THEORY 3-D_E

MATHS

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 & y axes is called x-y plane, similarly y & z axes form y-z plane and z & 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.

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 ˆi + y ˆj + z kˆ .

Distance formula Distance between any two points (x1, y1, z1) and (x2, 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 | 

AB = |(x 2 ˆi + y2 ˆj + z2 kˆ ) – (x 1 ˆi + y1 ˆj + z1 kˆ )|



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 =

x2  y2

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MATHS Example # 1 : 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



 ABC = 90°

 BC = 3 2 , & AC = 6

Similarly

Clearly AB2 +BC2 = AC2 Also AB = BC Hence ABC is right angled isosceles.

Example # 2 : 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

BC =

(0  2)2  ( 11  3)2  (3  1)2  84

AC =

( 4  2)2  (5  3)2  ( 5  1)2  84

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

( x  0) 2  ( y  0)2  ( z   ) 2 +



x 2  y 2  ( z   )2 = 2a –

( x  0 )2  ( y  0)2  ( z   )2 = 2a

x 2  y 2  ( z   )2



x 2 + y2 + z2 + 2 + 2z = 4a2 + x 2 + y2 + z2 + 2 – 2z – 4a



4z– 4a2 = – 4a



or,

z 2 2

x 2  y 2  ( z   )2

+ a2 – 2z = x 2 + y2 + z2 + 2 – 2z

a2

 2    x + y + z 1  2  = a2 – 2   a  2

x 2  y 2  ( z   )2

2

2

x2 a2   2

+

y2 a2   2

+

z2 a2

=1

This is the required locus. 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 xaxis, +ve y-axis and +ve z-axis respectively and are of lengths 2, 3, 2 respectively find out the vertices.

(2)

Show that the points (0, 4, 1), (2, 3, –1), (4, 5, 0) and (2, 6, 2) are the vertices of a square.

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MATHS (3)

Find the locus of point P if AP2 – BP2 = 18, where A  (1, 2, – 3) and B  (3, – 2, 1). Answers :

(1) (3)

(1, 2, 5), (3, 2, 5), (3, 2, 3), (1, 5, 5), (1, 5, 3), (3, 5, 3), (3, 5, 5). 2x – 4y + 4z – 9 = 0

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,  mx 2  nx 1 my 2  ny 1 mz 2  nz1  internally then coordinates of P are given as  , ,  mn mn   mn

Note :- Mid point  x 1  x 2 y1  y 2 z1  z 2  , ,   2 2   2

Centroid of a triangle

 x 1  x 2  x 3 y1  y 2  y 3 z1  z 2  z 3  , ,  G   3 3 3  

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 (x 2, y2, z2) C (x 3, y3, z3) and D (x 4, y4, z4) are the vertices of a tetrahedron, then coordinate

of its centroid (G) is given as

      

4

4



4

 x  y  z  i 1

i

4

,

i 1

4

i

i

,

i 1

 4   

Example # 4 : 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

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MATHS 

k  2 =–4 k 1

3k = – 6 k = –2

2k  3 3k  4 = 1, and = –10 k 1 k 1 Since k < 0, therefore C divides AB externally in the ratio 2 : 1 and points A, B, C are collinear. For this value of k,

Example # 5 : 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 :

AB =

42  52  32  5 2

AC =

12  12  4 2  3 2

A(5, 4, 6)

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  

or,

 23 12 19  D=  , ,   8 8 8 



AD =

2

2

23  12  19     5    4    6   8  8  8    

2

=

B (1, –1, 3)

D

C (4, 3, 2)

1530 unit 8

Example # 6 : 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  : 1, then     4  2  7   8  A  , , .  1  1   1

.... (1)

Let A divide RS in the ratio k : 1, then  k  2 2k  3 5k  4  , ,  A    k 1 k 1 k 1 

..... (2)

From (1) and (2), we have,   4 k  2   1 k 1

..... (3)

2  7 2k  3   1 k 1

..... (4)

  8 5k  4   1 k 1

..... (5)

From (3), –k –  + 4k + 4 = k + 2 + k + 2 or 2k + 3 – 3k – 2 = 0 ..... (6) From (4), –2k – 2 + 7k + 7 = 2k + 3 + 2k + 3 or 4k + 5 – 5k – 4 = 0 ..... (7) Multiplying equation (6) by 2, and subtracting from equation (7), we get

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MATHS –+k=0 or , =k Putting  = k in equation (6), we get 22 + 3 – 3 – 2 = 0 or,  = ± 1. But  –1, as the co-ordinates of P would be undefined and in this case PQ || RS, which is not true.   = 1 = k. Clearly  = k = 1 satisfies eqn. (5). Hence our assumption is correct 

  1 4  2  7 1 8  A  , ,  2 2   2

3 5 9 A   , ,  . 2 2 2

or,

Self practice problems : (4)

Find the ratio in which xy plane divides the line joining the points A (1, 2, 3) and B (2, 3, 6).

(5)

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).

(6)

(7)

(8)

8  Two vertices of a triangle are (4, –6, 3) and (2, –2, 1) and its centroid is  ,  1, 2  . Find the 3   third vertex.

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, where O is the origin.  1  Show that   , 2, 0  is the circumcentre of the triangle whose vertices are A (1, 1, 0),  2 

B (1, 2, 1) and C (– 2, 2, –1) and hence find its orthocentre. Answers :

(4) (7)

1 : 2 Externally (5)

(3, 4, 5)

(8)

(1, 1, 0)

75

(6)

(2, 5, 2)

Direction cosines and direction ratios (i)

(ii) (iii)

Direction cosines : Let  be the angles which a directed line makes with the positive directions of the axes of x, y and z respectively, then cos , cos  cos  are called the direction cosines of the line. The direction cosines are usually denoted by , m, n. Thus  = cos , m = cos  , n = cos . 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 ˆi  bˆj  ckˆ will be a vector parallel to the line L. If , m, n are direction cosines of line L, then  ˆi + m ˆj + n kˆ is a unit vector parallel to the line L.

