = area of ∆ABC. 4 4 Solved Example: P, Q are the mid-points of the non-parallel sides BC and AD of a trapezium ABCD. Show that ∆APD = ∆CQB. Solution. Let AB = b and AD = d Now DC is parallel to AB ⇒ there exists a acalar t sush that DC = t DB = t b ∴ AC = AD + DC = d + t b (ii)


Now

2∆ APD = AP × AD 1 1 = (1 + t) (b × d) (b + d + t b) × d = 2 2  1  Also 2∆ CQB = CQ × CB =  d − (d + tb  × [b − ( d + t b)] 2     1 1 = − d − t b × [− d + (1 − t ) b] = − (1 − t ) d × b + t b × d 2   2 1 1 2∆ APD = (1 − t + 2t ) b × d = (1 + t ) b × d = Hence the result. 2 2 Solved Example: Let u and v are unit vectors and w is a vector such that u × v + u = w and w × u = v then find the value of [u v w ] . Solution. Given u × v + u = w and w × u = v ⇒ (as, w × u = v ) (u × v + u ) × u= w × u ⇒ (u × v ) × u + u × u = v ⇒ (u . u) v − (v . u) u + u × u = v (using u . u = 1 and u × u = 0, since unit vector) ⇒ ⇒ v − ( v . u) u = v (u . v ) u = 0 ⇒ (as; u ≠ 0) .............(i) u. v = 0 ⇒ u . (v × w ) = u . ( v × (u × v + u)) (given w = u × v + u) = u . ( v × (u × v ) + v × u ) = u . (( v . v ) u − ( v . u) v + v × u) = u . (| v |2 u − 0 + v × u) (as; u . v = 0 from (i)) = | v | 2 (u . u ) – u . ( v × u ) = | v |2 | u |2 – 0 (as, [u v u] = 0) =1 (as; | u | = | v | = 1) ∴ [u v w ] = 1 Sol. Ex.: In any triangle, show that the perpendicular bisectors of the sides are concurrent. Solution. Let ABC be the triangle and D, E and F are respectively middle points of sides BC, CA and AB. Let the perpendicular of D and E meet at O join OF. We are required to prove that OF is ⊥ to AB. Let the position vectors of A, B, C with O as origin of reference be a , b and c respectively.. 1 1 1 ∴ OD = ( b + c ), OE = ( c + a ) and OF = (a + b ) 2 2 2 Also BC = c – b , CA = a – c and AB = b – a 1 Since OD ⊥ BC, (b + c ) . (c – b ) = 0 2 ⇒ b2 = c 2 ............(i) 1 Similarly (c + a) . (a + c ) = 0 2 ⇒ a2 = c 2 ............(ii) from (i) and (ii) we have a2 – b2 = 0 1 ⇒ (a + b ) . (b + a) = 0 ⇒ (b + a ) . (b – a) = 0 2 Solved Example: A, B, C, D are four points in space. using vector methods, prove that 2 2 2 2 2 2 AC + BD + AD + BC ≥ AB + CD what is the implication of the sign of equaility. Solution.: Let the position vector of A, B, C, D be a , b , c and d respectively then AC2 + BD2 + AD2 + BC2 = (c − a ) . (c − a ) + d − b . d − b + d − a . d − a + c − b . c − b 2 2 2 = | c |2 + | a | – 2 a . c + | d |2 + | b |2 – 2 d . b + | d |2 + | a | – 2 a . d + | c | + | b |2 – 2b . c 2 2 2 2 = | a | + | b |2 – 2 a . b + | c | + | d |2 – 2 c . d + | a | + | b |2 + | c | + | d |2 + 2 a . b + 2c . d – 2a . c – 2b . d – 2a . d – 2b . c = a − b . a − b + c − d . c − d + a − b − c − d ≥ AB2 + CD2 = AB2 + CD2 + a + b − c − d . a + b − c − d ≤ AB2 + CD2 ∴ AC2 + BD2 + AD2 + BC2 ≥ AB2 + CD2 for the sign of equality to hold, a + b − c − d = 0 a−c = d−b

(

(

⇒

) (

) ( (

) ( ) ( ) (

) (

)

) (

) (

)

AC and BD are collinear the four points A, B, C, D are collinear

) (

) (

)


1 1 respectively.. (b + d + t b) and 2 2 d


The position vectors of P and Q are

SHORT REVISION 1.

DEFINITIONS: A VECTOR may be described as a quantity having both magnitude & direction. A →

vector is generally represented by a directed line segment, say AB . A is called the initial point & B is →

→

called the terminal point. The magnitude of vector AB is expressed by  AB . ZERO VECTOR a vector of zero magnitude i.e.which has the same initial & terminal point, is called a ZERO VECTOR. It is denoted by O. UNIT VECTOR a vector of unit magnitude in direction of a vector a is called unit vector along a and is a denoted by aˆ symbolically aˆ = . a EQUAL VECTORS two vectors are said to be equal if they have the same magnitude, direction & represent the same physical quantity. COLLINEAR VECTORS two vectors are said to be collinear if their directed line segments are parallel disregards to their direction. Collinear vectors are also called PARALLEL VECTORS. If they have the same direction they are named as like vectors otherwise unlike vectors. Simbolically, two non zero vectors a and b are collinear if and only if, a = K b , where K ∈ R COPLANAR VECTORS a given number of vectors are called coplanar if their line segments are all parallel to the same plane. Note that “TWO VECTORS ARE ALWAYS COPLANAR”. →

2. 3.

4.

5.

POSITION VECTOR let O be a fixed origin, then the position vector of a point P is the vector OP . If a & b & position vectors of two point A and B, then , → AB = b − a = pv of B − pv of A . → → VECTOR ADDITION : If two vectors a & b are represented by OA & OB , then their → sum a + b is a vector represented by OC , where OC is the diagonal of the parallelogram OACB. a + b = b + a (commutative) (a + b) + c = a + (b + c) (associativity) a + 0=a =0 + a a + (− a ) = 0 = (− a )+ a MULTIPLICATION OF VECTOR BY SCALARS : If a is a vector & m is a scalar, then m a is a vector parallel to a whose modulus is m times that of multiplication is called S CALAR MULTIPLICATION. If a & b are vectors & m, n are scalars, then: a . This m ( a ) = ( a )m = m a m ( n a ) = n ( m a ) = ( mn ) a ( m + n )a = m a + n a m (a + b ) = m a + m b SECTION FORMULA : If a & b are the position vectors of two points A & B then the p.v. of a point which divides AB in the na + m b a+b ratio m : n is given by : r = . Note p.v. of mid point of AB = . m+n 2 DIRECTION COSINES : Let a = a1î + a 2ˆj + a 3kˆ the angles which this vector makes with the +ve directions OX,OY & OZ are called DIRECTION ANGLES & their cosines are called the DIRECTION COSINES . a a a , cos β = 2 , cos Γ = 3 . Note that, cos² α + cos² β + cos² Γ = 1 cos α = 1 a

6.

a

a

VECTOR EQUATION OF A LINE : Parametric vector equation of a line passing through two point A (a ) & B( b) is given by, r = a + t (b − a ) where t is a parameter. If the line passes through the point A (a ) & is parallel to the vector b then its equation is, r = a + t b

( )

r = a + t b + c

r = a + λ b & r = a + µ c is :

( )

& r = a + p c − b .

TEST OF COLLINEARITY : Three points A,B,C with position vectors a , b, c respectively are collinear, if & only if there exist scalars x , y, z not all zero simultaneously such that ; xa + yb + zc = 0 , where x + y + z = 0. 8. SCALAR PRODUCT OF TWO VECTORS : a . b = a b cos θ( 0 ≤ θ ≤ π ) , note that if θ is acute then a.b > 0 & if θ is obtuse then a.b < 0 2 a.a = a =a 2 ,a.b=b.a (commutative) a . (b + c) = a . b + a . c (distributive) ( a ≠ 0 b ≠ 0) a.b = 0 ⇔ a ⊥ b a.b î . î = ˆj. ˆj = kˆ . kˆ = 1 î . ˆj = ˆj. kˆ = kˆ . î = 0 ; projection of a on b = . b  a ⋅ b   Note: That vector component of a along b =  a ⋅ b  b and perpendicular to b = a –   b . 2  b2   b  a .b the angle φ between a & b is given by cos φ = 0≤φ≤π ab if a = a1î + a 2ˆj + a 3kˆ & b = b1î + b 2ˆj + b3kˆ then a .b = a1b1 + a2b2 + a3b3

7.

a =

a 12 + a 2 2 + a 32

b = b12 + b 2 2 + b 3 2

,

Note : (i) Maximum value of a . b =  a   b  (ii) Minimum values of a . b = a . b = −  a   b  (iii) Any vector a can be written as , a = a . i i + a . j j + a . k k .

( ) ( ) ( )

(iv)

a b A vector in the direction of the bisector of the angle between the two vectors a & b is + . Hence a b + bisector of the angle between the two vectors a & b is λ a + b , where λ ∈ R . Bisector of the exterior angle between a & b is λ a − b , λ ∈ R+ .

(

9. (i)

)

VECTOR PRODUCT OF TWO VECTORS :

(

)

If a & b are two vectors & θ is the angle between them then a × b = a b sin θ n , where n is the unit vector perpendicular to both a & b such that a , b & n forms a right handed screw system .

2 2 a .a Lagranges Identity : for any two vectors a & b ;(a x b) 2 = a b − (a . b) 2 = a .b

(ii)

a .b b.b

(iii)

Formulation of vector product in terms of scalar product: The vector product a x b is the vector c , such that (i) | c | = a 2 b 2 − (a ⋅ b) 2 (ii) c ⋅ a = 0; c ⋅ b =0 and (iii) a , b, c form a right handed system (iv) a × b = 0 ⇔ a & b are parallel (collinear) ( a ≠ 0 , b ≠ 0 ) i.e. a = K b , where K is a scalar.. a × b ≠ b × a (not commutative) ( ma )× b = a × ( mb ) = m (a × b) where m is a scalar . a × (b + c) = (a × b) + (a × c) (distributive) ˆ ˆ ˆ ˆ ˆ ˆ î × ˆj = kˆ , ˆj × kˆ = î , kˆ × î = ˆj i × i = j× j = k × k = 0 î ˆj kˆ (v) If a = a1î + a 2ˆj + a 3kˆ & b = b1î + b 2ˆj + b3kˆ then a × b = a1 a 2 a 3 b1 b 2 b3 (vi) Geometrically a × b = area of the parallelogram whose two adjacent sides are represented by a & b . a×b (vii) Unit vector perpendicular to the plane of a & b is nˆ = ± a×b r a×b A vector of magnitude ‘r’ & perpendicular to the palne of a & b is ± a×b a×b If θ is the angle between a & b then sin θ = a b (viii) Vector area If a , b & c are the pv’s of 3 points A, B & C then the vector area of triangle ABC =

( )

[

10.

]

1 a x b + b x c + cx a . The points A, B & C are collinear if a x b + b x c + cx a = 0 2 1 Area of any quadrilateral whose diagonal vectors are d 1 & d 2 is given by d1 xd 2 2

SHORTEST DISTANCE BETWEEN TWO LINES : If two lines in space intersect at a point, then obviously the shortest distance between them is zero. Lines which do not intersect & are also not parallel are called SKEW LINES. For Skew lines the direction of the shortest distance would be perpendicular to both the lines. The magnitude of the shortest distance →

vector would be equal to that of the projection of AB along the direction of the line of shortest distance, → → → → → i.e. LM = Pr ojection of AB on LM = Pr ojection of AB on p xq LM is parallel to p x q → AB . (p x q) ( b − a ) . ( p x q) = = pxq pxq

1.

The two lines directed along p & q will intersect only if shortest distance = 0 i.e. ( b − a ).(p x q) = 0 i.e. b − a lies in the plane containing p & q . ⇒ b − a p q = 0 .

(

)

[( ) ]

2.

b x(a 2 − a 1 ) If two lines are given by r1 = a 1 + Kb & r2 = a 2 + Kb i.e. they are parallel then , d = b

11.

SCALAR TRIPLE PRODUCT / BOX PRODUCT / MIXED PRODUCT :

The scalar triple product of three vectors a , b & c is defined as : a x b .c = a b c sin θ cos φ where θ is the angle between a & b & φ is the angle between a × b & c . It is also defined as [ a b c ] , spelled as box product . Scalar triple product geometrically represents the volume of the parallelopiped whose three couterminous edges are represented by a , b & c i . e . V = [ a b c ] In a scalar triple product the position of dot & cross can be interchanged i.e. a . ( b x c ) = ( a x b ) .c OR [ a b c ] = [ b c a ] = [ c a b ]

a . (b x c) = − a .( cx b) i. e. [ a b c ] = − [ a c b ]

1 2 3 If a = a1î + a 2ˆj + a 3kˆ ; b = b1î + b 2ˆj + b3kˆ & c = c1î + c 2ˆj + c3kˆ then [a b c] = b1 b 2 b 3 .

a a a

c1 c2 c3 In general , if a = a 1 l + a 2 m + a 3 n ; b = b1 l + b 2 m + b 3 n & c = c1 l + c2 m + c3 n a1 a 2 a 3 then a b c = b1 b 2 b 3 l m n ; where , m & n are non coplanar vectors . c1 c2 c3 If a , b , c are coplanar ⇔ [ a b c ] = 0 . Scalar product of three vectors, two of which are equal or parallel is 0 i.e. [ a b c ] = 0 , Note : If a , b , c are non − coplanar then [ a b c ] > 0 for right handed system & [ a b c ] < 0 for left handed system . [(a + b) c d ] = [ a c d ] + [ b c d ] [i j k] = 1 [ K a b c ] = K[ a b c ]

[

[ ]

1 [a b c] 6

1 [a + b + c + d] . 4

Note that this is also the point of concurrency of the lines joining the vertices to the centroids of the opposite faces and is also called the centre of the tetrahedron. In case the tetrahedron is regular it is equidistant from the vertices and the four faces of the tetrahedron . Remember that : a − b b − c c − a = 0 & a + b b+ c c+a = 2 a b c . VECTOR TRIPLE PRODUCT : Let a , b , c be any three vectors, then the expression a × ( b × c ) is a vector & is called a vector triple product . GEOMETRICAL INTERPRETATION OF a × ( b × c ) Consider the expression a × ( b × c ) which itself is a vector, since it is a cross product of two vectors a & ( b x c ) . Now a x ( b x c ) is a vector perpendicular to the plane containing a & ( b x c ) but b x c is a vector perpendicular to the plane b & c , therefore a x ( b x c ) is a vector lies in the plane of b & c and perpendicular to a . Hence we can express a x ( b x c ) in terms of b & c

[

13.

The positon vector of the centroid of a tetrahedron if the pv’s of its angular vertices are a , b , c & d are given by

*12.

The volume of the tetrahedron OABC with O as origin & the pv’s of A, B and C being a , b & c respectively is given by V =

]

]

[

]

[

]

i.e. a x ( b x c ) = xb + yc where x & y are scalars . a x ( b x c ) = (a . c) b − (a . b) c (a x b) x c = (a . c) b − ( b . c) a (a x b) x c ≠ a x (b x c) LINEAR COMBINATIONS / Linearly Independence and Dependence of Vectors : Given a finite set of vectors a , b , c ,...... then the vector r = x a + y b + z c + ........ is called a linear

combination of a , b , c ,...... for any x, y, z ...... ∈ R. We have the following results : (a) FUNDAMENTALTHEOREM IN PLANE : Let a ,b be non zero , non collinear vectors . Then any vector r coplanar with a ,b can be expressed uniquely as a linear combination of a ,b i.e. There exist some unique x,y ∈ R such that x a + yb = r . (b) FUNDAMENTAL THEOREM IN SPACE : Let a , b , c be non−zero, non−coplanar vectors in space. Then any vector r , can be uniquily expressed as a linear combination of a , b , c i.e. There exist some unique x,y ∈ R such that x a + y b + z c = r . (c) If x 1 , x 2 ,...... x n are n non zero vectors, & k1, k2, .....kn are n scalars & if the linear combination k 1 x 1 + k 2 x 2 + ........ k n x n = 0 ⇒ k 1 = 0 ,k 2 = 0 ..... k n = 0 then we say that vectors x 1 , x 2 ,...... x n are LINEARLY INDEPENDENT VECTORS . (d) If x 1 , x 2 ,...... x n are not LINEARLY INDEPENDENT then they are said to be LINEARLY DEPENDENT vectors . i.e. if k 1 x 1 + k 2 x 2 + ........ + k n x n = 0 & if there exists at least one kr ≠ 0 then x 1 , x 2 ,...... x n are said to be LINEARLY DEPENDENT . Note : If a = 3i + 2j + 5k then a is expressed as a LINEAR COMBINATION of vectors î , ˆj, kˆ . Also , a , î , ˆj, kˆ form a linearly dependent set of vectors. In general , every set of four vectors is a linearly dependent system. î , ˆj , kˆ are LINEARLY INDEPENDENT set of vectors. For K î + K ˆj + K kˆ = 0 ⇒ K = 0 = K = K . 1 2 3 1 2 3 Two vectors a & b are linearly dependent ⇒ a is parallel to b i.e. a x b = 0 ⇒ linear dependence of a & b . Conversely if a x b ≠ 0 then a & b are linearly independent . If three vectors a , b , c are linearly dependent, then they are coplanar i.e. [ a , b, c ] = 0 , conversely, if [ a , b, c ] ≠ 0 , then the vectors are linearly independent. 14. COPLANARITY OF VECTORS : Four points A, B, C, D with position vectors a , b , c , d respectively are coplanar if and only if there exist scalars x, y, z, w not all zero simultaneously such that x a+ y b + z c + w d = 0 where, x + y + z + w = 0. 15. RECIPROCAL SYSTEM OF VECTORS : If a , b , c & a ' ,b ' ,c ' are two sets of non coplanar vectors such that a . a '= b . b '= c . c '= 1 then the two systems are called Reciprocal System of vectors. Note : 16. (a) (b) 17. (a)

(b)

(c)

bx c a'= ; b' = abc

[ ]

cx a ; c' = abc

[ ]

axb abc

[ ]

EQUATION OF A PLANE : The equation ( r − r0 ).n = 0 represents a plane containing the point with p.v. r0 where n is a vector normal to the plane . r . n = d is the general equation of a plane. Angle between the 2 planes is the angle between 2 normals drawn to the planes and the angle between a line and a plane is the compliment of the angle between the line and the normal to the plane. APPLICATION OF VECTORS : Work done against a constant force F over a displacement s is defined as W = F.s The tangential velocity V of a body moving in a circle is given by V = w × r where r is the pv of the point P. The moment of F about ’O’ is defined as M = r × Fwhere r is the pv of P wrt ’O’. The direction of M is along the normal to the plane OPN such that r , F & M form a right handed system.

