1.
Substituting gives, 2(2λ + 4) + 3(–λ – 2) – (3λ + 2) = 2 ⇔ 4λ + 8 – 3λ – 6 – 3λ – 2 = 2 ⇔ –2λ = 2 λ = –1
(M1) (M1) (A1)
Intersection is (2, –1, –1)
(A1) (C4) [4]
2.
The system of equations will not have a unique solution if the determinant of the matrix representing the equations is equal to zero.
4 Therefore, 2
−1 2 0 =0
3
(M1)
1 −2 a 4 × 3a + 2a + 2 × (–4 – 3) = 0 14a = 14 a=1
⇔ ⇔
(M1) (M1) (A1) (C4) [4]
3.
(a)
A perpendicular vector can be found from the vector product ! ! ! i j k ! ! ! OP × OQ = 1 − 3 2 = i − 3 j − 5k
−2
(b)
Area ΔOPQ =
1
−1
1 OP OQ sin θ, 2
where θ is the angle between OP and OQ
1 OP × OQ 2 35 = 2
(M1)(A1) (C2)
(M1)
=
(A1) (C2) [4]
1
4.
(a)
(b)
Since the coordinates of the points P, Q and R are (4, 1, –1), (3, 3, 5) QR PR and (1, 0,!2c), !respectively, and are given by ! the vectors QR = −2i − 3 j + (2c − 5)k (M1)(A1) ! ! ! PR = −3i − j + (2c + 1)k (M1)(A1)
QR is perpendicular to PR if and only if QR ⋅ PR = 0 ie 6 + 3 + (2c – 5)(2c +1) = 0
(M1)
⇒ 4c2 – 8c + 4 = 0 ⇒ (c – 1)2 = 0 ⇒c=1
(M1) (A1)
! ! ! ! ! PR = −3i − j + 3k , PS = −3i + 3k ! ! ! i j k
PS × PR = − 3 0 3 − 3 −1 3 ! ! = 3i + 3k (c)
7
(M1)(M1) (M1) (A1)
4
The parametric equation of a line l which passes through the point (3, 3, 5) and is parallel to the vector PR is given by
! ! ! ! ! ! ! r = (3i + 3 j + 5k ) + t (−3i − j + 3k ) ! ! ! = 3(1 − t )i + (3 − t ) j + (5 + 3t )k (–∞ < t < ∞)
(M1)(M1) (A1)
3
Note: If –∞ < t < ∞ is not mentioned, do not penalise. Also note that some candidates may give the parametric equation of the line in the form x = 3(1 – t), y = 3 – t, z = (5 + 3t), –∞ < t < ∞.
2
(d)
Let P1 and P2 be points on the line l corresponding to t = 0 and t = 1, ! ! ! respectively.! r = x i + yj + zk , Hence, for x = 3(1 – t), y = (3 – t) and z = 5 + 3t. Putting t = 0 and t = 1, we get the coordinates of points P1 and P2 as (3, 3, 5) and (0, 2, 8), respectively. ! ! ! SP1 SP2 SP1 = 2i + 2 j + 3k Vectors !, and ! ! are given by SP2 = −i + j + 6k and . ! ! i j A vector perpendicular to both
SP1
and
SP2
is
SP1 × SP2 = 2
! ! ! 9 i − 15 j + 4 k =
(M1) ! k
2
3
−1 1
6
(M1)
(A1)
Let T(x, y, z) be any point of the plane π. Since S = (1, 1, 2), ! ! ! ! ST = ( x − 1)i + ( y − 1) j + ( z − 2)k is a vector in π. Hence TS ⋅ n = 0 ie 9(x – 1) – 15(y – 1) + 4(z – 2) = 0 ⇒ 9x – 15y + 4z – 2 = 0
(A1)
OR
! ! ! QS = (1 − 3)i + (1 − 3) j + (2 − 5)k
(M1)
The equation of the plane containing the line l and passing through the point S is determined by l and the vector SQ . Hence, the equation is: ⎛ 3 ⎞ ⎛ − 3 ⎞ ⎛ − 2 ⎞ ⎜ ⎟ ! ⎜ ⎟ ⎜ ⎟ r = ⎜ 3 ⎟ + λ ⎜ − 1 ⎟ + µ ⎜ − 2 ⎟ ⎜ 5 ⎟ ⎜ 3 ⎟ ⎜ − 3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(M1) (A1)
4
3
(e)
! PQ ⋅ n Shortest distance ! = n 15 = 322
⎛ − 1⎞ ⎛ 9 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 2 ⎟ . ⎜ − 15 ⎟ ⎜ 6 ⎟ ⎜ 4 ⎟ ⎝ ⎠ ⎝ ⎠ 322
(M1)(M1) (M1)(A1)
OR The distance of P from π is:
9(4) − 15(1) + 4(−1) − 2
=
9 2 + 15 2 + 4 2 15 322
(M1)(A1) (M1)(A1)
4
Note: Accept 0.836 (3 sf) [22]
5.