(iv)

If , m, n be the direction cosines and a, b, c be the direction ratios of a vector, then     

a a2  b2  c 2

,m 

b a2  b2  c 2

,n 

  2 2 2  a b c  c

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MATHS a =

or

b

a2  b2  c 2

,m=

c

a2  b2  c 2

,n=

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). If direction cosines of the line AB are , 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 x 2  x1 are, a = x 2  x 1, b = y2  y1 & c = z2  z1 and the direction cosines of line PQ are  = , | PQ | m=

(vii)

y 2  y1 z 2  z1 and n = . | PQ | | PQ |

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 # 7 : If a line makes angles , , with the co-ordinate axes, prove that sin2+ sin2+ sin2 = 2. Solution :   

Since a line makes angles , , with the co-ordinate axes, hence cos, cos, cosare its direction cosines cos 2+ cos 2+ cos 2= 1 (1 – sin2) + (1 – sin2) + (1 – sin2) = 1 sin2+ sin2+ sin2= 2.

Example # 8 : 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 2

or,

      – 2 = 0 m   m

or

  1 1 8  1 3   = 1, –2 m 2 2

[dividing by m 2]

 = 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

Case I. when   

±

1 1  1  ( 2 ) 2

2

2

,±

1 1  1  ( 2 ) 2

2

2

,±

2 1  12  ( 2) 2 2

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MATHS 1



,

1

6

2

,

6

or –

6

1 6

,–

1

2

,

6

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 Case II. When

2

±

,±

( 2 )  1  1 2

2 6

,

2

1 6

,

2

1

1 2

6

( 2)  12  12

2

2

or

1

, ±

(  2)  1  1 2

6

2

,

–1 6

,

–1 6

Self practice problems: (9)

Find the direction cosines of a line lying in the xy plane and making angle 30° with x-axis.

(10)

A line makes an angle of 60° with each of x and y axes, find the angle which this line makes with z-axis.

(11)

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.

(12)

A line makes angles , , ,  with the four diagonals of a cube, prove that cos 2 + cos 2+ cos 2+ cos 2=

Answers :

(9)

=

4 . 3

1 3 ,m=± , n=0 2 2

(10)

45°

(11)

a c , – b, 3 3

Angle between two line segments : If two lines have direction ratios a1, b1, c 1 and a2, b2, c 2 respectively, then we can consider two vectors parallel to the lines as a1 ˆi + b1 ˆj + c 1 kˆ and a2 ˆi + b2 ˆj + c 2 kˆ and angle between them can be given as. cos

=

a1a 2  b1b 2  c1c 2 a12

 b12  c12 a 22  b 22  c 22

.

(i)

The lines will be perpendicular if a1a2 + b1b2 + c 1c 2 = 0

(ii)

The lines will be parallel if

(iii)

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

Example # 9 : W hat



is

the

angle

c1 b a1 = 1 = b2 a2 c2

between

the

lines

whos e

dir ection

cos ines

ar e

3 1 3 3 1 3 , , and  , , . 4 4 2 4 4 2

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MATHS Solution :

Let  be the required angle, then cos = 12 + m 1m 2 + n1n2

 3   3   1   1   3   3        . =          4   4  4 4  2   2  =

3 1 3 1    16 16 4 2



 = 120°,

Example # 10 :Find the angle between any two diagonals of a cube. Solution :

The cube has four diagonals OE, AD, CF and GB The direction ratios of OE are a, a, a or 1, 1, 1 

1

its direction cosines are

3

,

Direction ratios of AD are – a, a, a 

3

1

,

3

or

1

its direction cosines are

1

3

,

1

. – 1, 1, 1.

,

3

1

.

3

Similarly, direction cosines of CF and GB respectively are

1

,

3

1

,

1

3

3

and

1 3

,

1 3

,

1 3

.

We take any two diagonals, say OE and AD Let  be the acute angle between them, then

 1   1  1   1   1   1  1   .   .   cos =            3  3 3  3  3  3  3 or,

 1  = cos –1   . 3

Example # 11 : 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 are (x 1, y1, z1), (x 2, y2, z2), (x 3, y3, x 3) respectively.

A (x1, y1, z1)

O (0, 0, 0) B (x2, y2, z2) Let

C (x3, y3, z3)

OA  BC and OB  CA .

We have to prove that OC  BA . Now, direction ratios of OA are x 1 – 0, y1 – 0, z1 – 0 direction ratios of BC are (x 3 – x 2), (y3 – y2), (z3 – z2). 

or,

x 1, y1, z1

OA  BC .

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MATHS 

x 1(x 3 – x 2) + y1(y3 – y2) + z1(z3 – z2) = 0 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), (z1 – z2))



..... (1)

Self practice problems: (13)

Find the angle between the lines whose direction cosines are given by  + m + n = 0 and 2 + m 2 – n2 = 0

(14)

Let

(15)

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 1n 2 – m 2n 1 , n 1 2 – n 2 1,  1m 2 –  2m 1

P (6, 3, 2), Q (5, 1, 4), R (3, 3, 5) are vertices of a  find Q.

Answers :

(13)

60°

(14)

90°

Projection of a line segment on a line (i)

If the coordinates of P and Q are (x 1, y1, z1) and (x 2, y2, z2), then the projection of the line segments PQ on a line having direction cosines , m, n is

(x 2  x1 )  m(y 2  y1 )  n(z2  z1 ) (ii)

    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) ˆi + (y2 – y1) ˆj + (z2 – z1) kˆ in place of a and

(iii) (iv)

  ˆi + m ˆj + n kˆ in place of b .   | r |, m | r | & n | r | are the projection of r in OX, OY & OZ axes.   r = | r | ( ˆi + m ˆj + n kˆ )

Example # 12 : 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 Direction ratios of the given line PQ are 2, 3, – 6 A 2 2  3 2  ( 6 ) 2 = 7 90° 90° 2 3 6 P L M Q  direction cosines of PQ are , ,– 7 7 7 Projection of AB on PQ = | (x 2 – x 1) + m(y2 – y1) + n(z2 – z1)| =

2 3 6 –466 (–1 – 1)  ( 4 – 2) – ( 2 – 3)  7 7 7 7



8 7

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MATHS Self practice problems : (16)

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.

(17)

The projections of a directed line segment on co-ordinate axes are – 2, 3, – 6. Find its length and direction cosines.

(18)

Find the projection of the line segment joining (2, – 1, 3) and (4, 2, 5) on a line which makes equal acute angles with co-ordinate axes. Answers :

(16)

5 22

(17)

7,

–2 3 –6 , , 7 7 7

7 (18)

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.

Equation of a plane (i)

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.

(ii)

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.

(iii)

The equation of a plane passing through the point (x 1, y1, z1) 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.

(iv)

Plane through three points: The equation of the plane through three noncollinear points

x (x 1, y1, z1), (x 2, y2, z2), (x 3, y3, z3) is

y

z 1

x1 y1 z1 1 x 2 y2 z2 1

=0

x 3 y3 z3 1

(v) (vi)

Note:

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

(a)

Vector equation of a plane normal to unit vector nˆ and at a distance d from the origin is . =d r n

(b)

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

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MATHS (c)

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.