Moment of the couple = ( r1 − r2 ) × F where r1 & r2 are pv’s of the point of the application of the forces F & − F . 3 -D COORDINATE GEOMETRY USEFUL RESULTS A General : (1) Distance (d) between two points (x1 , y1 , z1) and (x2 , y2 , z2) d = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 + (z 2 − z1 ) 2 (2) Section Fomula m 2 z1 + m1 z 2 m 2 x1 + m1 x 2 m 2 y1 + m1 y 2 x= ; y= ; z= m1 + m 2 m1 + m 2 m1 + m 2 ( For external division take –ve sign ) Direction Cosine and direction ratio's of a line (3) Direction cosine of a line has the same meaning as d.c's of a vector. (a) Any three numbers a, b, c proportional to the direction cosines are called the direction ratios i.e. l m n 1 = = =± a b c a 2 + b2 + c2 same sign either +ve or –ve should be taken through out. note that d.r's of a line joining x1 , y1 , z1 and x2 , y2 , z2 are proportional to x2 – x1 , y2 – y1 and z2 – z1 (b) If θ is the angle between the two lines whose d.c's are l1 , m1 , n1 and l2 , m2 , n2 cosθ = l1 l2 + m1 m2 + n1 n2 hence if lines are perpendicular then l1 l2 + m1 m2 + n1 n2 = 0 l1 m1 n1 if lines are parallel then l = m = n 2 2 2 l1 m1 n1

(d)

l2 m 2 n 2 = 0 l3 m 3 n 3 Projection of the join of two points on a line with d.c's l, m, n are l (x2 – x1) + m(y2 – y1) + n(z2 – z1) PLANE: (i) General equation of degree one in x, y, z i.e. ax + by + cz + d = 0 represents a plane. Equation of a plane passing through (x1 , y1 , z1) is a (x – x1) + b (y – y1) + c (z – z1) = 0 where a, b, c are the direction ratios of the normal to the plane. Equation of a plane if its intercepts on the co-ordinate axes are x1 , y1 , z1 is x y z + + = 1. x1 y1 z1 Equation of a plane if the length of the perpendicular from the origin on the plane is p and d.c's of the perpendicular as l , m, , n is l x + m y + n z = p Parallel and perpendicular planes – Two planes a1 x + b1 y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are perpendicular if a1 a2 + b1 b2 + c1 c2 = 0 a1 b1 c1 = = parallel if a 2 b 2 c 2 and a1 b1 c1 d1 = = = coincident if a 2 b2 c2 d 2 Angle between a plane and a line is the compliment of the angle between the normal to the plane and the Line : r = a + λ b  then cos(90 − θ) = sin θ = b . n . line . If Plane : r . n = d  | b | .| n |  where θ is the angle between the line and normal to the plane. note that if three lines are coplanar then

(4) B (ii)

(iii) (iv) (v)

(vi)

(vii)

Length of the perpendicular from a point (x1 , y1 , z1) to a plane ax + by + cz + d = 0 is p=

ax1+ by1+ cz1+ d a 2 + b2 + c2

(viii)

Distance between two parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is d1 − d 2

(ix)

a 2 + b2 + c2 Planes bisecting the angle between two planes a1x + b1y + c1z + d1 = 0 and a2 + b2y + c2z + d2 = 0 is given by a1x + b1y + c1z + d1 a12

(x)

+ b12

+ c12

=±

a 2 x + b2 y + c2z + d 2

a 22 + b 22 + c 22 Of these two bisecting planes , one bisects the acute and the other obtuse angle between the given planes. Equation of a plane through the intersection of two planes P1 and P2 is given by P1 + λP2 = 0

C (i)

STRAIGHT LINE IN SPACE Equation of a line through A (x1 , y1 , z1) and having direction cosines l ,m , n are x − x1 y − y1 z − z1 = = m n l and the lines through (x1 , y1 ,z1) and (x2 , y2 ,z2) x − x1 y − y1 z − z1 = = x 2 − x1 y 2 − y1 z 2 − z1 (ii) Intersection of two planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 together represent the unsymmetrical form of the straight line. x − x1 y − y1 z − z1 = = (iii) General equation of the plane containing the line is l m n A (x – x1) + B(y – y1) + c (z – z1) = 0 where Al + bm + cn = 0 . LINE OF GREATEST SLOPE AB is the line of intersection of G-plane and H is the horizontal plane. Line of greatest slope on a given plane, drawn through a given point on the plane, is the line through the point 'P' perpendicular to the line of intersetion of the given plane with any horizontal plane.

EXERCISE–1 Q.1 Q.2 Q.3

Q.4

If are non collinear vectors such that , p = ( x + 4 y)a + ( 2x + y + 1) b & that 3p = 2q . q = ( y − 2 x + 2)a + ( 2 x − 3 y − 1) b ,find x & y such (a) Show that the points a − 2 b + 3 c ; 2 a + 3 b − 4 c & − 7 b + 10 c are collinear . (b) Prove that the points A = (1,2,3), B (3,4,7), C (−3,−2,−5) are collinear & find the ratio in which B divides AC. → → Points X & Y are taken on the sides QR & RS , respectively of a parallelogram PQRS, so that QX = 4 XR → → →  21  → & RY = 4 YS . The line XY cuts the line PR at Z . Prove that PZ =   PR .  25  Find out whether the following pairs of lines are parallel, non-parallel & intersecting, or non-parallel & non-intersecting. a&b

(i)

(iii) Q.5 Q.6 Q.7

(

)

r1 = i + j +2 k + λ 3 i −2 j + 4 k r2 =2 i + j +3 k + µ −6 i + 4 j −8 k r1 = i + k + λ i + 3 j + 4 k r2 = 2 i + 3 j + µ 4 i − j + k

(

(

(

)

)

(ii)

(

)

r1 = i − j + 3 k + λ i − j + k r2 = 2 i + 4 j + 6 k + µ 2 i + j + 3 k

(

)

)

Let OACB be paralelogram with O at the origin & OC a diagonal. Let D be the mid point of OA. Using vector method prove that BD & CO intersect in the same ratio. Determine this ratio. A line EF drawn parallel to the base BC of a ∆ ABC meets AB & AC in F & E respectively. BE & CF meet in L. Use vectors to show that AL bisects BC. → ‘O’is the origin of vectors and A is a fixed point on the circle of radius‘a’with centre O. The vector OA

→ is denoted by a . A variable point ‘P’ lies on the tangent at A & OP = r . Show that a . r = a 2 . Hence if P ≡ (x,y) & A ≡ (x1,y1) deduce the equation of tangent at A to this circle. Q.8 (a) By vector method prove that the quadrilateral whose diagonals bisect each other at right angles is a rhombous. (b) By vector method prove that the right bisectors of the sides of a triangle are concurrent. Q.9 The resultant of two vectors a & b is perpendicular to a . If b = 2 a show that the resultant of 2 a & b is perpendicular to b . Q.10 a , b, c and d are the position vectors of the points A ≡ (x, y, z) ; B ≡ (y, – 2z, 3x) ; C ≡ (2z, 3x, – y) π ^ and D≡(1,–1, 2) respectively. If | a | = 2 3 ; a b = (a^c ) ; a ^d = and a ^ˆj is obtuse, then find x, y,, z. 2 Q.11 If r and s are non zero constant vectors and the scalar b is chosen such that r + b s is minimum, then show that the value of b s 2 + | r + b s |2 is equal to | r |2 . Q.12 Use vectors to prove that the diagonals of a trapezium having equal non parallel sides are equal & conversely. Q.13(a) Find a unit vector â which makes an angle (π/4) with axis of z & is such that â + i + j is a unit vector. 2 2  a   a−b  b (b) Prove that  −  =    a 2 b2   | a | | b |      Q.14 Given four non zero vectors a , b , c and d . The vectors a , b & c are coplanar but not collinear pair by ∧ π ∧ ∧ ∧ pair and vector d is not coplanar with vectors a , b & c and (a b) = (b c) = , (da) = α, (db) = β then 3 ∧ −1 prove that ( d c) = cos (cos β − cos α ) . Q.15 (a) Use vectors to find the acute angle between the diagonals of a cube . (b) Prove cosine & projection rule in a triangle by using dot product . Q.16 In the plane of a triangle ABC, squares ACXY, BCWZ are described , in the order given, externally to

( )

→

Q.17 Q.18 Q.19

Q.20 Q.21

Q.22

( )

( )

→

→

→

the triangle on AC & BC respectively. Given that CX = b , CA = a , CW = x , CB = y . Prove that → → a . y + x . b = 0 . Deduce that AW . BX = 0 . A ∆ OAB is right angled at O ; squares OALM & OBPQ are constructed on the sides OA and OB externally. Show that the lines AP & BL intersect on the altitude through 'O' . Given that u = î − 2ˆj + 3kˆ ; v = 2î + ˆj + 4kˆ ; w = î + 3ˆj + 3kˆ and ( u ·R − 10)î + ( v · R − 20)ˆj + ( w · R − 20) kˆ = 0. Find the unknown vector R . If O is origin of reference, point A ( a ) ; B( b ) ; C( c ) ; D(a + b) ; E ( b + c) ; F( c + a ); G (a + b + c) where a = a1î + a 2 ˆj + a 3kˆ ; b = b1î + b 2 ˆj + b 3 kˆ and c = c1î + c 2 ˆj + c3 kˆ then prove that these points are vertices of a cube having length of its edge equal to unity provided the matrix.  a1 a 2 a 3  b b b  2 3  is orghogonal. Also find the length XY such that X is the point of intersection of CM and  1  c1 c 2 c3  GP ; Y is the point of intersection of OQ and DN where P, Q, M, N are respectively the midpoint of sides CF, BD, GF and OB. (a) If a + b + c = 0 , show that a x b = b x c = c x a . Deduce the Sine rule for a ∆ ABC. (b) Find the minimum area of the triangle whose vertices are A(–1, 1, 2); B(1, 2, 3) and C(t, 1, 1) where t is a real number. (a) Determine vector of magnitude 9 which is perpendicular to both the vectors : 4î − ˆj + 3kˆ & − 2î + ˆj − 2kˆ π (b) A triangle has vertices (1, 1, 1) ; (2, 2, 2), (1, 1, y) and has the area equal to csc   sq. units. 4 Find the value of y. The internal bisectors of the angles of a triangle ABC meet the opposite sides in D, E, F ; use vectors to prove that the area of the triangle DEF is given by

(2abc) ∆ where ∆ is the area of the triangle. (a + b ) ( b + c ) ( c + a ) Q.23 If a , b , c ,d are position vectors of the vertices of a cyclic quadrilateral ABCD prove that : a x b + bxd + dxa b x c + cx d + d x b + =0 ( b − a ) . (d − a ) (b − c) . (d − c)

− ABC is 'a' . Point E and F are taken on the edges → AD and BD respectively such that E divides DA and F divides BD in the ratio 2:1 each . Then find the area of triangle CEF. 1 3ˆ j and x = a + (q 2 − 3) b , y = − p a + qb . If x ⊥ y , then express p Q.25 Let a = 3 î − ˆj and b = î + 2 2 as a function of q, say p = f (q), (p ≠ 0 & q ≠ 0) and find the intervals of monotonicity of f (q).

Q.24

T

h e

l e n g t h

o f

t h e

e d g e

o

f

t h e

r e g u l a r

t e t r a h e d r o n

→

D

EXERCISE–2

Q.1

Q.2

Q.3

Q.4 (i)

(ii) Q.5

1 A(a ) ; B( b ) ; C( c ) are the vertices of the triangle ABC such that a = ( 2î − r − 7 kˆ ) ; b = 3 r + ˆj − 4kˆ ; 2 ˆ ˆ ˆ ˆ . A vector is such that is parallel to î and ( r − 2î ) is parallel to ( r + p ) c = 22i − 11 j − 9 r p = 2j− k p . Show that there exists a point D (d ) on the line AB with d = 2 t î + (1 − 2 t )ˆj + ( t − 4) kˆ . Also find the shortest distance C from AB. The position vectors of the points A, B, C are respectively (1, 1, 1) ; (1, −1, 2) ; (0, 2, −1). Find a unit vector parallel to the plane determined by ABC & perpendicular to the vector (1, 0, 1) . ( a 1 − a ) 2 ( a 1 − b ) 2 ( a 1 − c) 2 2 2 2 Let (b1 − a ) (b1 − b) (b1 − c) = 0 and if the vectors α = î + aˆj + a 2 kˆ ; β = î + bˆj + b 2 kˆ ;

(c1 − a ) 2 (c1 − b) 2 (c1 − c) 2 2 2 γ = î + cˆj + c 2 kˆ are non coplanar, show that the vectors α1 = î + a1ˆj + a1 kˆ ;β1 = î + b1ˆj + b1 kˆ and ˆ 2 γ1 = i + c1ˆj + c1 kˆ are coplaner.. Given non zero number x1, x2, x3 ; y1, y2, y3 and z1, z2 and z3 such that xi > 0 and yi < 0 for all i = 1, 2, 3. Can the given numbers satisfy x1 x 2 x 3  x x + y1y 2 + z1z 2 = 0  1 2 y1 y 2 y 3 = 0 and  x 2 x 3 + y 2 y 3 + z 2 z 3 = 0  x 3 x1 + y 3 y1 + z 3z1 = 0 z1 z 2 z 3 If P = (x1, x2, x3) ; Q (y1, y2, y3) and O (0, 0, 0) can the triangle POQ be a right angled triangle? The pv's of the four angular points of a tetrahedron are : A j + 2 k ; B 3 i + k ; C 4 i + 3 j + 6 k

& D (i) (iii) (iv) Q.6

(2 i + 3 j + 2 k ) . Find :

Q.8 Q.9

)

(

)

(

)

the perpendicular distance from A to the line BC. (ii) the volume of the tetrahedron ABCD. the perpendicular distance from D to the plane ABC. the shortest distance between the lines AB & CD. → The length of an edge of a cube ABCDA1B1C1D1 is equal to unity. A point E taken on the edge AA 1 is 1 → 1 → → such that AE = . A point F is taken on the edge BC such that BF = . If O1 is the centre of 4 3 the cube, find the shortest distance of the vertex B1 from the plane of the ∆ O1EF. →

Q.7

(

The vector OP = î + 2ˆj + 2kˆ turns through a right angle, passing through the positive x-axis on the way.. Find the vector in its new position. Find the point R in which the line AB cuts the plane CDE where a = î + 2ˆj + kˆ , b = 2î + ˆj + 2kˆ , c = − 4 ˆj + 4kˆ , d = 2î − 2ˆj + 2kˆ & e = 4î + ˆj + 2kˆ . If a = a1î + a 2ˆj + a 3kˆ ; b = b1î + b 2ˆj + b 3kˆ and c = c1î + c 2 ˆj + c 3kˆ then show that the value of the a · î a · ˆj a · kˆ scalar triple product [ na + b nb + c nc + a ] is (n3 + 1) b · î b · ˆj b · kˆ c · î c · ˆj c · kˆ