(a)
An equation of the plane is 2x – y + 3z = 9. ⎛ 2 ⎞ ⎛ 1 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ OR r = ⎜ 1 ⎟ + λ⎜ 2 ⎟ + µ ⎜ 3 ⎟ ⎜ 2 ⎟ ⎜ 0 ⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ OR r = ⎜ − 1⎟ = ⎜ 1 ⎟.⎜ − 1⎟ = 9 ⎜ 3 ⎟ ⎜ 2 ⎟ ⎜ 3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(M1)(A1) (M1)(A1)
(M1)(A1) (C2)
4
(b)
(a, a – 1, a – 2) lies on the plane if 2a – (a – 1) + 3(a – 2) = 9 7 This gives a = 2 . OR (a, a – 1, a – 2) lies on the plane if a = 2 + λ, a – 1 = 1 + 2λ + 3µ and a – 2 = 2 + µ. Thus a – 1 = 1 + 2(a – 2) + 3(a – 4) 1 7 ⇒a=3 or 2 2
⎛ a ⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ OR (a, a – 1, a – 2) lies on the plane if ⎜ a − 1 ⎟.⎜ − 1⎟ = 9 ⎜ a − 2 ⎟ ⎜ 3 ⎟ ⎝ ⎠ ⎝ ⎠ ⇒ 2a – (a – 1) + 3(a – 2) = 9 1 ⇒a=3 2
(M1) (A1)
(M1) (A1)
(M1)
(A1) (C2) [4]
6.
AH = 5 cm, HC = 3 5 cm, AC = 2 13 cm
(A2)
Note: Award (A2) for all 3 correct, (A1) for 2 correct.
AH 2 + CH 2 − AC 2 cos AHˆ C = 2( AH )(CH ) =
(M1)
25 + 45 − 52
30 5 ˆ ie AHC = 74.4° (to the nearest one-tenth of a degree)
(A1) (C4) [4]
5
7.
A(0, 2, 2)
B(t + 5, 2t + 9, 2t + 6) v = i + 2j + 2k
8.
AB . v = 0
(M1)
⎛ t + 5 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ Therefore, ⎜ 2t + 7 ⎟.⎜ 2 ⎟ = 0 ⎜ 2t + 4 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ t + 5 + 4t + 14 + 4t + 8 = 0, giving t = –3 ⎛ 2 ⎞ ⎜ ⎟ Then AB = ⎜ 1 ⎟ , and the required distance AB = 3. ⎜ − 2 ⎟ ⎝ ⎠
(A1) (A1)
(a)
! ! ! ! ! AB = −i − 3 j + k , BC = i + j
! i (b)
(c)
(d)
(e)
! j
(A1) (C4) [4]
(A2)
! k
AB × BC = − 1 − 3 1
(M1)
1 ! ! ! = − i + j + 2k
(A2)
1
2
0
! ! 1 − i + j + 2k 2 1 1+1+ 4 = 2 6 = 2
Area of ΔABC =
! ! ! ! A normal to the plane is given by n = AB × BC = −i + j + 2k Therefore, the equation of the plane is of the form –x + y + 2z = g, and since the plane contains A, then –1 + 2 + 2 = g ⇒ g = 3. Hence, an equation of the plane is –x + y + 2z = 3. ! Vector n above is parallel to the required line. Therefore, x = 2 – t y = –1 + t z = –6 + 2t
3
(M1)
(A1)
2
(M1) (M1) (A1)
3
(M1) (A1)
2 6
(f)
Distance of a point (x0, y0, z0) from a plane ax + by + cz + d = 0 ax0 + by0 + cz 0 + d is given by a2 + b2 + c2 since –x + y + 2z – 3 = 0 and D is (2, –1, –6) − 2 − 1 − 12 − 3 18 = then; distance = 1+1+ 4 6 =3 6
(g)
(h)
! ! 1 ! Unit vector in the direction of n is e = ! × n n ! ! ! 1 ( −i + j + 2 k ) = 6 ! ( − e is also acceptable)
(M1) (M1)
(A1)
3
(M1) (A1)
2
Let H be the intersection of DE with the plane, then –2 + t + (–1 + t) + 2 (–6 + 2t) = 3 ⇒ 6t = 18 ⇒t=3 ⇒ H(–1, 2, 0) but H is the mid point of DE ⇒ E(–4, 5, 6)
(M1) (A1) (M1) (A1)
4 [21]
7
9.