(d)

Plane through origin: Equation of plane passing through origin is ax + by + cz = 0.

(e)

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  a 2  b2  c2

+

by  a 2  b2  c2

+

cz  a 2  b2  c2

=

d  a 2  b2  c2

Where (+) sign is to be taken if d > 0 and ( ) sign is to be taken if d < 0. (f)

Any plane parallel to the given plane ax + by + cz + d = 0 is ax + by + cz +  = 0. Distance between two parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is

(g)

| d1  d2 | 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 + b +  c (parametric form) where  &  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

 ax1  by1  cz1  d   ratio    ax 2  by 2  cz 2  d  (i)

The xyplane divides the line segment joining the points (x 1, y1, z1) and (x 2, y2, z2) in the ratio

 (j)

z1 . Similarly yzplane in z2



x1 and zx plane in x2



y1 y2

Coplanarity of four points The points A(x 1 y1 z1), B(x 2 y2 z2) C(x 3 y3 z3) and D(x 4 y4 z4) are coplaner then

x 2  x1 y 2  y1 z 2  z1 x 3  x1 y 3  y1 z3  z1 x 4  x1 y 4  y1 z 4  z1

=0

    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 # 13 : 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

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MATHS x + my + nz = p ..... (1) Here p = 10 Direction ratios of normal to the plane are 3, 2, 6 3 2  22  62 = 7

 =

Direction cosines of normal to the required plane are

3 2 6 ,m= ,n= 7 7 7

Putting the values of , m, n, p in (1), equation of required plane is or,

3 2 6 x + y + z = 10 7 7 7

3x + 2y + 6z = 70

Example # 14 : 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)

a b c   2  2  1 2 4  2

From (2) and (3), we have

or,

a b c   = 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 point D lies on the plane passing through A, B, C and hence points A, B, C and D are coplanar. Example # 15 : 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 + y2 + z2. Solution : Let P  (, , ), Q  (x 1, y1, z1) Direction ratios of OP are , ,  and direction ratios of OQ are x 1, y1, z1. Since O, Q, P are collinear, we have

     x1 y1 z1 = k (say)

..... (1)

As P (, , ) lies on the plane x + my + nz = p,  + m + n = p or k(x 1 + my1 + nz1) = p ..... (2) Given OP . OQ = p2 

 2  2   2

or,

k 2 ( x12  y12  z12 )

or,

k ( x12  y12  z12 ) = p2

x12  y12  z12 = p2 x12  y12  z12 = p2

On dividing (2) by (3), we get or,

..... (3) x 1  my 1  nz 1 x 12



y 12



z12



1 p

p (x 1 + my1 + nz1) = x12  y12  z12

Hence the locus of point Q is p (x + my + nz) = x 2 + y2 + z2.

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MATHS Example # 16 : A point P moves on a plane

x y z   = 1. A plane through P and perpendicular to OP meets s a b c

the 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 a b c

Let

P  (h, k, )

Then

h k    =1 a b c

OP =

h2  k 2   2

..... (2)

Direction cosines of OP are 

h h k  2

2

k

x

h k  2

,

2

k h k  2

2

2

,

 h  k 2  2 2

2



y

2

h k  2

2

2

z  h2  k 2   2

  h2  k 2   2   h2  k 2   2  h 2  k 2   2       0 , 0 , 0 , , 0 , 0 , 0 A   , C   , B    k h      

Let Q  (, , ), then  =

1 

2



From (3), h =



h k  2

hx + ky + z = (h2 + k 2 + 2)

or,

Now

h 2

Equation of the plane through P and normal to OP is

2



..... (1)

1 

2



1 

2



h2  k 2   2 h2  k 2   2 h2  k 2   2 ,= ,=  h k

h2  k 2   2 (h  k   ) 2

2

2 2



1 (h  k 2   2 ) 2

..... (3)

..... (4)

h2  k 2   2 

h h2  k 2  2  a a

Similarly

k h2  k 2   2  h2  k 2   2   and b b c c



h 2  k 2   2 h2  k 2   2 h2  k 2   2 h k       = 1 [from (2)] a b c a b c

or,

1 1 1 1 1 1 1       a b c h 2  k 2   2  2  2  2



Required locus of Q (, , ) is

[from (4)]

1 1 1 1 1 1    2  2  2 . ax by cz x y z

Self practice problems : (19)

Check whether given points 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)

(20)

Find the plane passing through point (– 3, – 3, 1) and perpendicular to the line joining the points (2, 6, 1) and (1, 3, 0).

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MATHS (21)

Find the equation of plane parallel to plane x + 5y – 4z + 5 = 0 and cutting intercepts on the axes whose sum is 150.

(22)

Find the equation of plane passing through (2, 2, 1) and (9, 3, 6) and perpendicular to the plane x + 3y + 3z = 8.

(23)

Find the equation of the plane parallel to ˆi  ˆj  kˆ and ˆi  ˆj and passing through (1, 1, 2).

(24)

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.

Answers :

(19)

yes, 4x – 3y + 2z = 3

(21)

x + 5y – 4z =

(23)

x + y – 2z + 2 = 0

3000 19

(20)

x + 3y + z + 11 = 0

(22)

3x + 4y – 5z = 9

(24)

17x + 2y – 7z = 26

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 y2 z2) 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 # 17 : 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, then points are on opposite sides of the plane.

A plane & a point ax'by'cz'd

(i)

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

(ii)

The length of the perpendicular from a point having position vector a to plane r . n = d



is p = (iii)

a2  b2  c 2

.

 

  | a.n  d | .  |n|

The coordinates of the foot of perpendicular from the point (x 1, y1, z1) to the plane ax + by + cz + d = 0 are

(ax1  by1  cz1  d) x  x1 y  y1 z  z1   =– a b c a 2  b2  c 2

(iv)

To find image of a point w.r.t. a plane. Let P (x 1, y1, z1) is a given point and ax + by + cz + d = 0 is given plane. Let (x, y, z) is the image of the point, then

(a)

x – x 1 = a, y – y1 = b, z – z1 = c  x = a + x 1, y = b + y1, z = c + z1

(b)

 x  x 1   y  y 1   z  z1   + b  + c  =0 a  2   2   2 

...... (i)

...... (ii)

from (i) put the values of x, y, z in (ii) and get the values of  and subtitute in (i) to get (x y z).