α & β if a x (b x c) + (a . b) b = (4 − 2β − sin α ) b + (β 2 − 1) c & ( c . c) a = c while b & c are non zero non vectors. collinear Q.11 If the vectors b , c ,d are not coplanar, then prove that the vector (a × b) × (c × d ) + (a × c ) × (d × b) + (a × d ) × ( b × c) is parallel to a . Q.12 a , b , c are non−coplanar unit vectors . The angle between b & c is α, between c & a is β and between

Q.10

F

i n d

t h e

s c a l a r s

(

)

a & b is γ . If A (a cos α ) , B b cos β , C ( c cos γ ) , then show that in ∆ ABC,

(

)

a x b x c

=

b x ( c x a )

=

(

c x a x b

)=

sin B sin C sin A b x c c x a a x b , n 2 = & n 3 = . n 1 = c x a a x b b x c

∏

(

)

a x b x c

∑ sin α cos β cos γ

n 1

where

Q.13 Given that a,b,p,q are four vectors such that a + b = µ p , b .q = 0 & ( b ) 2 = 1 , where µ is a scalar then prove that ( a .q ) p − ( p .q ) a = p .q . Q.14 Show that a = p x (q x r ) ; b = q x ( r x p) & c = r x ( p x q) represents the sides of a triangle. Further prove that a unit vector perpendicular to the plane of this triangle is n tan ( p ^ q) + n 2 tan (q ^ r ) + n 3 tan ( r ^ p) ± 1 where a , b , c , p , q are non zero vectors and n 1 tan ( p ^ q) + n 2 tan (q ^ r ) + n 3 tan ( r ^ p) pxq qx r rxp n n = = = n no two of p , q , r are mutually perpendicular & 1 ; 2 & 3 pxq qx r rxp

Q.15 Given four points P1, P2, P3 and P4 on the coordinate plane with origin O which satisfy the condition 3 O P n −1 + O P n +1 = O P n , n = 2, 3 2 (i) If P1, P2 lie on the curve xy = 1, then prove that P3 does not lie on the curve. (ii) If P1, P2, P3 lie on the circle x2 + y2 = 1, then prove that P4 lies on this circle. Q.16 Let a = α i + 2 j − 3 k , b = i + 2 α j − 2 k and c = 2 i − α j + k . Find the value(s) of α, if any, such that

α = 0. {(a × b) × (b × c)} × (c × a ) = 0. Find the vector product when

p.r p.s Q.17 Prove the result (Lagrange’s identity) (p × q) · ( r × s ) = & use it to prove the following. Let q.r q.s (ab)denote the plane formed by the lines a,b. If (ab) is perpendicular to (cd) and (ac) is perpendicular to (bd) prove that (ad) is perpendicular to (bc). p 2 b + (b . a ) a − p( bx a ) Q.18 (a) If px + ( x × a ) = b ; ( p ≠ 0) prove that x = . p (p 2 + a 2 ) (b) Solve the following equation for the vector p ; p x a + p . b c = b x c where a , b , c are non zero non coplanar vectors and a is neither perpendicular to b nor to c , hence show that  a bc   p × a + c  is perpendicular to b − c .   a ·c   Q.19 Find a vector v which is coplanar with the vectors i + j − 2 k & i − 2 j + k and is orthogonal to the vector − 2 i + j + k . It is given that the projection of v along the vector i − j + k is equal to 6 3 . Q.20 Consider the non zero vectors a , b , c & d such that no three of which are coplanar then prove that a b cd + c a b d = b a cd + d a b c . Hence prove that a , b , c & d represent the position vectors of

( )

[ ]

[ ] [ ] [ ] [ ]

b cd] + [ a b d] [ the vertices of a plane quadrilateral if =1 . a cd] + [ a b c] [

Q.21 The base vectors a1,a 2 ,a 3 are given in terms of base vectors b1,b 2 ,b 3 as, a1 = 2b1 + 3b 2 − b 3 ; a 2 = b1 − 2b 2 + 2b 3 & a 3 = −2b1 + b 2 − 2b 3 . If F = 3b1 − b 2 + 2b 3 , then express F in terms of

a 1, a 2 & a 3 .

()

Q.22 If A (a ) ; B b & C ( c) are three non collinear points , then for any point P ( p) in the plane of the

[ ]

(

)

a b c = p . a x b + bx c + cxa ∆ ABC , prove that ; (i) (ii) The vector v perpendicular to the plane of the triangle ABC drawn from the origin 'O' is given by a bc a ×b + b×c + c×a where ∆ is the vector area of the triangle ABC. v =± 2 4∆ Q.23 Given the points P (1, 1, –1), Q (1, 2, 0) and R (–2, 2, 2). Find (a) PQ×PR (b) Equation of the plane in (i) scalar dot product form (ii) parametric form (iii) cartesian form (iv) if the plane through PQR cuts the coordinate axes at A, B, C then the area of the ∆ABC Q.24 Let a , b & c be non coplanar unit vectors, equally inclined to one another at an angle θ. If a x b + b x c = p a + q b + r c . Find scalars p , q & r in terms of θ. Q.25 Solve the simultaneous vector equations for the vectors x and y . x + c × y = a and y + c × x = b where c is a non zero vector..

[ ](

)

EXERCISE–3

Q.1 Q.2 Q.3 Q.4 Q.5

Q.6

Find the angle between the two straight lines whose direction cosines l, m, n are given by 2l + 2m – n = 0 and mn + nl + lm = 0. If two straight line having direction cosines l, m, n satisfy al + bm + cn = 0 and f m n + g n l + h l m = 0 f g h are perpendicular, then show that + + = 0. a b c P is any point on the plane lx + my + nz = p. A point Q taken on the line OP (where O is the origin) such 2 that OP. OQ = p . Show that the locus of Q is p( lx + my + nz ) = x2 + y2 + z2. Find the equation of the plane through the points (2, 2, 1), (1, –2, 3) and parallel to the x-axis. Through a point P (f, g, h), a plane is drawn at right angles to OP where 'O' is the origin, to meet the r5 coordinate axes in A, B, C. Prove that the area of the triangle ABC is where OP = r.. 2f g h The plane lx + my = 0 is rotated about its line of intersection with the plane z = 0 through an angle θ.

Prove that the equation to the plane in new position is lx + my + z l 2 + m 2 tan θ = 0 Q.7 Find the equations of the straight line passing through the point (1, 2, 3) to intersect the straight line x + 1 = 2 (y – 2) = z + 4 and parallel to the plane x + 5y + 4z = 0. x −3 y −3 z = = at an Q.8 Find the equations of the two lines through the origin which intersect the line 2 1 1 π angle of . Q.9 A variable3plane is at a constant distance p from the origin and meets the coordinate axes in points A, B and C respectively. Through these points, planes are drawn parallel to the coordinates planes. Find the locus of their point of intersection. x + 2 2 y + 3 3z + 4 = = Q.10 Find the distance of the point P (– 2, 3, – 4) from the line measured parallel to 3 4 5 the plane 4x + 12y – 3z + 1 = 0. Q.11 Find the equation to the line passing through the point (1, –2, –3) and parallel to the line 2x + 3y – 3z + 2 = 0 = 3x – 4y + 2z – 4. Q.12 Find the equation of the line passing through the point (4, –14, 4) and intersecting the line of intersection Q.13 ofLet = (1, :0,3x – 1) ; Q–=z (1, 1) and = (2, the Pplanes + 2y = 51,and x –R2y – 2z1,=3)–1are atthree right points. angles. (a) Find the area of the triangle having P, Q and R as its vertices. (b) Give the equation of the plane through P, Q and R in the form ax + by + cz = 1. (c) Where does the plane in part (b) intersect the y-axis. (d) Give parametric equations for the line through R that is perpendicular to the plane in part (b). Q.14 Find the point where the line of intersection of the planes x – 2y + z = l and x + 2y – 2z = 5, intersects the plane 2x + 2y + z + 6 = 0.

Q.15 Feet of the perpendicular drawn from the point P (2, 3, –5) on the axes of coordinates are A, B and C. Find the equation of the plane passing through their feet and the area of ∆ABC. Q

. 1

6

Q.17 Q.18 Q.19 Q.20 Q.21

Q.22 Q.23

Q.24 Q.25

F

i n d

t h e

e q u a t i o

n s

t o

t h e

l i n e

w

h i c h

c a n

b e

d r a w

n

f r o m

t h e

p o

i n t

( 2

,

– 1

,

3 )

p e r p e n d i c u l a r

t o

t h e

l i n e s

x−4 y z+3 = = and at right angles. 4 5 3 x −1 y + 2 z = = Find the equation of the plane containing the straight line and perpendicular to the −3 2 5 plane x – y + z + 2 = 0. x +1 y − p z+2 x y−7 z+7 = = = = Find the value of p so that the lines and are in the same −3 −3 2 1 1 2 plane. For this value of p, find the coordinates of their point of intersection and the equation of the plane containing them. Find the equations to the line of greatest slope through the point (7, 2 , –1) in the plane x – 2y + 3z = 0 assuming that the axes are so placed that the plane 2x + 3y – 4z = 0 is horizontal. Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the area of triangles ABC, ACD and ADB be denoted by x, y and z sq. units respectively. Find the area of the triangle BCD. The position vectors of the four angular points of a tetrahedron OABC are (0, 0, 0); (0, 0, 2); (0, 4, 0) and (6, 0, 0) respectively. A point P inside the tetrahedron is at the same distance 'r' from the four plane faces of the tetrahedron. Find the value of 'r'. x + 6 y + 10 z + 14 = = The line is the hypotenuse of an isosceles right angled triangle whose opposite 5 3 8 vertex is (7, 2, 4). Find the equation of the remaining sides. Find the foot and hence the length of the perpendicular from the point (5, 7, 3) to the line x − 15 y − 29 5 − z = = . Also find the equation of the plane in which the perpendicular and the given 3 8 5 straight line lie. x −1 y − 2 z + 3 = = Find the equation of the line which is reflection of the line in the plane 9 −1 −3 3x – 3y + 10z = 26. x −1 y z x −3 y z−2 = = and parallel to the line = = Find the equation of the plane containing the line . 2 3 2 2 5 4 Find also the S.D. between the two lines. x −1 y − 2 z − 3 = = 2 3 4

EXERCISE–4

→

→

→

→

→

→

→

Q.1(a) Let OA = a , OB = 10 a + 2 b and OC = b where O, A & C are non−collinear points . Let p denote the area of the quadrilateral OABC, and let q denote the area of the parallelogram with OA and OC as adjacent sides. If p = kq, then k = _______ . (b) If A , B & C are vectors such that | B | = | C | , Prove that ;

[(A + B) x (A + C )] x(Bx C).(B + C) = 0

[ JEE ' 97, 2 + 5 ] 0

Q.2(a) Vectors x , y & z each of magnitude 2 , make angles of 60 with each other. If x × ( y × z ) = a , y × ( z × x ) = b and x × y = c then find x, y and z in terms of a , b and c . (b) The position vectors of the points P & Q are 5 i + 7 j − 2 k and − 3 i + 3 j + 6 k respectively. The vector A = 3 i − j + k passes through the point P & the vector B = − 3 i + 2 j + 4 k passes through the

point Q. A third vector 2 i + 7 j − 5 k intersects vectors A & B . Find the position vectors of the points of intersection. [ REE ' 97, 6 + 6 ] Q.3(a) Select the correct alternative(s) (i) If a = i + j + k , b = 4 i + 3j + 4 k and c = i + αj + βk are linearly dependent vectors & c = 3 , then: (A) α = 1, β = −1 (B) α = 1, β = ±1 (C) α = −1, β = ±1 (D) α = ±1, β = 1 (ii) For three vectors u , v , w which of the following expressions is not equal to any of the remaining three? (A) u . ( v x w ) (C) v . ( u x w ) (D) ( u x v ) . w (B) ( v x w ) . u (iii) Which of the following expressions are meaningful ?

(A) u . ( v x w ) (B) ( u . v ) . w (C) ( u . v ) w (D) u x ( v . w ) (b) Prove, by vector methods or otherwise, that the point of intersection of the diagonals of a trapezeum lies on the line passing through the mid−points of the parallel sides. (You may assume that the trapezeum is not a parallelogram.) (c) For any two vectors u & v , prove that [ JEE ' 98 , 2 + 2 + 2 + 8 + 8 ] 2 2 2 2 2 (i) (u. v) + | u × v| = | u| | v| & (ii) (1 + | u| )(1 +| v|2 ) = (1 − u. v)2 + | u + v + (u × v)|2 Q.4(a) If x × y = a , y × z = b, x. b = γ , x. y = 1 and y. z = 1 then find x, y & z in terms of a , b and γ . →

→

(b) Vectors AB = 3 i − j + k & CD = − 3 i + 2 j + 4 k are not coplanar. The position vectors of points A and C are 6 i + 7j + 4 k and − 9j + 2k respectively . Find the position vectors of a point P on the →

→

→

line AB & a point Q on the line CD such that PQ is perpendicular to AB and CD both. Q.5(a) Let a = 2î + ˆj − 2kˆ & b = î + ˆj . If c is a vector such that a · c =| c | , c − a = 2 2 and the angle between a × b and c is 30º, then a × b × c = (A) 2/3 (B) 3/2 (C) 2 (D) 3 ˆ ˆ ˆ (b) Let a =2i + j+ k , b = i + 2 j − k and a unit vector c be coplanar. If c is perpendicular to a , then c =

( )

( )

1 i − 2 j (D) ( i − j − k) ( ) 3 5 (c) Let a & b be two non-collinear unit vectors . If u = a − (a . b ) b & v = a x b , then v is : (A) u (B) u + u . a (C) u + u . b (D) u + u . (a + b )

(− j + k ) 2

1

(A)

(B)

(− i − j − k ) 3

1

(C)

1

(d) Let u & v be unit vectors . If w is a vector such that w + (w x u) = v , then prove that 1 ( u x v) . w ≤ 2 and the equality holds if and only if u is perpendicular to v . Q.6(a) An arc AC of a circle subtends a right angle at the centre O. The point B divides the arc in the ratio 1 : 2. → → → If OA = a & OB = b , then calculate OC in terms of a & b . (b) If a , b , c are non-coplanar vectors and d is a unit vector, then find the value of, a . d b x c + b . d ( c x a ) + c . d a x b independent of d . [ REE '99, 6 + 6 ] Q.7(a) Select the correct alternative : (i) If the vectors a , b & c form the sides BC, CA & AB respectively of a triangle ABC, then (A) a . b + b . c + c . a = 0 (B) a × b = b × c = c × a (C) a . b = b . c = c . a (D) a × b + b × c + c × a = 0 (ii) Let the vectors a , b , c & d be such that a × b × c × d = 0 . Let P1 & P2 be planes determined by the pairs of vectors a , b & c , d respectively . Then the angle between P1 and P2 is : (A) 0 (B) π/4 (C) π/3 (4) π/2 (iii) If a , b & c are unit coplanar vectors, then the scalar triple product

( )(

) ( )

( )(

)

( )( )

[ 2 a − b

]

2b−c 2c−a =

[ JEE ,2000 (Screening) 1 + 1 + 1 out of 35 ]

(D) 3 (A) 0 (B) 1 (C) − 3 Let ABC and PQR be any two triangles in the same plane . Assume that the perpendiculars from the points A, B, C to the sides QR, RP, PQ respectively are concurrent . Using vector methods or otherwise, prove that the perpendiculars from P, Q, R to BC, CA, AB respectively are also concurrent . [ JEE '2000 (Mains) 10 out of 100 ] Q.8. (i) If a = i + j − k , b = − i + 2 j + 2 k & c = − i + 2 j − k , find a unit vector normal to the vectors a + b and b − c . (ii) Given that vectors a & b are perpendicular to each other, find vector υ in terms of a & b satisfying (b)

the equations, υ . a = 0 , υ . b = 1 and

[ υ , a , b] = 1

(iii) (iv)

(

)

1 a , b & c are three unit vectors such that a × b × c = 2 a & b given that vectors b & c are non-parallel.

( b + c) . Find angle between vectors

A particle is placed at a corner P of a cube of side 1 meter . Forces of magnitudes 2, 3 and 5 kg weight act on the particle along the diagonals of the faces passing through the point P . Find the moment of these forces about the corner opposite to P . [ REE '2000 (Mains) 3 + 3 + 3 + 3 out of 100 ] Q.9(a) The diagonals of a parallelogram are given by vectors 2i + 3j − 6 k and 3i − 4 j − k . Determine its sides and also the area. (b) Find the value of λ such that a, b, c are all non-zero and −4 i + 5j a + (3i − 3j + k )b + (i + j + 3k )c = λ (ai + bj + ck ) [ REE '2001 (Mains) 3 + 3] Q.10(a) Find the vector r which is perpendicular to a = i − 2j + 5k and b = 2 i + 3j − k and r ⋅ 2i + j + k + 8 = 0.