Using Gaussian elimination, with the augmented matrix gives ⎛ 2 − 1 − 9 7 ⎞ ⎜ ⎟ 3 1 ⎟ ⎜ 1 2 ⎜ 2 1 − 3 k ⎟ ⎝ ⎠
⎛ 2 − 1 − 9 7 ⎞ ⎜ ⎟ 2r2 − r1 ⎜ 0 5 15 − 5 ⎟ r3 − r1 ⎜⎝ 0 2 6 k − 7 ⎟⎠
10.
(M1)
7 ⎞ ⎛ 2 − 1 − 9 ⎜ ⎟ − 5 ⎟ ⎜ 0 5 15 5r3 − 2r2 ⎜⎝ 0 0 0 5k − 25 ⎟⎠
(M1)
An infinite number of solutions exist only if 5k – 25 = 0 ⇒ k = 5.
(A1) (C3)
A vector that is normal to the plane is given by the vector product d1 × d2 where d1 and d2 are the direction vectors of the lines L1 and L2 respectively. i j k d1 × d2 = 2 1 − 2
(M2)
0 1 3 = 5i – 6j + 2k (or any multiple)
(A1) (C3)
[3]
[3]
11.
The direction vector, i + 2j + lk, for the line, is perpendicular to 6i – 2j + k, the normal of the plane. Therefore, (i + 2j +lk) . (6i – 2j + k) = 0 Therefore, 6–4+l=0 l = –2
(M1) (M1) (A1) (C3)
OR x = t + 1, y = 2t – 1, z = lt + 3 6t + 6 – 4t + 2 + lt + 3 = 11 2t + lt = 0 l = –2
(A1) (M1) (A1) [3]
8
12.
a .b ab 2 sin θ cos θ = (1)(1)
Method 1: Let the angle be α, then cos α =
= sin 2θ ⎛ π ⎞ − 2θ ⎟ ⎜ = cos ⎝ 2 ⎠
α=
π – 2θ or α = arccos (sin 2θ) 2
(M1)
(M1)
(A1) (C3)
Method 2: y
Q(sin , cos ) y=x v 1 1 u
α
P(cos , sin )
x Q is the image of P under a reflection in y = x α π = θ+ 2 4 π α= – 2θ 2
(M1) (A1) (A1) (C3) [3]
13.
(a)
(b)
Since det A ≠ 0, A–1 exists. Hence AB = C ⇒ B = A–1C
(i)
(ii)
⎛ 1 0 0 ⎞ ⎜ ⎟ DA = ⎜ 0 1 0 ⎟ ⎜ 0 0 1 ⎟ ⎝ ⎠ B = A–1C = DC ⎛ 1 ⎞ ⎜ ⎟ = ⎜ − 1⎟ ⎜ 2 ⎟ ⎝ ⎠
(M1) (C1)
2
(A1)
(M1) (A1)
3
9
(c)
The system of equations is
x + 2y + 3z = 5 2x – y + 2z = 7 3x – 3y + 2z = 10
⎛ x ⎞ ⎜ ⎟ or A ⎜ y ⎟ = C. ⎜ z ⎟ ⎝ ⎠ The required point = (1, –1, 2).
(M1) (A1)
2 [7]
14.
(a)
i j k 2 3 1 3 1 2 u×v= 1 2 3 =i . −j +k −1 2 2 2 2 −1 2 −1 2
(M2)(AG)
2
= 7i + 4j – 5k
(b)
⎛ λ + 2µ ⎞ ⎜ ⎟ w = ⎜ 2λ − µ ⎟ ⎜ 3λ + 2µ ⎟ ⎝ ⎠
(C1)
The line of intersection of the planes is parallel to u × v. (M1) . Now, w (u × v) = 7λ + 14µ + 8λ – 4µ –15λ – 10µ = 0 for all λ, µ. (M1)(C1) Therefore, w is perpendicular to the line of intersection of the given planes. (AG) OR The line of intersection of the planes is perpendicular to u and to v, so it will be perpendicular to the plane containing u and v, that is, to all vectors of the form λu + µv = w.
i 15.