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MATHS The coordinate of the image of point (x 1 , y1 , z1) w.r.t the plane ax + by + cz + d = 0 are given by (v)

x '  x 1 y '  y 1 z'  z1 (ax1  by1  cz1  d)   =–2 a b c a 2  b2  c 2

The distance between two parallel planes ax + by + cx + d = 0 and ax + by + cx + d’ = 0

| d  d' | is

a  b2  c 2 2

Example # 18 : Find the image of the point P (3, 5, 7) in the plane 2x + y + z = 0. Solution :



Given plane is 2x + y + z = 0 ..... (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  (, , ) Since R is the middle point of PQ –3=

2=

3  = – 9 2

5 2

 = – 1

7  = 1 2 Q = (– 9, – 1, 1). 4=



Example # 19 : Find the distance between the planes 2x – y + 2z = 4 and 6x – 3y + 6z = 2. Solution :

Given planes are 2x – y + 2z – 4 = 0 and 6x – 3y + 6z – 2 = 0

.....(1) .....(2)

a1 b1 c 1 We find that a  b  c 2 2 2 Hence planes (1) and (2) are parallel. Plane (2) may be written as 2x – y + 2z –

2 =0 3

..... (3)

4 

Required distance between the planes =

2 3

2  ( 1)  2 2

2

2



10 10  3.3 9

Example # 20 : 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 + y2 + z2 – ax – by – cz = 0

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MATHS Solution :



Let the equation of the variable plane be O(0, 0, 0) x + my + nz + d = 0 ..... (1) Plane passes through the fixed point (a, b, c)  a + mb + nc + d = 0 ..... (2) P(, , ) Let P (, , ) be the foot of perpendicular from origin to plane (1). Direction ratios of OP are – 0, – 0, – 0 i.e. , ,  From equation (1), it is clear that the direction ratios of normal to the plane i.e. OP are , m, n; , , and , m, n are the direction ratios of the same line OP    1 = = = (say)  m n k

  = k, m = k, n = k ..... (3) Putting the values of , m, n in equation (2), we get ka + kb + kc + d = 0 ..... (4) Since , , lies in plane (1)   + m + n + d = 0 ..... (5) Putting the values of , m, n from (3) in (5), we get k2 + k 2 + k2 + d = 0 ..... (6) or k2 + k 2 + k2 – ka – kb – kc = 0 [putting the value of d from (4) in (6)] or 2 +  2 + 2 – a – b – c = 0 Therefore, locus of foot of perpendicular P (, , ) is x 2 + y2 + z2 – ax – by – cz = 0 ..... (7) Self practice problems: (25)

Find the intercepts of the plane 3x + 4y – 7z = 84 on the axes. Also find the length of perpendicular from origin to this plane and direction cosines of this normal.

(26)

Find : (i) (ii) (iii)

Answers :

perpendicular distance foot of perpendicular image of (1, 0, 2) in the plane 2x + y + z = 5

(25)

a = 28, b = 21, c = – 12, p =

(26)

(i)

 4 1 13  (ii)  , ,  3 6 6 

1 6

84 74

;

3 74

,

4 74

,

7 74

5 1 7 (iii)  , ,  3 3 3

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 , the angle between them, is given by cos

=

aa' bb' cc' a 2  b2  c2

a ' 2  b ' 2  c' 2

Planes are perpendicular if aa + bb + cc  = 0 and planes are parallel if

a b c = = a' b' c'  

(ii)

n1 .n2      The angle  between the planes r . n1 = d1 and r . n 2 = d2 is given by, cos  = 

| n1 | | n2 |

 

Planes are perpendicular if n1 . n 2 = 0 & planes are parallel if

  n1 =  n 2 .

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MATHS 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 (ii)



b12

 c12

=±

a 2 x  b2 y  c2 z  d 2 a 22  b 22  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  b12  c12

= ±

a 2 x  b2 y  c2 z  d 2 a 22  b 22  c22

gives the bisector of

the angle which contains the origin. (iii)

Bisector of acute/obtuse angle: First make both the constant terms positive. Then a 1a 2 + b 1b 2 + c 1c 2 > 0  origin lies on obtuse angle a 1a 2 + b 1b 2 + c 1c 2 < 0



origin lies in acute angle

Family of planes (i)

Any plane passing through the line of intersection of nonparallel planes or equation of the plane through the given line in non symmetric form. a 1x + b 1y + c 1z + d 1 = 0 & a2x + b2y + c 2z + d2 = 0 is a1x + b1y + c 1z + d1 +  (a2x + b2y + c 2z + d2) = 0, where  R

(ii)

The equation of plane passing through the intersection of the planes r . n1 = d 1

 

 



& r . n 2 = d2 is r . (n1 +

  n 2 ) = d1  d2 where  is arbitrary scalar

Example # 21 : 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 Putting k =

2 in equation (3), we get 5x + y + 4z = 20 3

This is the equation of the required plane. Example # 22 : 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 1 + 1 + 1 – 6 + k (2 + 3 + 4 + 5) = 0

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MATHS 3 14 From (3) required plane is 20x + 23y + 26z – 69 = 0 or,

k=

Example # 23 : 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) 2 x  y  2 z  9

Equations of bisecting planes are or, or, and Now  

( 2)  ( 1)  ( 2) 2

2

2



3 x  4 y  12z  13 3 2  ( 4) 2  (12)2

13 [– 2x – y – 2z + 9] = ± 3 (3x – 4y + 12z + 13) 35x + y + 62z = 78, ..... (3) [Taking +ve sign] 17x + 25y – 10z = 156 ..... (4) [Taking – ve sign] a1a2 + b1b2 + c 1c 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 + c 1c 2 < 0, origin lies in the acute angle between the planes.

Example # 24 : 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 +  (cx – y + az) = 0 or, x (1 + c) – y (c + ) + z (– b + a) = 0 ..... (4) If planes (3) and (4) are the same, then equations (3) and (4) will be identical. 

1  c (c   ) b  a   b a 1

(i) (ii) (iii) From (i) and (ii), a + ac = – bc – b or,

=–

(a  bc ) (ac  b)

..... (5)

From (ii) and (iii), c +  = – ab + a2 or

=

From (5) and (6), we have or, or, or,

–(ab c) 1 a2

..... (6)

(a  bc ) (ab  c )  . ac  b (1  a 2 )

a – a3 + bc – a2bc = a2bc + ac 2 + ab2 + bc a2bc + ac 2 + ab2 + a3 + a2bc – a = 0 a2 + b2 + c 2 + 2abc = 1.