(

)

(

)

(b) Two vertices of a triangle are at − i + 3j and 2 i + 5 j and its orthocentre is at i + 2j . Find the position vector of third vertex. [ REE '2001 (Mains) 3 + 3] 2 2 2 Q.11 (a) If a , b and c are unit vectors, then a − b + b − c + c − a does NOT exceed (A) 4 (B) 9 (C) 8 (D) 6 (b) Let a = î − kˆ, b = x î + ˆj + (1 − x )kˆ and c = yî + x ˆj + (1 + x − y)kˆ . Then [a , b, c] depends on (A) only x (B) only y (C) NEITHER x NOR y (D) both x and y [ JEE '2001 (Screening) 1 + 1 out of 35] Q.12(a) Show by vector methods, that the angular bisectors of a triangle are concurrent and find an expression for the position vector of the point of concurrency in terms of the position vectors of the vertices. (b) Find 3–dimensional vectors v 1 , v 2 , v 3 satisfying v1 ⋅ v1 = 4, v 1 ⋅ v 2 = –2, v 1 ⋅ v 3 = 6, v 2 ⋅ v 2 = 2, v 2 ⋅ v 3 = –5, v 3 ⋅ v 3 = 29. (c) Let A(t ) = f1 (t )i + f2 (t )j and B( t ) = g1 ( t ) i + g 2 ( t ) j , t ∈ [0, 1], where f1, f2, g1, g2 are continuous functions. If A(t ) and B( t ) are nonzero vectors for all t and A(0) = 2 i + 3j , A(1) = 6 i + 2j , B(0) = 3i + 2j and B(1) = 2 i + 6 j , then show that A(t ) and B(t ) are parallel for some t. [ JEE '2001 (Mains) 5 + 5 + 5 out of 100 ] Q.13(a) If a and b are two unit vectors such that a + 2 b and 5 a – 4 b are perpendicular to each other then the angle between a and b is  2  1 (A) 450 (B) 600 (C) cos–1   (D) cos–1    7  3 (b) Let V = 2î + ˆj − kˆ and W = î + 3kˆ . If U is a unit vector, then the maximum value of the scalar triple product U V W is [JEE 2002(Screening), 3 + 3]

[

]

(C) 59 (D) 60 (B) 10 + 6 Q.14 Let V be the volume of the parallelopiped formed by the vectors a = a1î + a 2 ˆj + a 3kˆ , b = b1î + b 2 ˆj + b 3kˆ , c = c1î + c 2 ˆj + c3kˆ . If ar , br , cr , where r = 1, 2, 3, are non-negative real (A) –1

3

[JEE 2002(Mains), 5] + b r + c r ) = 3L, show that V < L3. r =1 Q.15 If a = î + aˆj + kˆ , b = ˆj + akˆ , c = aî + kˆ , then find the value of ‘a’ for which volume of parallelopiped formed by three vectors as coterminous edges, is minimum, is 1 1 1 (B) – (C) ± (D) none [JEE 2003(Scr.), 3] (A) 3 3 3 Q.16(i) Find the equation of the plane passing through the points (2, 1, 0) , (5, 0, 1) and (4, 1, 1). (ii) If P is the point (2, 1, 6) then find the point Q such that PQ is perpendicular to the plane in (i) and the mid numbers and

∑ (a r

point of PQ lies on it. [ JEE 2003, 4 out of 60] Q.17 If u , v , w are three non-coplanar unit vectors and α, β, γ are the angles between u and v , v and w , w and u respectively and x , y , z are unit vectors along the bisectors of the angles 1 2 β γ 2 α sec 2 sec 2 . α, β, γ respectively. Prove that [x × y y × z z × x ] = [u v w ] sec 16 2 2 2 [ JEE 2003, 4 out of 60 ] x −1 y + 1 z −1 x −3 y−k z = = = = intersect, then k = Q.18(a) If the lines and 2 3 4 1 2 1 2 9 (A) (B) (C) 0 (D) – 1 9 2 (b) A unit vector in the plane of the vectors 2î + ˆj + kˆ , î − ˆj + kˆ and orthogonal to 5î + 2ˆj + 6kˆ

2î − 5kˆ 6î − 5kˆ 3ˆj − kˆ 2î + ˆj − 2kˆ (D) (B) (C) (A) 29 61 10 3 ˆ ˆ ˆ ˆ (c) If a = i + j + k , a · b = 1 and a × b = j − k , then b = [ JEE 2004 (screening)] (A) î (B) î − ˆj + kˆ (C) 2ˆj − kˆ (D) 2î Q.19(a) Let a , b, c , d are four distinct vectors satisfying a × b = c × d and a × c = b × d . Show that a ·b + c·d ≠ a ·c + b·d . (b) T is a parallelopiped in which A, B, C and D are vertices of one face. And the face just above it has corresponding vertices A', B', C', D'. T is now compressed to S with face ABCD remaining same and A', B', C', D' shifted to A., B., C., D. in S. The volume of parallelopiped S is reduced to 90% of T. Prove that locus of A. is a plane. (c) Let P be the plane passing through (1, 1, 1) and parallel to the lines L1 and L2 having direction ratios 1, 0, –1 and –1, 1, 0 respectively. If A, B and C are the points at which P intersects the coordinate axes, find the volume of the tetrahedron whose vertices are A, B, C and the origin. b · a b · a Q.20(a) If a , b, c are three non-zero, non-coplanar vectors and b1 = b − 2 a , b 2 = b + a , |a| | a |2 c · a b1 · c c · a b · c c·a b·c c·a b·c c1 = c − 2 a + 2 b1,c 2 = c − 2 a − 2 b1,c3 = c − 2 a + 2 b1,c4 = c − 2 a − 2 b1 |a| | b1 | |a| |c| |c| |c| |c| |b| then the set of orthogonal vectors is (A) a , b1 , c3 (B) a , b1 , c 2 (C) a , b1 , c1 (D) a , b 2 , c 2

(

)

(

)

(

)

(

)

(b) A variable plane at a distance of 1 unit from the origin cuts the co-ordinate axes at A, B and C. If the 1 1 1 centroid D (x, y, z) of triangle ABC satisfies the relation 2 + 2 + 2 = k, then the value of k is x y z (A) 3 (B) 1 (C) 1/3 (D) 9 [JEE 2005 (Screening), 3] (c) Find the equation of the plane containing the line 2x – y + z – 3 = 0, 3x + y + z = 5 and at a distance of 1 from the point (2, 1, – 1). 6 (d) Incident ray is along the unit vector vˆ and the reflected ray is along the ˆ . The normal is along unit vector aˆ outwards. Express unit vector w ˆ in terms of aˆ and vˆ . w [ JEE 2005 (Mains), 2 + 4 out of 60 ] Q.21(a) A plane passes through (1, –2, 1) and is perpendicular to two planes 2x – 2y + z = 0 and x – y + 2z = 4. The distance of the plane from the point (1, 2, 2) is (A) 0 (B) 1 (C) 2 (D) 2 2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (b) Let a = i + 2 j + k , b = i − j + k and c = i + j − k . A vector in the plane of a and b whose projection 1 , is [JEE 2006,3 marks each] on c is 3

(A) 4î − ˆj + 4kˆ (B) 3î + ˆj − 3kˆ (C) 2î + ˆj − 2kˆ (D) 4î + ˆj − 4kˆ (c) Let A be vector parallel to line of intersection of planes P1 and P2 through origin. P1 is parallel to the vectors 2 ˆj + 3 kˆ and 4 ˆj – 3 kˆ and P2 is parallel to ˆj – kˆ and 3 î + 3 ˆj , then the angle between vector A and 2 î + ˆj – 2 kˆ is π π π 3π (A) (B) (C) (D) [JEE 2006, 5] 2 4 6 4 (d) Match the following (i) Two rays in the first quadrant x + y = | a | and ax – y = 1 intersects each other in the interval a ∈ (a0, ∞), the value of a0 is (A) 2 (ii) Point (α, β, γ) lies on the plane x + y + z = 2. Let a = α î + β ˆj + γ kˆ , kˆ × ( kˆ × a ) = 0, then γ = (B) 4/3 1

(iii)

∫ (1 − y 0

0

2

) dy +

∫ (y

2

− 1) dy

(C)

1

∫ 0

(iv) If sinA sinB sinC + cos A cosB = 1, then the value of sin C = (D) 1 (e) Match the following ∞  1  (i) ∑ tan −1  2  = t , then tan t = (A) 0  2i  i =1 (ii) Sides a, b, c of a triangle ABC are in A.P. a b c , cos θ 2 = and cos θ1 = , cos θ3 = , b+c a+c a+b θ θ then tan 2 1 + tan 2 3 = (B) 1 2 2 (iii) A line is perpendicular to x + 2y + 2z = 0 and passes through (0, 1, 0). The perpendicular distance of this line from the origin is

0

1

5 3 (D) 2/3

1 − x dx +

∫

1 + x dx

−1

[JEE 2006, 6]

(C)

[JEE 2006, 6]

ANSWER KEY

EXERCISE–1 Q.1

x = 2 , y = −1

Q.4 Q.5

(i) parallel (ii) the lines intersect at the point p.v. − 2 i + 2 j (iii) lines are skew 2:1 Q.7 xx1 + yy1 = a2 Q.10 x = 2, y = – 2, z = – 2

Q.13 (a) Q.19

−1 1 1 i − j+ k 2 2 2

11 3

Q.25 p =

(b) externally in the ratio 1 : 3

Q.2

Q.20 (b)

Q.15

3 Q.21 2

(a) arc cos

1 3

Q.18

− î + 2ˆj + 5kˆ

5a 2 (a) ± 3( î − 2ˆj − 2kˆ ), (b) y = 3 or y = – 1 Q.24 sq. units 12 3

q (q 3 − 3) ; decreasing in q ∈ (–1, 1), q ≠ 0 4

EXERCISE–2 Q.1

2 17

Q.5

(i)

Q.8

Q.2

±

1 3 3

6 3 14 (ii) 6 (iii) 10 (iv) 7 5

p.v. of R = r = 3i + 3k

Q.10

(î + 5ˆj − kˆ )

Q.4 11

Q.6

6

α =n π +

Q.7

170

4 ˆ 1 ˆ 1 ˆ i− j− k 2 2 2

( −1) π , n ∈ I &β = 1 2 n

(

)

Q.16 α = 2/3 ; if α = 0 then vector product is − 60 2 i + k  b. b c  a bc b.c b Q.18 (b) p = a + c × b + −  a .b a .b  (a . c ) a . b  Q.21 F = 2a1 + 5a 2 + 3a 3

[ ] ( ( )

NO, NO

) ( ) ( ) ( ) ( )

(

Q.19 9 − j + k

)

4 22 Q.23 (a) 2î − 3ˆj + 3kˆ , (b) (i) – 4, (ii) rˆ = î + ˆj − kˆ + λ (ˆj + kˆ ) + µ(−3î + ˆj + 3kˆ ) , (iii) – 4, (iv) 9

Q.24 p = − or

1 1 + 2 cosθ

p=

; q= 1

1 + 2 cos θ

2 cos θ 1 + 2 cos θ

; q=−

; r= − 2 cosθ

1 + 2cosθ

1 1 + 2 cosθ

; r=

1 1 + 2 cos θ

a + ( c. a ) c + b × c b + ( c. b) c + a × c , y= Q.25 x = 1 + c2 1 + c2

EXERCISE–3 Q.1

θ = 900

Q.4

y + 2z = 4

Q.7

x −1 y − 2 z − 3 = = 2 2 −3

Q.8

x y z = = 1 2 −1

Q.11

x −1 y + 2 z + 3 = = 6 13 17

Q.13 (a)

x y z = = −1 1 − 2

or

1 1 1 1 + 2+ 2 = 2 2 x y z p

Q.12

x − 4 y + 14 z − 4 = = 3 10 4

Q.10

17 2

 3  3 2x 2 y z + − = 1; (c)  0, , 0  ; (d) x = 2t + 2 ; y = 2t + 1 and z = – t + 3 ; (b) 2 3 3 3  2 

Q.14 (1, –2, – 4)

x y z 19 x − 2 y +1 z−3 + + = 1 , Area = = = sq. units Q.16 2 3 −5 2 − 10 11 2 Q.18 p = 3, (2, 1, –3) ; x + y + z = 0

Q.15

Q.17

2x + 3y + z + 4 = 0 x −7 y−2 z +1 = = Q.19 −4 22 5 Q.22

Q.9

(x 2 + y2 + z 2 )

Q.20

Q.21

2 3

x −7 y−2 z−4 x −7 y−2 z−4 = = = = ; 3 6 2 2 6 −3

Q.23 (9, 13, 15) ; 14 ; 9x – 4y – z = 14

x − 4 y +1 z − 7 = = −1 9 −3

Q.24

Q.25 x – 2y + 2z – 1 = 0; 2 units

EXERCISE–4 Q.1 (a) 6

Q.2 (a) x = a × c ; y = b × c ; z = b + a × c or b × c − a Q.3 (a) (i) D (ii) C (iii) A, C Q.4 (a) x =

a×b a×b γ −a × γ 2 a×b  γ 



a×b a×b a×b γ +b× γ ; y= ; z= 2 γ  a × b

(a) (i) B

Q.8

(i) + i ;

Q.9

(a)

(

(ii) A (iii) A b axb (ii) 2 + 2 ; b (a × b)

)

(b) P ≡ (3, 8, 3) & Q ≡ (−3, −7, 6)

 γ   



Q.5 (a) B (b) A (c) A, C Q.7

(b) (2, 8, − 3) ; (0, 1, 2)

Q.6

(iii)

2π ; 3

(a) c = − 3 a + 2b

(iv) | M| =

[a b c]

(b)

7

1 1 1 5i − j − 7k , (− i + 7j − 5k ); 1274 sq. units (b) λ = 0, λ = –2 + 2 2 2

29

5 17 Q.10 (a) r = −13i + 11j + 7k ; (b) î + ˆj 7 7 Q.11 (a) B (b) C Q.12 (b) v 1 = 2i , v 2 = − i ± j, v 3 = 3i ± 2 j ± 4 k

Q.13 Q.18 Q.20 Q.21

(a) B (a) B, (a) B, (a) D;

; (b) C Q.15 D Q.16 (i) x + y – 2z = 3 ; (ii) (6, 5, –2) (b) B, (c) A Q.19 (c) 9/2 cubic units ˆ = vˆ – 2( aˆ · vˆ ) aˆ (b) D ; (c) 2x – y + z – 3 = 0 and 62x + 29y + 19z – 105 = 0, (d) w (b) A; (c) B, D; (d) (i) D, (ii) A, (iii) B, C, (iv) D; (e) (i) B, (ii) D, (iii) C

EXERCISE–5 Part : (A) Only one correct option 1. The locus of a point P which moves such that PA2 – PB2 = 2k2 where A and B are (3, 4, 5) and (– 1, 3 – 7) respectively is (A) 8x + 2y + 24z – 9 + 2k 2 = 0 (B) 8x + 2y + 24z – 2k2 = 0 (C) 8x + 2y + 24z + 9 + 2k2 = 0 (D) none of these The position vectors of three points A, B, C are i + 2 j + 3 k , 2 i + 3 j + k & 3 i + j + 2 k . A unit vector perpendicular to the plane of the triangle ABC is:

2.

 

(A)  −

(

1   i + j+k 3

)

(

 1   i − j+k  3

(B) 

)

4.

5. 6. 7.

a + λ b meets the plane r . nˆ = p in the point P whose position vector is  a . nˆ   p − a . nˆ   a . nˆ   p − a . nˆ  (A) a +  (B) a +  (D) a −   b  b (C) a −  b . nˆ  b  b    b . nˆ   b . nˆ   b . nˆ 

A straight line r =

9.

Equation of the angle bisector of the angle between the lines

x −1 y − 2 z−3 x −1 y − 2 z−3 = = & = = is 1 1 −1 1 1 1 z−3 x −1 y−2 = = (B) 3 1 2

x −1 y−2 = ;z –3=0 2 2 y−2 z−3 (C) x – 1 = 0 ; = 1 1

(A)

(D) None of these

The distance of the point, (− 1, − 5, − 10) from the point of intersection of the line,

10.

x − 2 y +1 z − 2 = = 3 4 12

and the plane, x − y + z = 5, is: (A) 10 (B) 11 (C) 12 (D) 13 If a plane cuts off intercepts OA = a, OB = b, OC = c from the coordinate axes, then the area of the triangle ABC =

11.

13.

(D) none

8.

12.