(a)
j
k
The vector product, p × q = 3 2
1
(M2) (R1) (C1)
4 [6]
(M1)
1 3 −2 = –7i + 7j + 7k
(b)
Area of parallelogram = ⏐p × q⏐ =
147 or 7 3 or 12.1 units2
(A1) [3]
10
16.
x + 3 y −1 z −1 = = = λ, then x = 2λ – 3, y = –λ + 1, z = 2λ + 1 2 −1 2 Substituting into P gives; 4λ – 6 – 3λ + 3 – 2λ – 1 = –5 ⇒λ=1
(M1) (A1)
Therefore x = –1, y = 0, z = 3
(A1)
Let
Therefore the point of intersection is (–1, 0, 3) [3]
17.
Using an elimination method, 3x – 2y + z = –4 x – y – z = –2 4x – y = – 6 4x + 6y = 8
(M1)
7y = 14
(A1)
Therefore x = –1, y = 2, z = 3
(M1)
OR Using matrices, ⎛ 3 − 2 1 ⎞⎛ x ⎞ ⎛ − 4 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ 1 1 − 1⎟⎜ y ⎟ = ⎜ − 2 ⎟ ⎜ 2 3 0 ⎟⎠⎜⎝ z ⎟⎠ ⎜⎝ 4 ⎟⎠ ⎝
3 1 ⎞⎛ − 4 ⎞ ⎛ x ⎞ ⎛ 3 ⎜ ⎟ 1 ⎜ ⎟⎜ ⎟ ⎜ y ⎟ = ⎜ − 2 − 2 4 ⎟⎜ − 2 ⎟ ⎜ z ⎟ 14 ⎜ 1 − 13 5 ⎟⎜ 4 ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ (using a graphic display calculator) Therefore x = –1, y = 2, z = 3
(M1)
(A1)
[3]
18.
(a)
i
j
a×b= 2
1
k − 2 = (–1 – 2)i – (–2 + 4)j + (–2 – 2)k
2 −1 −1 = –3i – 2j – 4k
(M1)(A1) (AG)
11
i (b)
j
k
(a × b) × c = − 3 − 2 − 4 = (–4 + 8)i – (–6 + 4)j + (–6 + 2)k
1
2
2
= 4i + 2j – 4k . b c = 2 – 2 – 2 = – 2, and so – (b . c)a = 2a = 4i + 2j – 4k = (a × b) × c
(A1) (A1) (M1)(AG) [5]
19.
(a)
(b)
(c)
20.
OP = OA + OB = 4i – 3k
⇒ P = (4, 0, –3)
(A1)
OQ = OA + OC = 3i + 3j
⇒ Q = (3, 3, 0)
(A1)
OR = OB + OC = 3i + j + k
⇒ R = (3, 1, 1)
(A1)
OS = OP + OC = 5i + 2j – k
⇒ S = (5, 2, 1)
(A1)
OA × OB = –3i – 2j – 4k: (from part (i)). An equation of the plane is 3x + 2y + 4z = 0
(M1)(A1)
V = ⏐(–3i – 2j – 4k).(I + 2j + 2k)⏐ = ⏐–3 – 4 – 8⏐= 0 Notes: Accept alternative forms ⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ − 3 ⎞ ⎛ x ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ eg r = λ ⎜ 1 ⎟ + µ ⎜ − 1⎟ or ⎜ − 2 ⎟ . ⎜ y ⎟ = 0. ⎜ − 2 ⎟ ⎜ − 1⎟ ⎜ 4 ⎟ ⎜ z ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
[8]
For the line of intersection: –4x + y + z = –2 3x – y + 2z = –1 –x + 3z = –3
(M1)
–8x + 2y + 2z = –4 3x – y + 2z = –1 11x – 3y =3
(M1)
The equation of the line of intersection is x =
3y + 3 = 3z + 3 (or equivalent) (A1) (C3) 11
12
OR
⎧ y + z = −2 Let x = 0 ⇒ ⎨ ⎩− y + 2 z = −1 ⇒ 3z = –3, z = –1, y = –1 ⇒ (0, –1, –1)
(M1)
⎧− 4 x + y = −2 Let z = 0 ⇒ ⎨ ⎩ 3 x − y = −1 ⇒ –x = –3, x = 3, y = 10 ⇒ (3, 10, 0)
(M1)
⎛ 0 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ The equation of the line of intersection is r = ⎜ − 1⎟ + λ⎜11⎟ (or equivalent) ⎜ − 1⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠
21.