Self practice problems: (27)

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  19  . angle between the faces OAB and ABC will be cos –1   35 

(28)

Find the equation of plane passing through the line of intersection of the planes 4x – 5y – 4z = 1 and 2x + y + 2z = 8 and the point (2, 1, 3).

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MATHS (29)

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 between them.

(30)

Show that the origin lies in the acute angle between the planes x + 2y + 2z – 9 = 0 and 4x – 3y + 12z + 13 = 0

(31)

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. Answers :

10 3

(28)

32x – 5y + 8z – 83 = 0,  =

(29)

2x + 7y – 5z = 21, 11x + 19y + 31z = 18; 2x + 7y – 5z = 21

Area of a triangle : Let A (x 1, y1, z1), B (x 2, y2, z2), C (x 3, y3, z3) be the vertices of a triangle, then

where

x

1 = 2

y1

z1 1

x1 1

x1

y1 1

y2 y3

z2 z3

x 2 1 and z = x 2 x3 1 x3

y2 1 y3 1

z1 1 1 , y = z 2 2 1 z3

 = (2x  2y  2z )





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

i x 2  x1

j y 2  y1

k z2  z1

x 3  x1

y 3  y1

z3  z1

Example # 25 : Through a point P (h, k, ) a plane is drawn at right angles to OP to meet the co-ordinate axes

p5 , where O is the origin. 2hk

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

Solution :

h2  k 2   2 = p

OP =

h

Direction cosines of OP are

h k  2

2

2

k

,

h k  2

2

2

,

 h  k 2  2 2

Since OP is normal to the plane, therefore, equation of the plane will be,

h h k  2

or, 

2

2

x

k h k  2

2

2

y

 h k  2

2

2

z  h2  k 2   2

hx + ky + z = h2 + k 2 + 2 = p2

..... (1)

2   p2   p2    , 0, 0   0,   0, 0, p  , 0 A  h  , B   k  , C         

Now area of ABC,  = A 2 xy  A 2 yz  A 2 zx Now Axy = area of projection of ABC on xy-plane = area of AOB

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MATHS

= Mod of

Similarly, Ayz =



2 =

=

1 2

p2 h

0

0

p k

1 =

0

0

1

1

2

1 p4 2 | hk |

1 p4 1 p4 and Azx = 2 | k | 2 | h |

1 p8 1 p8 1 p8   4 h 2k 2 4 k 2  2 4  2h 2 p8 2 2 2

4h k 

2

2

p10

2

( + k + h ) =

2 2 2

4h k 

or

=

p5 . 2hk

Volume of a tetrahedron : Volume of a tetrahedron with vertices A (x 1, y1, z1), B( x 2, y2, z2), C (x 3, y3, z3) and D (x 4, y4, z4) is

V = modulus of

x1

y1

z1 1

1 x2 6 x3 x4

y2 y3

z2 1 z3 1

y4

z4 1

A LINE 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 + c 1z + d1 = 0 and a2x + b2y + c 2z + 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

x  x1 y  y1 z  z1 = = = r. This form is called symmetric form. A general point on the line a b c

is given by (x  + ar, y + br, z + cr). (iii)

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 +  b where  is a scalar..

(iv)

The equation of the line passing through the points (x 1, y1, z1) and (x 2, y2, z2) is

x  x1 y  y1 z  z1 = = x 2  x1 y 2  y1 z 2  z1 (v)

Vector equation of a straight line passing through two points with position vectors

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  a &b

20

MATHS   is r = a + (vi)

   ( b  a ).

Reduction of cartesion form of equation of a line to vector form & vice versa x  x1 y  y1 z  z1 = = a b c

Note:

 r = (x 1 ˆi + y1 ˆj + z1 kˆ ) +



 (a ˆi + b ˆj + c kˆ ).

Straight lines parallel to co-ordinate axes: Straight lines (i) Through origin

Equation y = mx and z = nx

Straight lines (v) Parallel to x axis

Equation y = p, z = q

y = 0 and z = 0

(vi)

Parallel to yaxis

x = h, z = q

(vii)

Parallel to zaxis

x = h, y = p

(ii)

x axis

(iii)

y axis

x = 0 and z = 0

(iv)

z axis

x = 0 and y = 0

Example # 26 : 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)   a = OA = 3ˆi + 4ˆj – 7kˆ , Now  b = OB = ˆi – ˆj + 6 kˆ       Equation of the line through A( a ) and B( b ) is r = a + t ( b – a )



or r = 3 ˆi + 4 ˆj – 7 kˆ + t (–2 ˆi – 5 ˆj + 13 kˆ ) Equation in cartesian form : Equation of AB is

or,

..... (1)

x3 y4 z7   3 1 4 1  7  6

x3 y4 z7   2 5  13

Example # 27 : Find the co-ordinates of those points on the line

x 1 y  2 z  3   which is at a distance of 2 3 6

3 units from point (1, – 2, 3). Solution :

Given line is

x 1 y  2 z  3   2 3 6

..... (1)

Let P  (1, –2, 3) Direction ratios of line (1) are 2, 3, 6 

Direction cosines of line (1) are

2 3 6 , , 7 7 7

Equation of line (1) may be written as

x 1 y  2 z  3   2 3 6 7 7 7

..... (2)

3 6 2  Co-ordinates of any point on line (2) may be taken as  r  1 , r  2 , r  3  7 7 7 

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MATHS Let

3 6 2  Q   r  1, r  2, r  3  7 7 7 

Distance of Q from P = | r | According to question | r | = 3 Putting the value of r, we have 5 39   13  Q  , , 7 7   7



r=±3

23 3  1 ,  Q  , 7 7 7

or

Example # 28 : 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   3 2 1

..... (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 

9r – 6 + 4r – 4 + r + 3 = 0



Therefore, direction ratios of PQ are –

Equation of line PQ is

1 7 , 1, – 2 2

x –1 y – 0 z – 2   –1 2 –7

Example # 29 : Show that the two lines



14r = 7

or,

or,

r=

1 2

– 1, 2, – 7

x 1 y z2   1 2 7

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. Solution :

Given lines are

and



x 1 y  2 z  3   2 3 4

x  4 y 1 z  0   5 2 1

..... (1)

..... (2)

Any point on line (1) is P (2r + 1, 3r + 2, 4r +3) and any point on line (2) is Q (5 + 4, 2 + 1, ) Lines (1) and (2) will intersect if P and Q coincide for some value of  and r. 2r + 1 = 5 + 4  2r – 5 = 3 ..... (3) 3r + 2 = 2 + 1  3r – 2 = – 1 ..... (4) 4r + 3 =   4r –  = – 3 ..... (5) Solving (3) and (4), we get r = – 1,  = – 1 Clearly these values of r and  satisfy eqn. (5) Now P  (– 1, – 1, – 1) Hence lines (1) and (2) intersect at (– 1, – 1, – 1).