)

The square of the perpendicular distance of a point P (p, q, r) from a line through A(a, b, c) and whose direction cosine are , m, n is (A) Σ {(q – b) n – (r – c) m} 2 (B) Σ {(q + b) n – (r + c) m}2 (C) Σ {(q – b) n + (r – c) m}2 (D) none of these A variable plane passes through a fixed point (1, 2, 3). The locus of the foot of the perpendicular drawn from origin to this plane is: (A) x 2 + y2 + z 2 − x − 2y − 3z = 0 (B) x 2 + 2y 2 + 3z 2 − x − 2y − 3z = 0 (C) x 2 + 4y 2 + 9z 2 + x + 2y + 3 = 0 (D) x 2 + y2 + z2 + x + 2y + 3z = 0 The equation of the plane which bisects the angle between the planes 3x − 6y + 2z + 5 = 0 and 4x − 12y + 3z − 3 = 0 which contains the origin is (A) 33x − 13y + 32z + 45 = 0 (B) x − 3y + z − 5 = 0 (C) 33x + 13y + 32z + 45 = 0 (D) None The distance of the point of intersection of the line x – 3 = (1/2) (y–4) = (1/2) (z–5) and the plane x + y + z = 17 from the point (3, 4, 5) (A) 2 (B) 3 (C) 1/3 (D) 1/2 The lines x = ay + b, z = cy + d and x = a’y + b’, z = c’y + d’ will be mutually perpendicular provided (A) (a + a’)(b + b’) (c + c’) (B) aa’ + cc’ + 1 = 0 (C) aa’ + bb’ + cc’ + 1 = 0 (D) (a + a’) (b + b’) (c + c’) + 1 = 0

3.

(A)

(

 1   i + j−k  3

(C) 

1 2

b 2 c2 + c2a 2 + a 2 b 2

(B)

1 (bc + ca + ab) 2

(C)

1 abc 2

(D)

1 2

(b − c)2 + (c − a)2 + (a − b)2

The angle between the lines whose direction cosines satisfy the equations + m + n = 0 and 2 = m 2 + n2 is π π π π (A) (B) (C) (D) 6 3 2 4 If a1, b1, c1 and a2 , b2, c2 are the direction ratios of two lines and θ is the angle between the lines then tan θ is equal to Σ(b1c 2 − b 2c 1 )2

Σ(b1c 2 − b 2c 1 )2

Σ(b1c 2 + b 2c 1 )2

(D) none of these a1a 2 + b1b 2 + c 1c 2 a1a 2 + b1b 2 + c 1c 2 a1b1 + a 2b 2 + c 1c 2 14. A point moves so that the sum of the squares of its distances from the six faces of a cube given by x = ± 1, y = ± 1, z = ± 1 is 10 units. The locus of the point is (A) x 2 + y2 + z 2 = 1 (B) x 2 + y 2 + z 2 = 2 (C) x + y + z = 1 (D) x + y + z = 2 15. In the adjacent figure ‘P’ is any arbitrary interior point of the triangle ABC such that the lines AA1, BB1 and PA 1 PB1 PC1 CC1 are concurrent at P. Value of AA + BB + CC is always equal to . 1 1 1 (A)

(A) 1 16.

(B)

(B) 2

(C) 3

(C)

(D) None of these

The plane ax + by + cz = d, meets the coordinate axes at the points A, B and C respectively. Area of triangle ABC is equal to (A)

d2 a 2 + b 2 + c 2 | abc |

(B)

d2 a 2 + b 2 + c 2 2 | abc |

(C)

d2 a 2 + b 2 + c 2 4 | abc |

(D) None of these

17.

18.

19.

20. 21.

The length of projection, of the line segment joining the points (1, –1, 0) and (–1, 0, 1), to the plane 2x + y + 6z = 1, is equal to 255 237 137 155 (A) (B) (C) (D) 61 61 61 61 Two systems of rectangular axes have the same origin. If a plane cuts them at distances a, b, c and a1, b1, c1 from the origin, then 1 1 1 1 1 1 1 1 1 1 1 1 (B) 2 − 2 + 2 = 2 − 2 + 2 (A) 2 + 2 + 2 = 2 + 2 + 2 a1 b1 c 1 a1 b1 c 1 a b c a b c 2 2 2 (C) a2 + b2 + c2 = a1 + b1 + c1 (D) a2 – b2 + c2 = a12 − b12 + c12 The angle between the plane 2x – y + z = 6 and a plane perpendicular to the planes x + y + 2z = 7 and x – y = 3 is : π π π π (A) (B) (C) (D) 3 6 4 2 The non zero value of ‘a’ for which the lines 2x – y + 3z + 4 = 0 = ax + y – z + 2 and x – 3y + z = 0 = x + 2y + z + 1 are co-planar is : (A) – 2 (B) 4 (C) 6 (D) 0 The equation of the plane through the point (–1, 2 , 0) and parallel to the lines

x y +1 z − 2 x − 1 2y + 1 z + 1 = = and = = is 0 2 −1 1 3 −1

22.

(A) x + 2y + 3z - 1 = 0 (B) x – 2y + 3z + 5 = 0 (C) x + y – 3z + 1 = 0 (D) x + y + 3z – 1 = 0 The equation of the plane bisecting the acute angle between the planes 2x + y + 2z = 9 and 3x – 4y + 12z + 13 = 0 is : (A) 11x + 33y – 34z – 172 = 0 (B) 11x + 33y – 34z – 182 = 0 (C) 41x – 7y + 86z – 52 = 0 (D) 41x – 7y + 86z – 62 = 0

23.

The base of the pyramid AOBC is an equilateral triangle OBA with each side equal to 4 2 , ' O ' is the →

| |

origin of reference, AO is perpendicular to the plane of ∆ OBC and A O = 2 . Then the cosine of the angle between the skew straight lines one passing through A and the mid point of OB and the other passing through O and the mid point of BC is : (A) − 24.

(A)

1

(B) 0

2

(C)

1

(D)

6

2

The coplanar points A , B , C , D are (2 − x , 2 , 2) , (2 , 2 − y , 2) , (2 , 2 , 2 − z) and (1 , 1 , 1) respectively . Then :

1 1 1 + + =1 x y z

(B) x + y + z = 1

1 1 1 = 1(D) none of these + + 1− z 1− y 1− x

(C) →

25.

1

→

Let the centre of the parallelopiped formed by PA = i + 2 j + 2 k ; PB = 4 i − 3 j + k ; →

PC = 3 i + 5 j − k is given by the position vector (7, 6, 2). Then the position vector of the point P is: (A) (3, 4, 1)

(B) (6, 8, 2) →

26.

(C) (1, 3, 4)

(D) (2, 6, 8) →

1 → Taken on side A C of a triangle ABC, a point M such that A M = A C . A point N is taken on the 3 →

→

→

→

→

side CB such that BN = CB then, for the point of intersection X of A B & MN which of the following holds good? → → → 3 → 1 → AB (C) X N = MN (D) X M = 3 XN 4 3 If the acute angle that the vector, α i + β j + γ k makes with the plane of the two vectors 2 i + 3 j − k & i − j + 2 k is cot −1 2 then:

→

(A) X B =

27.

1 → AB 3

(A) α (β + γ) = β γ

→

(B) A X =

(B) β (γ + α) = γ α

(C) γ (α + β) = α β

(D) α β + β γ + γ α = 0

→

28.

Locus of the point P, for which OP represents a vector with direction cosine cos α =

1 2

( ' O ' is the origin) is: (A) A circle parallel to y z plane with centre on the x − axis (B) a cone concentric with positive x − axis having vertex at the origin and the slant height equal to the magnitude of the vector (C) a ray emanating from the origin and making an angle of 60º with x − axis →

29.

| |

a disc parallel to y z plane with centre on x − axis & radius equal to O P sin 60º x − x2 y − y2 z − z2 Equation of the plane passing through A(x 1, y1, z1) and containing the line = = is d1 d2 d3

(D)

x − x1

y − y1

z − z1

x − x 1 y 2 − y 1 z 2 − z1 =0 (A) 2 d1 d2 d3

x − d1 y − d2

(C)

x1 x2

y1 y2

z − d3 z1 z2

x − x2

y − y2

z − z2

x − x2 (B) 1 d1

y1 − y 2 d2

z1 − z 2 =0 d3

x

=0

x − x2 (D) 1 d1

y y1 − y 2 d2

z z1 − z 2 =0 d3

30.

The equations of the line of shortest distance between the lines x y z x−2 y −1 z−2 = = and = = are 2 −3 1 3 −5 2 x − (62 / 3) y + 31 z − (31/ 3) (A) 3(x – 21) = 3y + 92 = 3z – 32 (B) = = 1/ 3 1/ 3 1/ 3 x − 21 y + (92 / 3) z − (32 / 3) x−2 y+3 z −1 (C) = = (D) = = 1/ 3 1/ 3 1/ 3 1/ 3 1/ 3 1/ 3

31.

A line passes through a point A with p.v. 3 i + j − k & is parallel to the vector 2 i − j + 2 k . If P is a point on this line such that AP = 15 units, then the p.v. of the point P is:

32.

33.

34.

35.

(A) 13 i + 4 j − 9 k (B) 13 i − 4 j + 9 k (C) 7 i − 6 j + 11 k (D) − 7 i + 6 j − 11 k The equations of the planes through the origin which are parallel to the line 5 x −1 y+3 z +1 = = and distant from it are 3 2 −1 −2 (A) 2x + 2y + z = 0 (B) x + 2y + 2z = 0 (C) 2x – 2y + z = 0 (D) x – 2y + 2z = 0 The value(s) of k for which the equation x 2 + 2y2 – 5z2 + 2kyz + 2zx + 4xy = 0 represents a pair of planes passing through origin is/are (A) 2 (B) – 2 (C) 6 (D) – 6 y 2 x The equation of lines AB is = = . Through a point P(1, 2, 5), line PN is drawn perpendicular −3 6 2 to AB and line PQ is drawn parallel to the plane 3x + 4y + 5z = 0 to meet AB is Q. Then 78 156  9   52  ,  (A) coordinate of N is  , − (B) the coordinates of Q is  3, − , 9  49 49 49 2     x −1 y−2 z−5 y−2 z−5 x −1 (C) the equation of PN is = = (D) the equation of PQ is = = 3 − 176 − 89 − 13 8 4 x − 15 y − 29 z−5 Let a perpendicular PQ be drawn from P (5, 7, 3) to the line = = when Q is the 3 8 −5 foot. Then (A) Q is (9, 13, – 15) (B) PQ = 14 (C) the equation of plane containing PQ and the given line is 9x – 4y – z – 14 = 0 (D) none

EXERCISE–6

1.

2. 3.

4. 5.

6.

Find the equation of the plane which contains the origin and the line of intersection of the planes r . a = p and r . b = q z − c′ x−a y−b z−c x − a ′ y − b′ If the lines = = intersect at a point then the coordinate of the = = and b c a′ b′ c′ a point of intersection. The locus of a point which is a equidistant from the two given points with position vectors  1  a and b is the plane  r − (a + b ) . ( a – b ) = 0 bisecting the line joining the points normally.. 2   The foot of the perpendicular from (a, b, c) on the line x = y = z is the point (r, r, r) where 3r = a + b + c. Match the following : Column A Column B (a) Sum of the square of the direction (P) 0 cosines of line is (b) All the points on the z-axis have (Q) 1 their x and y coordinate equal to (c) Distance between the points (1, 3, 2) (R) 9 and (2, 3, 1) is (d) Shortest distance between the lines (S) 2 x+4 x−6 y−2 z−2 y z +1 = = and = = is 3 1 −2 2 −2 −2 Show that the angle between the straight lines whose direction cosines are given by the equations + m + n = 0 and amn + bn + cm = 0 is

π 1 1 1 if + + = 0. 3 a b c

7.

8.

9. 10.

Prove that the two lines whose direction cosines are given by the relations.p + qm + rn = 0 & a2 + bm 2 + cn2 = 0 are perpendicular if, p2(b + c) + q2 (c + a) + r2 (a + b) = 0 and parallel if

p2 q 2 r 2 + + =0. a b c

Find the plane π passing through the points of intersection of the planes 2x + 3y− z +1= 0 and x + y − 2z + 3 = 0 and is perpendicular to the plane 3x − y − 2z = 4. Find the image of point (1, 1, 1) in plane π. Given parallel planes r . (2 î − λ ˆj + kˆ ) = 3 and r . (4 î + ˆj − µ kˆ ) = 5 for what values of α, planes

r . (µ î − α ˆj + 3 kˆ ) = 0 & r . (α î − 3 ˆj + 2λ kˆ ) = 0 would be perpendicular..

The edges of a rectangular parallelepiped are a, b, c; show that the angles between the four diagonals are given by cos−1

11.

12. 13.

a 2 ± b2 ± c2 . a 2 + b2 + c2

Prove that the line of intersection of the planes r . ( î + 2 ˆj + 3 kˆ ) = 0 and r . (3 î + 2 ˆj + kˆ ) = 0 is r = t( î − 2 ˆj + kˆ ). Show that the line is equally inclined to î and kˆ and makes an angle (1/2) sec−1 3 with. ˆj .

x −1 y + 1 x +1 = =z& = (y − 2); z = 2 3 3 2

Find the shortest distance between the lines

Show that the line L whose equation is, r = (2 î − 2 ˆj + 3 kˆ ) + λ ( î − ˆj + 4 kˆ ) is parallel to the plane π ˆ whose vector r . ( i + 5 ˆj + kˆ ) = 5. Find the distance between them.

2 A sphere has an equation r − a + r − b = 72 where a = î + 3 ˆj − 6 kˆ and b = 2 î + 4 ˆj + 2 kˆ . Find: (i) the centre of the sphere (ii) the radius of the sphere (iii) perpendicular distance from the centre of the sphere to the plane r . 2 î + 2ˆj − kˆ = − 3. 15. Find the equation of the sphere which is tangential to the plane x − 2y − 2z = 7 at (3, −1, −1) and passes through the point (1, 1, −3). 16. P1 and P2 are planes passing through origin. L1 and L2 also passes through origin. L1 lies on P1 not on P2 and L2 lies on P2 but not on P1. Show that there exists points A, B, C and whose permutation A′ .B′.C′ can be chosen such that [IIT - 2004] (i)A is on L1, B on P1 but not on L1 and C not on P1.(ii)A′ in on L2, B′ on P2 but not on L2 and C′ not on P2. 2

14.

(

17.

)

A parallelopiped ‘S’ has base points A, B, C and D and upper face points A′, B′, C′ and D′. This parallelopiped is compressed by upper face A′B′C′D′ to form a new parallelopiped ‘T’ having upper face points A′′, B′′, C′′ and D′′. Volume of parallelpiped T is 90 percent of the volume of parallelopiped S. Prove that the locus of ‘A′′’ is a plane. [IIT - 2004]

EXERCISE–5 1.

C

2.

A

3.

A

4.

A

5.

D

6.

B

7.

B

8.

B

9.

10.

D

11.

A

13.

B

14.

16.

B

19.

EXERCISE–6 1.

r . (a q − pb) = 0

A

2.

( a + a′, b + b′, c + c ′ )

12.

C

3.

True

B

15.

A

5.

(a) → (Q), (b) → (P), (c) → (S), (d) → (R)

17.

B

18.

A

D

20.

A

21.

A

8.

 12 − 78 57  , ,  7x + 13y + 4z – 9 = 0 ;   117 117 117 

22.

C

23.

D

24.

A

25.

A

26.

C

27.

A

9.

α = + 3 12.

28.

B

29.

AB

30.

ABC

31.

AB

32.

AD

33.

BC

14.

(i) (0, 5, 5)

34.

ABCD 35.

BC

15.

True

4.

3 59

13.

8 3 (x – 2) 2 + (y – 1)2 + (z – 1)2 = 5

(ii) 9

(iii)

10 3 3

EXERCISE–7 Part : (A) Only one correct option 1.

The lengths of the diagonals of a parallelogram constructed on the vectors p = 2 a + b & q = a − 2 b ,

where a & b are unit vectors forming an angle of 60º are: (A) 3 & 4

(B)

(C)

7 & 13

(D) none

5 & 11

2

2.

   a − b  = 2  a 2 b   

2  a a − b b   (C)  (D) none  a b    A, B, C & D are four points in a plane with pv's a , b , c & d respectively such that a − d . b − c = b − d . ( c − a ) = 0. Then for the triangle ABC, D is its: 2  a − b  (B)   a b   

2 2 (A) a − b

3.

(

)(

) (

(A) incentre 4. 5.

)

(B) circumcentre

(C) orthocentre

(D) centroid

{(

)}

2π 2 Vectors a & b make an angle θ = . If a = 1, b = 2 then a + 3 b x 3 a − b = 3 (A) 225 (B) 250 (C) 275 (D) 300 Consider a tetrahedron with faces f 1, f 2, f 3, f 4. Let a 1 , a 2 , a 3 , a 4 be the vectors whose magnitudes are

) (

respectively equal to the areas of f 1, f 2, f 3, f 4 & whose directions are perpendicular to these faces in the outward direction. Then, (A) a 1 + a 2 + a 3 + a 4 = 0 (B) a 1 + a 3 = a 2 + a 4 (C) a 1 + a 2 = a 3 + a 4 (D) none

6.

(A) a . b = 0, b . c = 0

7.

9.