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MATHS Self practice problems: (32)

Find the equation of the line passing through point (1, 0, 2) having direction ratio 3, – 1, 5. Prove that this line passes through (4, – 1, 7).

(33)

Find the equation of the line parallel to line

x  2 y 1 z 7   and passing through the point 3 1 9

(3, 0, 5). (34)

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

Answers :

(32)

x –1 y z–2   3 –1 5

(33)

x3 y z5   3 1 9

(34)

 13 23  , 0  ,  5 5 

Reduction of non-symmetrical form to symmetrical form : Let equation of the line in nonsymmetrical form be a1x + b1y + c 1z + d1 = 0, a2x + b2y + c 2z + 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 a 1 + b 1m + c 1n = 0, a2 + b2m + c 2n = 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

Alternative method

i j k The vector a1 b1 c 1 = i (b1c 2 – b2c 1) + j (c 1a2 – c 2a1) + k (a1b2 – a2b1) will be parallel to the line of a 2 b2 c 2 intersection of the two given planes. hence : m: n = (b1c 2 – b2c 1): (c 1a2 – c 2a1): (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.

x  x1 y  y1 z0 = = or b1c2  b2 c1 c1a 2  c2a1 a1b2  a 2 b1 Note:

If 

b1d 2  b2 d1 d a  d 2a1 y 1 2 z0 a1b2  a 2 b1 a1b 2  a 2 b1 = = a1b2  a 2 b1 b1c2  b 2 c1 c1a 2  c2a1

x

 0, take a point on yzplane as (0, y1, z1) and if m  0, take a point on xzplane as (x1, 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 + c 1z + d1 = 0 and a2x + c 2z + d2 = 0 Example # 30 : 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)

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MATHS and 8x + 12y – 13z – 32 = 0 ..... (2) Let , m, n be the direction ratios of the line of intersection : then 4+ 4m – 5n = 0 and 8+ 12m – 13n = 0 

 m n    52  60  40  52 48  32

or,

 m n   8 12 16

or,

..... (3)

 m n   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 (, , 0). Then P lies on planes (1) and (2)  4+ 4– 12 = 0 or,  + – 3 = 0 ..... (5) and 8+ 12– 32 = 0 or, 2+ 3– 8 = 0 ..... (6) Solving (5) and (6), we get   1   89 68 32

or, 

  1   1 2 1  = 1,  = 2

Hence equation of line of intersection in symmetrical form is

x 1 y  2 z  0   . 2 3 4

Example # 31 : 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

and

 

x  3 y  4  0  4y  z  5  0

x  3 y  11  0  2y  z  6  0 

..... (1)

..... (2)

Let 1, m 1, n1 and 2, m 2, n2 be the direction cosines of lines (1) and (2) respectively 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 01 + 4m 1 – n1 = 0 ..... (4) Solving equations (3) and (4), we get

or,

1 m1 n  1 = 30 0  ( 1) 4  0

 1 m1 n1   = k (let). 3 1 4

 and

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) 2m 2 – n2 = 0 ..... (6)



2 = – 3m 2

or,

2 = m2 3

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MATHS and 



n2 = 2m 2

or,

n2 = m 2. 2

 2 m2 n2   = t (let). 3 1 2 If  be the angle between lines (1) and (2), then cos = 12 + m 1m 2 + n1n2 = (3k) (– 3t) + (k) (t) + (4k) (2t) = – 9kt + kt + 8kt = 0  = 90°.

Self practice problems: (35)

Find the equation of the line of intersection of the plane 4x + 4y – 5z = 12, 8x + 12y – 13z = 32.

(36)

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. Answers :

y2 x 1 z0 = = (36) 3 2 4

(35)

(3, 7, – 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 () 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  ) + m (mr + b  ) + n (nr + c ) = 0 i.e. r = ( a)  + ( b) m + ( 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 known, then the length of perpendicular is given by AP = (r  a   )  ( mr  b  )  ( nr  c   ) 2

by (ii)

2

2

the equation of perpendicular is given

y  z x = = mr  b   nr c  r  a  

 Vector Form : Equation of a line passing through a point having position vector  and         perpendicular to the lines r = a1 +  b1 and r = a 2 +  b 2 is parallel to b1 x b 2 . So the vector       equation of such a line is r =  +  ( b1 x b 2 ). Position vector  of the image of a point  in

       2 (a  ) . b       b   . Position vector of the a straight line r = a +  b is given by  = 2 a   2  |b|        (a ) . b   foot of the perpendicular on line is f = a    b . The equation of the perpendicular 2  |b|    is r =  +

      (a   ) .b    (a  )    2    |b| 

 b . 

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MATHS To find image of a point w. r. t a line Let L 

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 (i)

(ii)

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

x1  x y1  y z1  z  x2  y2  z2 2 2 2 = = = a b c from (ii) get the value of x, y, z in terms of  as x = 2a + 2x 2 – x 1, y = 2b + 2y2 – y1, z = 2c + 2z2 – z1 now put the values of x, y, zin (i) get  and resubtitute the value of  to to get (x y z).

Example # 32 : Find the length of the perpendicular from P (2, – 3, 1) to the line

Solution :

x 1 y 3 z  2   2 3 1

Given line is

x 1 y 3 z  2   . 2 3 1

..... (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 

or,

14r + 15 = 0



  22  3  13  , ,  Q  14 14   7



PQ =

r=

15 14

531 units. 14

Second method : Given line is

x 1 y 3 z  2   2 3 1

P  (2, – 3, 1)

P (2, –3, 1) 2

Direction ratios of line (1) are

14

3

,

14

,–

1 14

RQ = length of projection of RP on AB 2

=

14

( 2  1) 

3 14

( 3  3) 

1 14

(1  2) 

A 15

R (–1, 3, –2)

Q

B

14

PR2 = 32 + 62 + 32 = 54 

PQ =

PR 2  RQ 2 =

54 

225  14

531 units. 14

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26

MATHS Self practice problems: (37)

Find the length and foot of perpendicular drawn from point (2, –1, 5) to the line x  11 y  2 z  8   . Also find the image of the point in the line. 10 4  11

(38)

Find the image of the point (1, 6, 3) in the line

(39)

Find the foot and hence the length of perpendicular from (5, 7, 3) to the line

x y 1 z  2   . 1 2 3

x  15 y  29 z  5   . Find also the equation of the perpendicular.. 3 8 5

Answers :

14 , N  (1, 2, 3),  (0, 5, 1)

(37) (39)

(38)

(1, 0 , 7)

x5 y7 z3   2 3 6

(9, 13, 15) ; 14 ;

Angle between a plane and a line: (i)

If  is the angle between line



sin

 =  

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  is the angle between a line r = ( a +  b ) and r . n = d then sin =     .

(iii)

Condition for perpendicularity

 m n = = a b c

  b xn = 0

(iv)

Condition for parallel

a + bm + cn = 0

  b.n = 0

| b | | n |

Condition for a line to lie in a plane (i)

Cartesian form: Line

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. (ii)





Vector form: Line r = a +

        b would lie in the plane r . n = d if b . n = 0 & a . n = d

Coplanar lines : (i)

If the given lines are

x  y  z   x  ' y  ' z   ' = = and = = , then condition  m n ' m' n'

   '   '    '  m n for intersection/coplanarity is = 0 & equation of plane containing ' m' n'

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MATHS x  y  z   '

the above two lines is

(ii)

m m'

n n'

x   y   z   =0

or



m

n

'

m'

n'

=0

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 &

x +  y + z +  = 0 = x + y + z +  a b c a' b' c' They are coplanar if    ' ' '

d d' =0  '

Alternative method get vector along the line of shortest distance as

 u =

ˆi 

ˆj m

kˆ n

 m n

 Now get unit vector along u , let it be uˆ  Let v = ( – ) ˆi + ( – ) ˆj + ( – ) kˆ  S. D. = uˆ . v Example # 33 : Find the distance of the point (1, 0, – 3) from the plane x – y – z = 9 measured parallel to the line Solution :

x2 y2 z6   . 2 3 6

Given plane is x – y – z = 9 Given line AB is

..... (1)

x2 y2 z6   2 3 6

..... (2)

Equation of a line passing through the point Q(1, 0, – 3) and parallel to line (2) is x 1 y z  3   = r.. 2 3 6

..... (3)

Co-ordinates of any 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), 

B (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.

"manishkumarphysics.in"

Q (1, 0, – 3) A P

28

MATHS Example # 34 : Find the equation of the plane passing through (1, 2, 0) which contains the line x  3 y 1 z  2   . 3 4 2

Solution :

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   3 4 2



..... (2)

If plane (1) contains the given line, then 3a + 4b – 2c = 0 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 Solving equations (3) and (4), we get

or,

..... (3)

..... (4)

a b c   8  2 8  6  3  16

a b c   = k (say). 6 2 13

..... (5)

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.

Example # 35 : Find the equation of the projection of the line

Solution :

x 1 y 1 z  3   2 1 4 Given plane is x + 2y + z = 9 Let the given line AB be

x 1 y 1 z  3   on the plane x + 2y + z = 9. 2 1 4 ..... (1) ..... (2)

B

A

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

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MATHS 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 and

9x  2y  5z  4  0 x  2y  z  9  0

.....(i)   .....(ii)

..... (7)

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

 m n   4 – 7 10 Let any point on line (7) is (, , 0)  9 – 2 + 4 = 0  + 2 – 9 = 0 

1 17 , = 2 4 So equation of line is 

=

1 17 y– 2  4  z–0 4 –7 10

x–

Example # 36 : Show that the lines

x–7 y z7 x 3 y 1 z  2     and are coplanar. Also find the –3 1 2 2 3 1

equation of the plane containing them. Solution :

Given lines are

and



x 3 y 1 z  2   = r (say) 2 3 1

x7 y z7   = R (say) 3 1 2

..... (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) 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.

x  3 y 1 z  2 2 3 1 =0 Equation of the plane containing lines (1) and (2) is 3 1 2 or, or, or,

(x – 3) (– 6 – 1) – (y + 1) (4 + 3) + (z + 2) (2 – 9) = 0 – 7 (x – 3) – 7 (y + 1) – 7 (z + 2) = 0 x–3+y+1+z+2=0 or, x + y + z = 0.

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MATHS Self practice problems: (40)

Find the values of a and b for which the line

x2 y3 z6   is perpendicular to the plane a 4 2

3x – 2y + bz + 10 = 0.

(41)

Prove that the lines

x 1 y  2 z  3 x2 y3 z4     and are coplanar. Also find the 2 3 3 3 4 5

equation of the plane in which they lie.

(42)

Find the plane containing the line

y 3 z4 x2 = = and parallel to the line 3 5 2

x 1 y 1 z  1 = = 1 2 1 (43)

Show that the line

y2 x4 x 1 z3 y 1 = = & = = z are intersecting each other. Find 3 5 2 4 2

their intersection point and the plane containing the line. (44)

  Show that the lines r = (– ˆi – 3 ˆj – 5 kˆ ) +  (–3 ˆi – 5 ˆj – 7 kˆ ) & r (2 ˆi + 4 ˆj + 6 kˆ ) + µ

( ˆi +4 ˆj + 7 kˆ ) are coplanar and find the plane containing the line. Answers :

(40) (42)

a = –6, b = 1 (41) 13x + 3y – 7z – 7 = 0 (43)

(44)

r . ˆi  2ˆj  kˆ = 0





3x – y – z + 2 = 0 (– 1, – 1, – 1) & 5x – 18y + 11z – 2 = 0

Skew Lines : (i)

The straight lines which are not parallel and noncoplanar i.e. nonintersecting are called

' '  '  m n  0, skew lines. If  =  ' m' n' x– y– z–  x –  y –  z –      & are skew lines.  m n  m n Shortest distance: Suppose the equation of the lines are then lines (ii)

x  y  z   x  ' y  ' z   '   = = and  m n ' m' n'

S.D. =

(  ' ) (mn'm' n)  (  ' ) (n  n'  )  (    ' ) (m'' m)

 (mn'm' n) ' '  '    m n = ' m' n'

2

 (mn  mn)

2

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MATHS (iii)

(iv)