(B) c . a = 0, a . b = 0

(C) a . c = 0, b . c = 0

(D) a . b = b . c = c . a = 0

a .a a .b b .a b.b If a = i + j + k , b = i − j + k , c = i + 2 j − k , then the value of c.a c.b (A) 2

8.

For non−zero vectors a , b , c , a x b . c = a b c holds if and only if;

(B) 4

(C) 16

a.c b.c = c.c

(D) 64

If a , b & c are any three vectors, then a × b × c = a × b × c is true if: (A) b & c are collinear (B) a & c are collinear (C) a & b are collinear ( r . i) (i × r) + ( r . j) ( j × r) + ( r . k) (k × r) = (A) 0 (B) r (C) 2 r (D) 3 r : ,

(

) (

)

(D) none

10.

A point taken on each median of a triangle divides the median in the ratio 1 3 reckoning from the vertex. Then the ratio of the area of the triangle with vertices at these points to that of the original triangle is: (A) 5: 13 (B) 25: 64 (C) 13: 32 (D) none

11.

Given a parallelogram ABCD. If AB = a, AD = b & AC = c, then DB . AB has the value:

→

12. 13.

→

→

(B)

(A) p + q + r = 0 (B) p2 + q2 + r 2 − pq − qr − rp = 0 (C) p3 + q3 + r3 − 3 pqr = 0 (D) none of these If p & s are not perpendicular to each other and r x p = q x p & r . s = 0, then r =

(A) p . s 14.

→

a 2 + 3 b 2 − c2 a 2 − b 2 + 3 c2 (C) (D) none 2 2 The points whose position vectors are p i + q j + r k ; q i + r j + p k & r i + p j + q k are collinear if:

(A)

3 a 2 + b 2 − c2 2

→

 q . s (B) q −   p  p . s

If a, b, c are pth, qth, rth terms of an H.P. and

 q . p (C) q +   p  p . s

(D) q + µ p for all scalars µ

i j k u = (q − r) i + (r − p) j + (p − q) k , υ = + + , then: a b c (A) u , υ are parallel vectors (B) u , υ are orthogonal vectors (C) u . υ = 1 (D) u × υ = i + j + k

15.

16.

17.

18.

19.

If p , q are two noncollinear and nonzero vectors such that (b − c ) p × q + (c − a ) p + (a − b )q = 0 , where a, b, c are the length of the sides of a triangle, then the triangle is (A) right angled (B) obtuse angled (C) equilateral (D) isoceles

 2α 2β 2γ  cos ec 2 + cos ec 2 + cos ec 2  =   (A) 1 (B) 2 (C) 3 (D) none of these , If r x b = cx b & r . a = 0 where a = 2 i + 3 j − k b = 3 i − j + k & c = i + j + 3 k , then r is equal to: (A) 2 i − j + k (B) 2 i + j − k (C) 2 − i + j + k (D) 2 i + j + k The value of d b c a + d c a b + d a b c − d a b c is equal to: (A) 0 (B) 2 a b c d (C) – 2 a b c d (D) none of these

(

[

)

( ) ] [ ] [ ] [ ]

(

[

]

)

[

(

)

]

The three vectors i + j , j + k , k + i taken two at a time form three planes. The three unit vectors drawn perpendicular to these three planes form a parallelopiped of volume:

1 3

(B) 4

(C) →

(B) QRS

→

→

3 3 4

(D)

→

→

4 3 3

→

P Q × R S − Q R × P S + R P × Q S is equal to 4 times the area of the

For any four points P, Q, R, S, triangle: (A) PQR

21.

If cos α i + j + k , i + cos β j + k & i + j + cos γ k (α ≠ β ≠ γ ≠ 2 n π) are coplanar then the value of

(A) 20.

(C) PRS

(D) PQS

If a , b , c are three non − coplanar & p , q , r are reciprocal vectors, then: → → → →  →   →   a + m b + n c .  p + mq + n r is equal to:    

(A) 2 + m 2 + n2

(B) m + m n + n

(C) 0

→

22.

 

In a quadrilateral ABCD, A C is the bisector of the  A B

 

→

| |

→

= 5 A D then cos  BA (A) −

14

∧

21

(B) −

7 2

∧

(D) none of these → → 2π  A D which is , 15 A C = 3 A B 3  →

| |

| |

→  C D is: 

2

(C)

7 3 →

23.

→

(D)

7

2 7 14

→

| | | |

In the isosceles triangle ABC A B = BC = 8, a point E divides AB internally in the ratio 1: 3, then the →

→

→

| |

cosine of the angle between C E & C A is (where CA = 12)

3 7 3 8 3 7 −3 8 (B) (C) (D) 8 17 8 17 If p = 3 a − 5 b ; q = 3 a + b ; r = a + 4 b ; s = − a + b are four vectors such that ∧ q = 1 and r ∧ s = 1 then cos a ∧ b is: sin p (A) −

24.

(

)

(

(

)

)

19

25.

26.

( )

Let a & b be two non−collinear unit vectors. If u = a − a . b b & v = a x b , then v is [IIT - 1999]

(B) u + u . a

(C) u + u . b

( )

(D) u + u . a + b b.a b.a If a , b , c are three non-zero, non coplanar vectors and b1 = b – 2 a , b2 = b + 2 a , |a| |a| (A) u

27.

19

(B) 0 (C) (A) − (D) 1 5 43 5 43 If p, q, r be three mutually perpendicular vectors of the same magnitude. If a vector x satisfies the equation p × ( (x − q) × p ) + q × ( (x − r ) × q ) + r ( (x − p ) × r ) = 0 , then x is given by [IIT - 1997] 1 1 1 1 (A) (B) (C) (D) ( ( ( (2p + q − r ) p + q − 2r ) p + q+ r) p + q+ r) 3 3 2 2

b1.c c.a b.c c.a b.c c.a c 1 = c – 2 a + 2 b1 , c 2 = c – 2 a – 2 b1 , c 3 = c – 2 a + 2 b1 , | b1 | |c| |c| |a| |c| |a| c.a b.c c 4 = c – 2 a – b1 , then the set of orthogonal vectors is [IIT - 2005] |c| | b |2 (A) (a, b1, c 3 ) (B) (a, b1, c 2 ) (C) (a, b1, c 1 ) (D) (a, b 2 , c 2 ) Let A be vector parallel to line of intersection of planes P1 and P2 through origin, P1 is parallel to the vectors 2 î + 3 kˆ and 4 ˆj – 3 kˆ and P2 is parallel to ˆj – kˆ and 3 î + 3 ˆj , then the angle between vector [IIT - 2006] A and 2 î + ˆj – 2 kˆ is

28.

π π π (B) (C) 6 2 4 Part : (B) May have more than one options correct

(A)

(D)

3π 4

29.

If a , b , c & d are linearly independent set of vectors & K1 a + K 2 b + K 3 c + K 4 d = 0 then: (A) K1 + K2 + K3 + K4 = 0 (B) K1 + K3 = K2 + K4 = 0 (C) K1 + K4 = K2 + K3 = 0 (D) none of these

30.

Given three vectors a , b , c such that they are non − zero, non − coplanar vectors, then which of the following are coplanar.

(A) a + b , b + c , c + a (B) a − b , b + c , c + a (C) a + b , b − c , c + a (D) a + b , b − c , c − a 31.

32.

Let p = 2 i + 3 j + a k , q = b i + 5 j − k & r = i + j + 3 k . If p , q , r are coplanar and p . q = 20, a & b have the values: (A) 1, 3 (B) 9, 7 (C) 5, 5 (D) 13, 9 If z1 = a i + b j & z 2 = c i + d j are two vectors in i & j system where z1 = z 2 = r & z1 . z 2 = 0 then w1 = a i + c j & w 2 = b i + d j satisfy:

(A) w1 = r

33.

(B) w 2 = r

(C) w1 . w 2 = 0

(

34.

35.

(D) none of these

If a & b are two non colinear unit vectors & a , b , x a − y b form a triangle, then:  ∧   a b (A) x = − 1; y = 1 &  a + b  = 2 cos    2     ∧   ∧  (B) x = − 1; y = 1 & cos  a b +  a + b  cos a , − a + b  = − 1    

)

 ∧   ∧   a b  a b (C)  a + b  = − 2 cot  (D) none  cos  2  & x = − 1, y = 1 2         The value(s) of α ∈ [0, 2π] for which vector a = i + 3 j + (sin 2α ) k makes an obtuse angle with the α α Z-axis and the vectors b = (tan α ) i − j + 2 sin k and c = ( tan α ) i + ( tan α ) j − 3 cos ec k are 2 2

orthogonal, is/are: −1 (A) tan −1 3 (B) π − tan −1 2 (D) 2 π − tan −1 2 (C)π + tan 3 A parallelogram is constructed on the vectors p & q . A vector which coincides with the altitude of the

parallelogram & perpendicular to the side p expressed in terms of the vectors p & q is:

36.

q . p p (A) q − 2 ( p)

(p x q ) x p

q.p (C) 2 p − q p

p2

p x ( p x q) (D) p2

Identify the statement(s) which is/are incorrect? (A) (B) (C)

(D) 37.

(B)

[ ( )] ( ) (

2 a x a× a×b = a×b a If a , b , c are non coplanar vectors and v . a = v . b = v . c = 0 then v must be a null vector If a and b l ie i n a pl ane norm al to t he pl ane contai ni ng t he v ect ors c and d then a × b x c × d = 0 If a , b , c and a ′ , b ′ , c′ are reciprocal system of vectors then a . b ′ + b . c′ + c . a ′ = 3

(

)(

)

)

(

)

If a = i + 2 j + 4 k , b = 2 i − 3 j + k , c = i + 4 j − 4 k , then the vector a × b × c is orthogonal to:

(B) b (C) c (D) a + b + c If a , b , c are non-zero, non-collinear vectors such that a vector p = a b ∧ b c and a vector q = a c cos π − a ∧ c b then p + q is cos 2 π − a (A) parallel to a (B) perpendicular to a (C) coplanar with b & c (D) none of these (A) a

38.

39.

(

))

(

( (

))

Which of the following statement(s) is/are true?

[ ]

If n . a = 0, n . b = 0 & n . c = 0 for some non zero vector n , then a b c = 0 there exist a vector having direction angles α = 30º & β = 45º locus of point for which x = 3 & y = 4 is a line parallel to the z - axis whose distance from the z axis is 5 the vertices of a regular tetrahedron are OABC where ' O ' is the origin. The vector

(A) (B) (C) (D)

→

→

→

OA + OB + O C is perpendicular to the plane ABC.

40.

In a ∆ ABC, let M be the mid point of segment AB and let D be the foot of the bisector of ∠ C. Then the

Area ∆ CDM is: Area ∆ ABC 1 a−b 1 1 a−b 1 A−B A+B A−B A+B (A) (B) (C) tan cot (D) cot tan 4 a+b 4 2 a+b 2 2 2 2 2 , The vectors a , b , c are of the same length & pairwise form equal angles. If a = i + j & b = j + k the pv's of c can be: 1 4 1  1 4 1 4  4 1 (D)  − , , −  (C)  , − ,  (A) (1, 0, 1) (B)  − , , −   3 3 3 3 3 3  3 3 3 ratio

41.

EXERCISE–8

1.

Through the middle point M of the side AD of a parallelogram ABCD the straight line BM is drawn cutting AC at R and CD produced at Q prove that QR = 2RB

2.

Show that the perpendicular distance of the point c from the line joining a & b is,

b× c + c×a + a× b . b−a

(

)

3.

If α = i + 2 j + 3 k ; β = 2 i − j + k ; γ = 3 i + 2 j + k and α × β × γ = p α + q β + r γ then find the values of p, q, r

4.

, then find the value of If a = 2 i − 3 j + k , b = i − j + 2 k , c = 2 i + j − k & d = 3 i − j − 2 k

( a x b ) x ( a x c ). d

((q × c) × (p × b)) = b × ((p × c) × (q × a )) + c × ((p × a ) × (q × b))

5.

Show that a ×

6.

It is given that x = ; y = ; z =

bx c [ a b c]

cx a [a b c]

x .(a + b) + y .( b + c ) + z .( c + a ) .

axb where a , b , c are non − coplanar vectors. Show [ a b c]

that x , y , z also forms a non − coplanar system. Find the value of 7. 8.

The median AD of a triangle ABC is bisected at E and BE is produced to meet the side AC in F. Prove that AF = (1/3) AC and EF = (1/4) BF. Points X and Y are taken on the sides QR and RS, respectively of a parallelogram PQRS, so that QX = 4XR and RY = 4YS. The line XY cuts the line PR at Z. Find the ratio PZ: ZR.

9.

Forces P , Q act at O & have a resultant R . If any transversal cuts their line of action at A,B,C respectively,, then show that

10.

11. 12.

P Q R + = . OA OB OC

In a tetrahedron, if two pairs of opposite edges are perpendicular, then show that the third pair of opposite edges is also perpendicular & in this case the sum of the squares of two opposite edges is the same for each pair. Also show that the segment joining the mid points of opposite edges bisect one another. Use vectors to prove that the diagonals of a trapezium having equal non parallel sides are equal & conversely.

Given four non zero vectors a , b , c and d . The vectors a , b & c are coplanar but not collinear pair by

∧ ∧ ∧ ∧ π pair and vector d is not coplanar with vectors a , b & c and ( a b ) = ( b c ) = , (d a ) =α, (d b) =β , 3 ∧

c) = cos −1 (cos β − cos α) .

prove that ( d 13.

If p , q & r are three non-coplanar vectors, prove that,

a×b =

14.

15.

p p . a p.b

1

q q . a q.b

r r .a r.b

[q × r , r × p , p × q] Consider the non zero vectors a , b , c & d such that no three of which are coplanar then prove that a b cd + c a b d = b a cd + d a b c . Hence prove that a , b , c & d represent the position vectors

[ ] [ ] [ ] [ ]

b cd] + [ a b d] [ of the vertices of a plane quadrilateral if and only if = 1. a cd] + [ a b c] [

( )

Solve the following equation for the vector p ; p x a + p . b c = b x c where a , b , c are non zero non coplanar

[ ]

 a b c   p x a + c is perpendicular vectors and a is neither perpendicular to b nor to c , hence show that  a .c    to b − c .

16.

If a, b, c & a', b', c ' are reciprocal system of vectors then prove that:

17.

a + b +c + + = ( a ' x b ' ) ( b ' x c ' ) ( c ' x a ' ) (ii) . [abc ] Let A = 2i + k; B = i + j + k & C = 4i − 3j + 7k. Determine a vector R satisfying R x B = C x B & R . A = 0

18.

For any two vectors u & v , prove that

19.

Let ABC and PQR be any two triangles in the same plane. Assume that the perpendiculars from the points A, B, C to the sides QR, RP, PQ respectively are concurrent. Using vector methods or otherwise, prove that the perpendiculars from P, Q, R to BC, CA, AB respectively are also concurrent. [IIT - 2000] Find 3 – dimensional vectors v 1, v 2 , v 3 satisfying [IIT - 2001] v 1 . v 1 = 4, v 1 . v 2 = –2, v 1 . v 3 = 6, v 2 . v 2 = 2, v 2 . v 3 = –5, v 3 . v 3 = 29.

(i) [a b c] [a' b' c ' ] = 1

20.

21.

[IIT - 1998]

(a) ( u. v) 2 + | u × v|2 = | u|2 | v|2 &

(b) (1 + | u|2 )(1 +| v|2 ) = (1 − u. v) 2 + | u + v + ( u × v)|2

ˆ , vˆ , w ˆ be three non-coplanar unit vectors with angles between uˆ & vˆ is α, between vˆ & w ˆ is β If u

ˆ & uˆ is γ. If a , b , c are the unit vectors along angle bisectors of α, β , γ respectively,, and between w

[

]

then prove that, a × b , b × c , c × a =

1 α β γ [ uˆ vˆ wˆ ]2 sec2  2  sec2  2  sec2  2  . [IIT - 2003] 16

EXERCISE–7 1. B

2. B

3. C

4. D

5. A

6. D

7. C

8. B

9. A

10. B

11. A

12. B

13. B

14. B

15. C

16. B

17. C

18. A

3. p = 0; q = 10; r = − 3

19. D

20. B

21. A

22. C

23. C

24. C

4. – 98

25. B

26. A

27. B

28. D

29. ABC

30. BCD

31. AD 32. ABC

35. BD 36. ACD 40. BC 41. AD

33. AB 34. BD

37. AD 38. BC 39. ACD

EXERCISE–8 6. 3

20. v 1 = 2î , v 2 = −î ± ˆj , v 3 = 3 î ± 2ˆj ± 4kˆ are some

possible values

Assertion- Reason Some questions (Assertion–Reason type) are given below. Each question contains Statement – 1 (Assertion) and Statement – 2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. So select the correct choice :Choices are : (A)Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. (B)Statement – 1 is True, Statmnt – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1. (C) Statement – 1 is True, Statement – 2 is False. (D) Statement – 1 is False, Statement – 2 is True. 452.