      Vector Form: For lines r = a 1 +   b1 & r = a 2 +   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 +

 b &

      (a 2  a1 ) x b r = a 2 +  b is d =  . |b|

Example # 37 : Find the shortest distance and the vector equation of the line of shortest distance between the lines given by





r  3ˆi  8ˆj  3kˆ   3ˆi  ˆj  kˆ

Solution :



r  3ˆi  7ˆj  6kˆ    3ˆi  2ˆj  4kˆ



 Given lines are r  3ˆi  8ˆj  3kˆ   3ˆi  ˆj  kˆ

and





r  3ˆi  7ˆj  6kˆ    3ˆi  2ˆj  4kˆ





and





..... (1)



..... (2)

Equation of lines (1) and (2) in cartesian form is

A

x3 y8 z3   AB : = 3 1 1

and

CD :





B

M

D

90°

x3 y7 z–6   = –3 2 4

90° C

Let L  (3 + 3, –  + 8,  + 3) and M  (– 3 – 3, 2 – 7, 4 + 6) Direction ratios of LM are 3 + 3 + 6, –  – 2 + 15,  – 4 – 3. Since 

L

LM  AB 3 (3 + 3 + 6) – 1 (– – 2 + 15) + 1 ( – 4 – 3) = 0 or, 11 + 7 = 0 ..... (5) Again LM  CD – 3 (3 + 3 + 6) + 2 (–  – 2 + 15) + 4 ( – 4 – 3) = 0 or, – 7 – 29 = 0 ..... (6) Solving (5) and (6), we get  = 0,  = 0 L  (3, 8, 3), M  (– 3, – 7, 6) Hence shortest distance LM = 

( 3  3 ) 2  ( 8  7 ) 2  (3  6 ) 2



Vector equation of LM is r  3ˆi  8ˆj  3kˆ  t 6ˆi  15ˆj  3kˆ Note : Cartesian equation of LM is

=

270 = 3 30 units



x3 y8 z3   . 6 15 3

Example # 38 : Prove that the shortest distance between any two opposite edges of a tetrahedron formed by

Solution :

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

3 a is

x+y+z= 3a ..... (iv) Clearly planes (i), (ii) and (iii) meet at O(0, 0, 0) Let the tetrahedron be OABC

2 a.

, 0) O (0, 0

C (0, 0,

3 a)

P

A

Q

D

Let the equation to one of the pair of opposite edges OA and BC be

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MATHS y + z = 0, x + z = 0

..... (1)

x + y = 0, x + y + z = 3 a ..... (2) equation (1) and (2) can be expressed in symmetrical form as

and,

 

x0 y0 z0   1 1 1

..... (3)

x0 y0 z 3 a   1 1 0

..... (4)

d. r. of OA and BC are respectively (1, 1, – 1) and (1, – 1, 0). Let PQ be the shortest distance between OA and BC having direction cosines (, m, n) PQ is perpendicular to both OA and BC. +m–n=0 and –m=0

 m n   = k (say) 1 1 2

Solving (5) and (6), we get, also, 



2 + m 2 + n2 = 1

1

k 2 + k 2 + 4k 2 = 1  k = 

 = 

1

, m = 

6

1 6

6

, n = 

1 6

0.

1 6

2

 3a.

C

B

2 6

Shortest distance between OA and BC i.e. PQ = The length of projection of OC on PQ = | (x 2 – x 1)  + (y2 – y1) m + (z2 – z1) n | = 0.

O

A

=

2 a.

A

90° Q

B

90°

C

6

P

O

Self practice problems: (45)

Find the shortest distance between the lines

x2 y4 z5 x 1 y  2 z  3   and .   3 4 5 2 3 4

Find also its equation. (46)

Prove that the shortest distance between the diagonals of a rectangular parallelopiped whose coterminous sides are a, b, c and the edges not meeting it are

bc b c 2

2

,

ca c a 2

2

,

ab a  b2 2

1 Answers :

(45)

6

, 6x – y = 10 – 3y = 6z – 25

Sp here General equation of a sphere is given by x 2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0, where ( u, v, w) is the centre and u 2  v 2  w 2  d is the radius of the sphere. Example # 39 : 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)

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MATHS H Let H be the centre of the required sphere. Given H  (1, 2, 3) 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 # 40 : Find the equation of the sphere if it touches the plane r .( 2ˆi  2ˆj  kˆ ) = 0 and the position vector of its centre is 3ˆi  6ˆj  4kˆ Solution :

 Given plane is r .( 2ˆi  2ˆj  kˆ ) = 0

..... (1)

 Let H be the centre of the sphere, then OH = 3ˆi  6ˆj  4kˆ = c (say) Radius of the sphere = length of perpendicular from H to plane (1)  | c.(2ˆi  2ˆj  kˆ ) | | (3ˆi  6ˆj  4kˆ ).(2ˆi  2ˆj  kˆ ) | | 6  12  4 | 2  = = = = a (say) ˆ ˆ ˆ ˆ ˆ ˆ 3 3 | 2i  2 j  k | | 2i  2 j  k | 



Equation of the required sphere is | r  c | = a

or or

or

2 | xˆi  yˆj  zkˆ  (3ˆi  6ˆj  4kˆ ) |  3

or

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

or

(x – 3)2 + (y – 6)2 + (z + 4)2 =

4 9

9 (x 2 + y2 + z2 – 6x – 12y + 8z + 61) = 4 9x 2 + 9y2 + 9z2 – 54x – 108y + 72z + 545 = 0

Example # 41 : 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) 2v  8 6 From (7), 4w = 3 + 2v

From (6), u =

Putting the values of u, v and w in (5), we get

..... (8) ..... (9)

2v  8  2v – 3 – 2v  1 2

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MATHS 



2v + 8 – 4v – 6 – 4v = 2

From (8),

u=

08 4  6 3



From (9), 4w = 3

v=0

3 4

w=

From (3), d = 2v – 1 = 0 – 1 = – 1 From (1), equation of required sphere is x 2 + y2 + z2 – or

3 8 x+ z–1=0 3 2

6x 2 + 6y2 + 6z2 – 16x + 9z – 6 = 0

Example # 42 : 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  Self practice problems : (46)

Find the value of k for which the plane x + y + z = 2

2

3 k touches the sphere

2

x + y + z – 2x – 2y – 2z – 6 = 0. (47)

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

(48)

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).

(49)

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.

(50)

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 Answers :

a b c   = 2. x y z

(46) (48)

(47) x 2 + y2 + z2 – 2x + 6y – 4z + 10 = 0 3 ±3 2 2 x + y + z – 6x – 4y + 10 z + 12 = 0

(49)

4 2 units, centre (21, 14 , –7)

2

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