Let a, b, c be three non-coplanar vectors then ( b − c).[(c − a) × (a − b)] = 0

453.

Statement 1: b − c can be expressed as linear combination of c − a and a − b . Statement 2: Given non-coplanar vectors one vector can be expressed as a linear combination of other two. A vector has components p and 1 with respect to a rectangular cartesian system. If the axes are rotated through an angle α about the origin in the anticlockwise sense. Statement–1 : If the vector has component p + 2 and 1 with respect to the new system then p = –1 Statement–2 : Magnitude of vector original and new system remains same

454.

Let | a | = 4, | b | = 2 and angle between a and

b is π/6

Statement–2 : (a × b) 2 = | a |2

Statement–1 : (a × b) 2 = 4

455.

456.

457.

458.

Statement–1 :  b × c c × a a × b  = 0 Statement–2 : If a , b , r are linearly dependent vectors then they are coplanar. Statement–1 : If a + b = a − b then a is parallel to b . Statement–2 : If a + b = a − b then a.b = 0. Let r be a non-zero vector satisfying r.a = r.b = r.c = 0 for given non–zero vectors a, b and c . Statement–1 : a, b and c are coplanar vectors. Statement–2 : r is perpendicular to the vectors a, b and c . Let a and r be two non–collinear vectors. Statement–1 : vector a × ( a × r ) is a vector in the plane of a and r , perpendicular to a . Statement–2 : a × a × b = 0 , for any vector b . Statement–1 : If three points P, Q, R have position vectors a , b , c respectively and , then the points P, Q, R must be collinear. Statement–2 : If for three 2a + 3b − 5c = 0 points A, B, C; AB = λ AC , then the points A, B, C must be collinear. Statement–1 : Let a and b be two non collinear unit vectors. If u = a − a.b b and v = a × b then | v |=| u | . π 1 ˆ ˆ ˆ Statement–2 : The vector with the vector 2i − 2 j + k is makes an angle of 3 3 5iˆ − 4ˆj + 3kˆ . Statement-1: If u & v are unit vectors inclined at an angle α and x is a unit vector bisecting the angle between u+v them, then x = α 2 cos 2

(

459.

460.

)

( )

(

(

461.

)

)

Statement-2: If ∆ABC is an isosceles triangle with AB = AC = 1, then vector representing bisector of angle A is

AB + AC given by AD = 2

462.

Statement-1: The direction ratios of line joining origin and point (x, y, z) must be x, y, z. Statement-2: If P is a point (x, y, z) in space and OP = r, then direction cosines of OP are

x y z , , . r r r

463.

Statement-1: If the vectors 2iˆ − ˆj + kˆ , î + 2ˆj − 3kˆ and 3iˆ − λˆj + 5kˆ are coplanar, then |λ|2 is equal to 16.

464.

Statement-2: The vectors a , b and c are coplanar iff a, (b × c ) = 0 Statement-1: A line L is perpendicular to the plane 3x – 4y + 5z = 10

1 > 5 2 5 2 2 Statement-1 : The points with position vectors a − 2b + 3c, − 2a + 3b − c , 4a − 7b + 7c are collinear. Statement-2: The position vectors a − 2b + 3c, − 2 a + 3b − c, 4a − 7b + 7c are linearly dependent vectors. 1 Statement-1: If a, b, c are three unit vectors such that a × (b × c) = b then the angle between a & b is π/2 2 1 Statement-2: If a × (b × c) = b, then a .b = 0. 2 Statement-2: Direction co-sines of L be <

465.

466.

467.

468.

469. 470.

471.

472.

3

,−

4

,

Statement-1: If cosα, cosβ, cosγ are the direction cosine of any line segment, cos2α + cos2β + cos2γ = 1. Statement-2: If cosα, cosβ, cosγ are the direction cosine of line segment, cos2α + cos2β + cos2γ = −1. Statement-1: The direction cosines of one of the angular bisector of two intersecting lines having direction cosines as l1 , m1, n1, & l2, m2, n2 is proportional to l1 + l2, m1+ m2, n1 + n 2. Statement-2: The angle between the two intersecting lines having direction cosines as l1, m1, n1 & l2, m2, n2 is given by cosθ = l1 l2 + m1m2 + n1n2.

Statement-2: a.b = 0 ⇒ either a = 0 or b = 0 or a ⊥ b a ⊥b Statement-1: A × B = B × A Statement-2: A × B =| A || B | sinθ nˆ , when θ is angle, when your fingers curls from A to B Statement-1 : A vector ⊥ r the plane of (1, -1, 0), (2, 1, -1) & (-1, 1, 2) is 6iˆ + 6kˆ Statement-2 : A × B always gives a vector perpendicular to plane of A & B Statement-1 : Angle between planes r.n1 = q1 & r.n 2 = q 2 . (acute angle) is given by cosθ = n1 .n 2 Statement-1: If a.b = 0 ⇒

Statement-2 : Angle between the planes in same as acute angle formed by their normals.

473.

Statement-1: In ∆ABC, AB + BC + CA = 0

Statement-1: a = 3i + pj + 3k and b = 2i + 3 j + qk are parallel vectors it p = 9/2 and q = 2. a a a Statement-2 : If a = a1 i + a 2 j + a 3 k and b = b1 i + b 2 j + b3 k are parallel 1 = 2 = 3 b1 b 2 b3 Statement-2 : If OA = a, OB = b then AB = a + b

474.

475.

Statement-1: The direction ratios of line joining origin and point (x, y, z) must be x, y, z

x y z , , r r r | [a − c bd] | Statement-1: The shortest distance between the skew lines r = a + αb and r = c + β d is | b×d | Statement-2: If P is a point (x,y, z) in space and OP = r then directions cosines of OP are

476.

Statement-2: Two lines are skew lines if three axist no plane passing through them. Statement-1: a = î + pjˆ + 2kˆ , b = 2iˆ + 3jˆ + qkˆ are parallel vectors of p = 3/2 and q = 4.

477.

Statement-2: a = a1î + a 2 ˆj + a 3 kˆ and b = b1î + b 2 ˆj + b3 kˆ are parallel if

a1 a 2 a 3 = = . b1 b 2 b3

ˆ b = 3jˆ + 4kˆ and c = 8iˆ − 3jˆ are coplanar then c = 4a − b . Statement-1: If a = 2iˆ + k,

478.

Statement-2: A set of vectors a1 , a 2 ...a n is said to be linearly independent if every relation of the form l1 a 1 + l2

a 2 + … + ln a n = 0 implies that l1 = l2 = …. = ln = 0 (scalars). 479.

(a − c).(b × d) Statement-1: The shortest distance between the skew lines r = a + αb and r = c + β d is | b×d |

480.

Statement-2: Two lines are skew lines if there exists no plane passing through them. Statement-1: The curve which is tangent to a sphere at a given point is the equation of a plane. Statement-2: Infinite number of lines touch the sphere at a given point.

481.

482.

483.

Statement-2: If OA = a, OB = b, then AB = a + b (∆ law of addition). 3 Statement-1: a = î + pjˆ + 2kˆ and b = 2iˆ + 3jˆ + qkˆ are parallel vectors if P = , q = 4 2 a a a Statement-2: If a = a1î + a 2 ˆj + a 3 kˆ and b = b1î + b 2 ˆj + b3 kˆ are parallel then 1 = 2 = 3 . b1 b 2 b3 ˆ ˆ ˆ ˆ ˆ ˆ Statement-1: If a = 2i + k a , a , a .....a , b = 3j + 4k and c = 8i − 3j are coplanar then c = 4a − b Statement-1: In ∆ABC AB + BC + CA = O

1

3

3

n

Statement-2: A set of vectors is said to be linearly independent if every relation of the form l1a + l2 a 2 + ..... + ln a n = 0 ⇒ l1 = l2 = ….. = ln = 0.

r = a1 + αb1 and r = a 2 + β b 2 is

484.

Statement-1:

485.

[b1 b 2 (a 2 − a1 )] Statement-2: Two lines are skew lines if there exists no plane passing through them. (b1 × b 2 ) ˆ + ˆj(kˆ × î) + k(i ˆ ˆ × ˆj) = 3 Statement-1 : The value of expression î(ˆj × k)

The shortest distance between the skew lines

ˆ = [i.j.k] ˆ ˆ ˆ =1 Statement-2 : î(ˆj × k) 486.

a×b Statement-1: A relation between the vectors r, a and b is r × a = b ⇒ r = Statement-2 : r.a = 0 a.a

3-Dimension 487.

488.

x −1 y + 3 z − 2 x − 2 y −1 z + 3 and = = = = −3 2 1 1 2 −3 Statement–1 : The given lines are coplanar Statement–2 : The equation 2x1 – y1 = 1, x1 + 3y1 = 4, 3x1 + 2y1 = 5 are consistent. Statement–1 : The distance between the planes 4x – 5y + 3z = 5 and 4x – 5y + 3z + 2 = 0 is 3 . Statement–2 The distance between ax + by + cz + d1 = 0 and ax + by + cz + d2 5 2 The equation of two straight line are

= 0 is

489.

d1 − d 2

. a + b 2 + c2 x −1 y +1 z − 3 Given the line and the plane π : x - 2y - z = 0 = = −1 3 2 2

Statement-1: L lies in π 490.

Statement-2: L is parallel to π

The image of the point (1, b, 3) in the Statement-1: Line

x y −1 z − 2 will be (1, 0, 7) = = 1 2 3

Statement-2: Length of the perpendicular from the point A( α ) on the line r = a + tb, is given by d =

| (a − α) × b | |b|

Answer 452. 459. 466. 473. 480. 487.

C A A C A A

453. 460. 467. 474. 481. 488.

A C B A C D

454. 461. 468. 475. 482. 489.

D A B A A C

455. 462. 469. 476. 483. 490.

D A D B B B

456. 463. 470. 477. 484.

D A D A B

457. 464. 471. 478. 485.

A A A B A

458. 465. 472. 479. 486.

C A A B A

Que from Compt. Exams Co-ordinate Geometry of Three Dimensions 1.

2. 3.

The direction cosines of a line segment AB are − 2 / 17 , 3 / 17 , − 2 / 17 . If AB = 17 and the co-ordinates of A are (3, – 6, 10), then the co-ordinates of B are (a) (1, –2, 4) (b) (2, 5, 8) (c) (–1, 3, –8) (d) (1, – 3, 8) The projection of any line on co-ordinate axes be respectively 3, 4, 5 then its length is [MP PET 1995; RPET 2001] (a) 12 (b) 50 (c) (d) None of these 5 2 If centroid of the tetrahedron OABC , where A, B, C are given by (a, 2, 3),(1, b , 2) and (2, 1, c) respectively be then distance of P(a, b, c) from origin is equal to (c) (b) (a) 107 14 107 / 14 If P ≡ (0, 1, 0 ), Q ≡ (0, 0, 1) , then projection of PQ on the plane x + y + z = 3 is

(d)

4.

(a) (b) 3 (c) 3 The points A(4 , 5 , 1), B(0 ,−1,−1), C(3, 9, 4 ) and D(−4 , 4 , 4 ) are

2

5.

2

(d)

(1, 2, –1),

None of these [EAMCET 2002]

[Kurukshetra CEE 2002]

8.

(a) Collinear (b) Coplanar (c) Non- coplanar (d) Non- Collinear and non-coplanar The angle between two diagonals of a cube will be [MP PET 1996, 2000; RPET 2000, 02; UPSEAT 2004] (a) sin −1 1 / 3 (b) cos −1 1 / 3 (c) Variable (d) None of these x − 8 y + 19 z − 10 and The equations of the line passing through the point (1,2,–4) and perpendicular to the two lines = = 3 − 16 7 x − 15 y − 29 z −5 , will be [AI CBSE 1983] = = 3 8 −5 x −1 y − 2 z + 4 x −1 y − 2 z + 4 x −1 y − 2 z + 4 (a) (b) (d) None of these (c) = = = = = = 2 3 6 3 2 8 −2 3 8 If three mutually perpendicular lines have direction cosines (l1 , m 1 , n1 ), (l 2 , m 2 , n 2 ) and (l 3 , m 3 , n 3 ) , then the line having direction

9.

cosines l1 + l 2 + l 3 , m 1 + m 2 + m 3 and n 1 + n 2 + n 3 make an angle of ..... with each other (a) 0 ° (b) 30 ° (c) (d) 60 ° 90 ° The straight lines whose direction cosines are given by al + bm + cn = 0, fmn + gnl + hlm = 0 are perpendicular, if

6. 7.

(a)

10.

11.

f g h + + =0 a b c

a + f

b + g

c = 0 (c) h

af = bg = ch (d)

a = f

b = g

c h

If the straight lines x = 1 + s, y = −3 − λs, z = 1 + λs and x = t / 2, y = 1 + t, z = 2 − t , with parameters s and t respectively, are co-planar, then λ equals [AIEEE 2004] (a) 0 (b) –1 (c) –1/2 (d) –2 The co-ordinates of the foot of perpendicular drawn from point P(1, 0, 3) to the join of points A(4 , 7, 1) and B(3, 5, 3) is [RPET 01]  5 7 17  2 5 7 (c) (b)  , ,   , ,  3 3 3  3 3 3 x −1 y +1 z −1 x −3 y −k z If the lines and = = = = intersect, then k = 2 3 4 1 1 1

(a) (5, 7, 1)

12.

(b)

(d)

5 2 7  , ,  3 3 3 [IIT Screening 2004]

2 9 (b) (c) 0 (d) None of these 9 2 A square ABCD of diagonal 2a is folded along the diagonal AC so that the planes DAC and BAC are at right angle. The shortest distance between DC and AB is [Kurukshetra CEE 1998]

(a)

13.

14.

(b) 2a / 3 (c) (a) 2a 2a / 5 (d) ( 3 / 2)a A line with direction cosines proportional to 2,1, 2 meets each of the lines x = y + a = z and x + a = 2 y = 2 z . The co-ordinates of each of the points of intersection are given by [AIEEE 2004] (a) (2a, a, 3 a), (2 a, a, a) (b) (3 a, 2 a, 3 a), (a, a, a) (c) (3 a, 2 a, 3 a), (a, a, 2 a) (d) (3 a, 3 a, 3 a), (a, a, a)

15.

The equation of the planes passing through the line of intersection of the planes 3 x − y − 4 z = 0 and x + 3 y + 6 = 0 whose distance from the origin is 1, are (a) x − 2 y − 2 z − 3 = 0 , 2 x + y − 2 z + 3 = 0 (b) x − 2y + 2z − 3 = 0 , 2 x + y + 2z + 3 = 0 (c) x + 2 y − 2 z − 3 = 0 , 2 x − y − 2 z + 3 = 0 (d) None of these

16.

The co-ordinates of the points A and B are (2, 3, 4) (–2, 5,– 4) respectively. If a point P moves so that PA 2 − PB 2 = k where k is a constant, then the locus of P is (a) A line (b) A plane (c) A sphere (d) None of these The equation of the plane passing through the points (1,−3,−2) and perpendicular to planes x + 2 y + 2 z = 5 and [AISSE 1987] 3 x + 3 y + 2 z = 8 , is (a) 2 x − 4 y + 3 z − 8 = 0 (b) 2 x − 4 y − 3 z + 8 = 0 (c) (d) None of these 2x + 4 y + 3z + 8 = 0 A variable plane at a constant distance p from origin meets the co-ordinates axes in A, B, C . Through these points planes are drawn parallel to co-ordinate planes. Then locus of the point of intersection is 1 1 1 1 1 1 1 (b) x 2 + y 2 + z 2 = p 2 (c) x + y + z = p (a) (d) + + = 2 + + =p x y z x2 y2 z2 p P is a fixed point (a, a, a) on a line through the origin equally inclined to the axes, then any plane through P perpendicular to OP, makes intercepts on the axes, the sum of whose reciprocals is equal to 3 3a (a) a (b) (c) (d) None of these 2a 2 The equation of the plane through the intersection of the planes x + 2 y + 3 z − 4 = 0 , 4 x + 3 y + 2 z + 1 = 0 and passing through the origin will be [MP PET 1998] (a) x + y + z = 0 (b) 17 x + 14 y + 11 z = 0 (c) 7 x + 4 y + z = 0 (d) 17 x + 14 y + z = 0 π with plane x + y = 3 , are [AIEEE 2002] The d.r’s of normal to the plane through (1, 0, 0 ), (0 , 1, 0 ) which makes an angle 4

17.

18.

19.

20.

21.

22.

23.

24.

25.

(b) 1,1, 2 (c) 1, 1, 2 (d) (a) 1, 2 ,1 2 , 1, 1 Two systems of rectangular axes have the same origin. If a plane cuts them at distance a, b, c and a', b', c' from the origin, then 1 1 1 1 1 1 1 1 1 1 1 1 (a) (b) + + + + + =0 + − + + − =0 a 2 b 2 c 2 a' 2 b ' 2 c ' 2 a 2 b 2 c 2 a' 2 b ' 2 c ' 2 1 1 1 1 1 1 1 1 1 1 1 1 (c) (d) − − + − − =0 + + − − − = 0 [AIEEE 2003] a 2 b 2 c 2 a' 2 b ' 2 c ' 2 a 2 b 2 c 2 a' 2 b ' 2 c ' 2 x −1 y +1 z If 4 x + 4 y − kz = 0 is the equation of the plane through the origin that contains the line = = , then k = 2 3 4 (a) 1 (b) 3 (c) 5 (d) 7 [MP PET 1992] x y z The distance of the point (1, –2, 3) from the plane x − y + z = 5 measured parallel to the line = = , is 2 3 −6 (a) 1 (b) 6/7 (c) 7/6 (d) None of these [AI CBSE 1984] x −3 y−4 z −5 and the plane x + y + z = 17 from the point (3, 4, 5) The distance of the point of intersection of the line = = 1 2 2 is given by (a) 3

26.

27.

28.

(b) 3/2 (c) (d) None of these 3 x −b +c y −b z −b −c x −a+d y −a z −a−d The lines and = = are coplanar and then equation to the plane in = = β −γ β β +γ α −δ α α +δ which they lie, is (a) x + y + z = 0 (b) x − y + z = 0 (c) (d) x − 2y + z = 0 x + y − 2z = 0 x−3 y−4 z −5 lies in the plane 4 x + 4 y − kz − d = 0 . The values of k and d are The line = = 2 3 4 (a) 4, 8 (b) –5, – 3 (c) 5, 3 (d) – 4, – 8 x −4 y −2 z −k lies in the plane 2 x − 4 y + z = 7 , is The value of k such that [IIT Screening 2003] = = 1 1 2

(a) 7 (b) – 7 (c) No real value (d) 4 The shortest distance from the plane 12 x + 4 y + 3 z = 327 to the sphere x 2 + y 2 + z 2 + 4 x − 2 y − 6 z = 155 is 4 (a) 26 (b) 11 (c) 13 (d) 39 13 29. 30.

31.

[AIEEE 2003]

The radius of the circle in which the sphere x 2 + y 2 + z 2 + 2 x − 2 y − 4 z − 19 = 0 is cut by the plane x + 2y + 2 z + 7 = 0 is (a) 1 (b) 2 (c) 3 (d) 4 [AIEEE 2003] The equation of motion of a rocket are: x = 2 t, y = −4 t, z = 4 t where the time 't' is given in seconds, and the co-ordinates of a moving point in kilometers. What is the path of the rocket? At what distance will be the rocket be from the starting point 0(0, 0, 0) in 10 seconds (a) Straight line, 60 km (b) Straight line, 30 km (c) Parabola, 60 km (d) Ellipse, 60 km The plane lx + my = 0 is rotated an angle α about its line of intersection with the plane z = 0 , then the equation to the plane in its new position is (a) lx + my ± z (l 2 + m 2 ) tan α = 0

(b) lx − my ± z (l 2 + m 2 ) tan α = 0 (c) lx + my ± z (l 2 + m 2 ) cos α = 0 (d)

lx − my ± z (l 2 + m 2 ) cos α = 0

32.

The distance between two points P and Q is d and the length of their projections of PQ on the co-ordinate planes are d1 , d 2 , d 3 .

33.

Then d12 + d 22 + d 32 = kd 2 where ‘k’ is (a) 1 (b) 5 (c) 3 (d) 2 If P1 and P2 are the lengths of the perpendiculars from the points (2,3,4) and (1,1,4) respectively from the plane

34.

(b) 7 P 2 − 23 P + 16 = 0 (c) P 2 − 17 P + 16 = 0 (d) P 2 − 16 P + 7 = 0 The edge of a cube is of length ‘a’ then the shortest distance between the diagonal of a cube and an edge skew to it is

3 x − 6 y + 2 z + 11 = 0 , then P1 and P2 are the roots of the equation (a)

P 2 − 23 P + 7 = 0

(a) a 2

(b) a

(c)

(d)

2 /a

a/ 2

Que from Compt. Exams Co-ordinate Geometry of Three Dimensions 1 6 11 16 21 26 31

d b b b b c a

2 7 12 17 22 27 32

c a b a d c a

3 8 13 18 23 28 33

a a b a c a d

4 9 14 19 24 29 34

c a b d a c b

5 10 15 20 25 30 35

b d a b a c d

Que from Compt. Exams Vector Algebra 1. 2.

Three forces of magnitudes 1, 2, 3 dynes meet in a point and act along diagonals of three adjacent faces of a cube. The resultant force is [MNR 1987] (a) 114 dyne (b) 6 dyne (c) 5 dyne (d) None of these The vectors b and c are in the direction of north-east and north-west respectively and |b|=|c|= 4. The magnitude and direction of the vector d = c – b, are [Roorkee 2000] (a) 4 2 , towards north

(b) 4 2 , towards west

(c)

4, towards east

(d)

4, towards south

3.

If a, b and c are unit vectors, then | a − b | 2 + | b − c | 2 + | c − a | 2 does not exceed[IIT Screening 2001] (a) 4 (b) 9 (c) 8 (d) 6

4.

The vectors AB = 3 i + 5 j + 4 k and AC = 5 i − 5 j + 2 k are the sides of a triangle ABC. The length of the median through A is [UPSEAT 2004]

5.

(a) 13 unit (b) 2 5 unit (c) 5 unit (d) 10 unit Let the value of p = (x + 4 y ) a + (2 x + y + 1) b and q = (y − 2 x + 2) a + (2 x − 3 y − 1) b , where a and b are non-collinear vectors. If 3 p = 2 q, then the value of x and y will be

[RPET 1984; MNR 1984]

6.

7.

8.

(a) – 1, 2 (b) 2, – 1 (c) 1, 2 (d) 2, 1\ The points D, E, F divide BC, CA and AB of the triangle ABC in the ratio 1 : 4, 3 : 2 and 3 : 7 respectively and the point K divides AB in the ratio 1 : 3 , then ( AD + BE + CF ) : CK is equal to [MNR 1987] (a) 1 : 1 (b) 2 : 5 (c) 5:2 (d) None of these If two vertices of a triangle are i − j and j + k , then the third vertex can be [Roorkee 1995] (a) i + k (b) i − 2 j − k (c) (d) i−k 2i − j (e) All the above 15 k If a of magnitude 50 is collinear with the vector b = 6 i − 8 j − , and makes an acute angle with the positive 2 direction of z-axis, then the vector a is equal to [Pb. CET 2004]

9.

(a) 24 i − 32 j + 30 k (b) −24 i + 32 j + 30 k (c) (d) 16 i − 16 j − 15 k −12 i + 16 j − 30 k If three non-zero vectors are a = a1 i + a 2 j + a 3 k, b = b 1 i + b 2 j + b 3 k and c = c 1 i + c 2 j + c 3 k. If c is the unit vector

π perpendicular to the vectors a and b and the angle between a and b is , then 6

(c) α = cos θ , β = sin θ , γ

2

= cos 2θ

(d)

b2

b3

c1

c2

c3

(b)

α = β = cos θ , γ 2 = − cos 2θ

(d)

None of these

is equal to

None of these

The points O, A, B, C, D are such that OA = a , OB = b , angle between BD and AC is π π (b) (a) 3 4

13.

b1

2

The vector a + b bisects the angle between the vectors a and b, if (a) | a | =| b | (b)| a | =| b | or angle between a and b is zero

(c) | a | = m | b | 12.

a3

(b)

(a) α = β = cos θ , γ 2 = cos 2θ 11.

a2

3 (Σa12 ) (Σb12 ) (Σc12 ) (Σa12 ) (Σb12 ) (c) 1 (d) 4 4 Let the unit vectors a and b be perpendicular and the unit vector c be inclined at an angle θ to both a and b. If [Orissa JEE 2003] c = α a + β b + γ (a × b ), then

(a) 0

10.

a1

(c)

OC = 2 a + 3 b and OD = a − 2 b . If | a | = 3| b |, then the

π 6

(d)

None of these

If A = i + 2 j + 3 k, B = −i + 2 j + k and C = 3 i + j, then the value of t such that A + t B is at right angle to vector C, is [RPET 2002]

14.

15.

(a) 2 (b) 4 (c) 5 (d) 6 Let b = 4 i + 3 j and c be two vectors perpendicular to each other in the xy-plane. All vectors in the same plane having projections 1 and 2 along b and c respectively, are given by [IIT 1987] 2 11 2 11 2 11 2 11 (b) 2 i + j, − i + (c) (a) 2 i − j, i + j j 2i + j, − i − j (d) j 2i − j, − i + 5 5 5 5 5 5 5 5 Let a = 2i − j + k, b = i + 2 j − k and c = i + j − 2 k be three vectors. A vector in the plane of b and c whose projection on a is of magnitude

2 / 3 is [IIT 1993; Pb. CET 2004]

16.

17.

(a) 2 i + 3 j − 3 k (b) 2i + 3 j + 3 k (c) (d) − 2i − j + 5 k 2i + j + 5 k A vector a has components 2p and 1 with respect to a rectangular cartesian system. The system is rotated through a certain angle about the origin in the anti-clockwise sense. If a has components p+1 and 1 with respect to the new system, then [IIT 1984] 1 1 (c) (d) (a) p = 0 (b) p = 1 or − p = −1 or p = 1 or −1 3 3 If u = 2 i + 2 j − k and v = 6 i − 3 j + 2 k, then a unit vector perpendicular to both u and v is [MP PET 1987] 1 1 18  1 k  (c) (7 i − 10 j − 18 k )  i − 2j − 5 5 473  17  If a = 2 i + k , b = i + j + k and c = 4 i − 3 j + 7 k . If d × b = c × b and d . a = 0 , then d will be

(a) i − 10 j − 18 k

18. 19.

(b)

(d)

None of these

[IIT 1990]

(a) i + 8 j + 2 k (b) i − 8 j + 2 k (c) (d) −i + 8 j − k −i − 8 j + 2 k If a × r = b + λa and a . r = 3, where a = 2 i + j − k and b = −i − 2 j + k , then r and λ are equal to 7 2 6 7 2 5 6 2 6 (b) r = i + j, λ = (c) (d) None of these (a) r = i + j, λ = r = i + j, λ = 6 3 5 6 3 6 7 3 5

20.

Let the vectors a, b, c and d be such that (a × b ) × (c × d ) = 0 . Let P1 and P2 be planes determined by pair of vectors a, b and c, d respectively. Then the angle between P1 and P2 is [IIT Screening 2000; MP PET 2004] (a) 0 o

21. 22.

23.

(b)

π

4 If a = i + j + k , a . b = 1 and a × b = j − k, then b =

(a) i (b) i − j + k The position vectors of the vertices of a quadrilateral formed by joining the middle points of its sides is 1 (a) | a ×b + b ×d + d ×a| 4 1 (c) a ×b + b ×c +c ×d +d×a 4

π

(c)

(d)

3

2

[IIT Screening 2004]

2i (c) 2 j − k (d) ABCD are a, b, c and d respectively. Area of the quadrilateral [Roorkee 2000]

1 b ×c + c ×d + a ×d + b ×a 4 1 b ×c +c ×d + d×b 4

(b) (d)

The moment about the point M (−2, 4 , − 6 ) of the force represented in magnitude and position by AB where the points A and B have the co-ordinates (1, 2, − 3) and (3, − 4 , 2) respectively, is (a) 8i – 9j – 14k (b) 2i – 6j + 5k (c) – 3i + 2j – 3k

24.

π

[MP PET 2000]

(d)

– 5i + 8j – 8k 1 1 1 If the vectors ai + j + k , i + b j + k and i + j + ck (a ≠ b ≠ c ≠ 1) are coplanar, then the value of + + = 1−a 1−b 1−c [BIT Ranchi 1988; RPET 1987;IIIT 1987; DCE 2001; MP P ET 2004; ORISSA JEE 2005]

1 1 (c) (d) 1 2 2 If α (a × b ) + β (b × c ) + γ (c × a ) = 0 and at least one of the numbers α , β and γ is non-zero, then the vectors a, b and c are (a) Perpendicular (b) Parallel (c) Coplanar (d) None of these The volume of the tetrahedron, whose vertices are given by the vectors −i + j + k , i − j + k and i + j − k with reference to the fourth vertex as origin, is 5 3 2 cubic unit (b) cubic unit (d) None of these cubic unit (c) (a) 3 5 3 ˆ is a unit vector such that a . d ˆ = 0 = [b c dˆ ], then dˆ is equal to [IIT 1995] Let a = i − j, b = j − k, c = k − i. If d (a) – 1

25.

26.

27.

i + j+k i + j − 2k (c) (b) ± (d) ± ± k 3 3 6 The value of 'a' so that the volume of parallelopiped formed by i + aj + k , j + a k and a i + k becomes minimum is

(a) ±

28.

(b) −

i + j−k

[IIT Screening 2003]

(a) – 3

(b) 3

1

(c)

(d)

3

3

29.

If b and c are any two non-collinear unit vectors and a is any vector, then (a . b ) b + (a . c ) c +

a . (b × c ) (b × c ) = | b ×c| [IIT 1996]

(a) a 30.

(b) b

(c)

If a, b, c are non-coplanar unit vectors such that a × (b × c ) =

c b +c

(d)

0

, then the angle between a and b is [IIT 1995]

2

31.

32.

33.

34.

π

(b) 4 2 [(a × b ) × (b × c ) (b × c ) × (c × a ) (c × a ) × (a × b )] =

(c)

3π 4

(a) [a b c ] 2

(c)

[a b c ] 4 (d)

(a)

π

(b) [a b c ] 3

(d)

π

None of these

Unit vectors a, b and c are coplanar. A unit vector d is perpendicular to them. If (a × b ) × (c × d ) =

1 1 1 i − j + k and 6 3 3

the angle between a and b is 30 o , then c is [Roorkee Qualifying 1998] (i − 2 j + 2 k ) (−i + 2 j − 2 k ) (−i + 2 j + k ) (2 i + j − k ) (b) (d) (c) (a) 3 3 3 3 [DCE 1999] The radius of the circular section of the sphere | r | = 5 by the plane r . (i + j + k ) = 3 3 is (a) 1 (b) 2 (c) 3 (d) 4

If x is parallel to y and z where x = 2 i + j + αk , y = αi + k and z = 5 i − j , then α is equal to (a) ± 5

(b) ± 6

(c)

± 7

(d)

None of these

[J & K 2005]

35.

The vector c directed along the internal bisector of the angle between the vectors a = 7 i − 4 j − 4 k and b = −2 i − j + 2 k with | c | = 5 6 , is 5 5 5 5 (b) (d) (c) (i − 7 j + 2 k ) (i + 7 j + 2 k ) (−5 i + 5 j + 2 k ) (5 i + 5 j + 2 k ) 3 3 3 3 The distance of the point B (i + 2 j + 3 k ) from the line which is passing through A (4 i + 2 j + 2 k ) and which is parallel to

(a)

36.

the vector C = 2 i + 3 j + 6 k is 37.

38.

[Roorkee 1993]

(a) 10 (b) 10 Let a, b, c are three non-coplanar vectors such that r1 = a − b + c , r2 = b + c − a , r3 = c + a + b ,

(c)

100

(d)

None of these

r = 2 a − 3 b + 4 c . If r = λ1 r1 + λ 2 r2 + λ 3 r3 , then (a) λ1 = 7 (b) λ1 + λ 3 = 3 (c) λ1 + λ 2 + λ 3 = 4 (d) λ3 + λ2 = 2 Let a = 2 i + j + k , b = i + 2 j − k and a unit vector c be coplanar. If c is perpendicular to a, then c = [IIT 1999; Pb. CET 2003; DCE 2005]

40.

1

1

1 (−i − j − k ) (i − 2 j) (− j + k ) (i − j − k ) (c) (d) (b) 2 3 5 3 Let p, q, r be three mutually perpendicular vectors of the same magnitude. If a vector x satisfies equation [IIT 1997 Cancelled] p × {(x − q ) × p} + q × {(x − r) × q} + r × {(x − p) × r} = 0, then x is given by 1 1 1 1 (b) (d) (c) (a) (p + q − 2 r) (p + q + r) (2 p + q − r) (p + q + r) 2 3 3 2 The point of intersection of r × a = b × a and r × b = a × b , where a = i + j and b = 2i − k is [Orissa JEE 2004] (a) 3 i + j − k (b) 3 i − k (c) (d) None of these 3i + 2 j + k

(a)

39.

1

Que from Compt. Exams Vector Algebra 1 6 11 16 21 26 31 36

c b b b a b c b

2 7 12 17 22 27 32 37

b e d b c c a,c b,c

3 8 13 18 23 28 33 38

b b c d a c d a

4 9 14 19 24 29 34 39

c d d b d a c b

5 10 15 20 25 30 35 40

b b a,c a c c a a

Maths Study Material - Three Dimentional Geometry

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