(A1) (C3) [3]
Let d1 and d2 be the direction vectors of the two lines. Then the normal to the plane is
i
j
d1 × d2 = 1 − 2
k 1
3 −3 5 = –7i – 2j + 3k (or equivalent)
(M1) (A1)
Then equation of the plane is of the form –7x – 2y + 3z = c or r.(–7i – 2j + 3k) = c Using the point (1, 1, 2) which is in the plane gives the equation of the plane –7x – 2y + 3z = –3 or r.(–7i – 2j + 3k) = –3 or ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ r = ⎜ 1 ⎟ + λ⎜ – 2 ⎟ + µ ⎜ − 3 ⎟ (or equivalent) ⎜ 2 ⎟ ⎜ 1 ⎟ ⎜ 5 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(A1) (C3) [3]
13
22.
(a)
Given the points A(–1, 2, 3), B(–1, 3, 5) and C(0, –1, 1), ⎛ 0 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ AB = 1 , AC = ⎜ ⎟ ⎜ − 3 ⎟ then ⎜ 2 ⎟ ⎜ − 2 ⎟ ⎝ ⎠ ⎝ ⎠ and
AB = 5 , AC = 14
The size of the angle between the vectors ⎛ ⎞ ⎜ AB ⋅ AC ⎟ ⎛ − 7 ⎞ ⎟⎟ ⎟ = arccos⎜⎜ θ = arccos ⎜ ⎝ 5 14 ⎠ ⎜ AB AC ⎟ ⎝ ⎠
(A1)
AB
and
AC is given by (M1)
θ = 147° (3 sf) or 2.56 radians
(b)
(c)
(A1)
1 1 AB × AC AB AC sinθ or 2 2 ⎛ 21 ⎞⎟ Area = 2.29 units2 ⎜⎜ accept 2.28, 2.30 and 2 ⎟⎠ ⎝ Area =
(i)
(A1)
4
(M1) (A1)
2
The parametric equations of l1 and l2 are l1: x = 2, l2: x = –1 + µ,
y = –1 + λ, y = 1 – 3µ,
z = 2λ z = 1 – 2µ
(A1) (A1)
Note: At this stage accept answers with the same parameter for both lines. (ii)
To test for a point of intersection we use the system of equations: 2 = –1 + µ 1 –1 + λ = 1 – 3µ 2 2λ = 1 – 2µ 3 (M1) Then µ = 3, λ = –7 from 1 and 2 Substituting into 3 gives RHS = –14, LHS = –5 Therefore the system of equations has no solution and the lines do not intersect.
(A1) (M1)
5
14
(d)
The shortest distance is given by
(e − d ) ⋅ (l1 × l 2 ) ( l1 × l 2 )
where d and e
are the position vectors for the points D and E and where l1 and l2 are the direction vectors for the lines l1 and l2. i j k
2 = 4i + 2 j − k Then l1 × l2 = 0 1 1 −3 −2
(e − d ) ⋅ (l1 × l 2 ) And
( l1 × l 2 )
=
(M1)(A1)
(−3i + 2 j + k ) ⋅ (4i + 2 j − k ) 21
(M2)
9 =
21 or 1.96
(A1)
5 [16]
23.
The direction of the line is v = 2i – 2j + k and ⎪v⎪= 3. Therefore, the position vector of any point on the line 6 units from A is 3i – 2k ± 2v = 7i – 4j or –i + 4j – 4k, giving the point (7, –4, 0) or (–1, 4, –4).
(A1) (M1) (A1) (C3) [3]
24.
(a)
⎛ 1 3 − 2 ⎞⎛ x ⎞ ⎛ − 6 ⎞ ⎜ ⎟⎜ ⎟ ⎜ ⎟ The system is ⎜ 2 1 3 ⎟⎜ y ⎟ = ⎜ 7 ⎟ ⎜ 3 − 1 1 ⎟⎜ z ⎟ ⎜ 6 ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ −1
⎛ x ⎞ ⎛ 1 3 − 2 ⎞ ⎛ − 6 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⇒ ⎜ y ⎟ = ⎜ 2 1 3 ⎟ ⎜ 7 ⎟ = ⎜ − 1⎟ . ⎜ z ⎟ ⎜ 3 − 1 1 ⎟ ⎜ 6 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Therefore, the solution is x = 1, y = –1, z = 2.
(M1) (G2)
15
OR The system of equations is: 1 2 3
3 1 –1
–2 3 1
–6 7 6
R2 ← R2 – 2R1 R3 ← R3 – 3R1
1 0 0
3 –5 –10
–2 7 7
–6 19 24
R3 ← 2R3 – R2
1 0 0
3 –5 0
–2 7 –7
–6 19 –14
Back substitution gives x = 1, y = –1, z = 2.
(M2) (C1)
OR x = 1, y = –1, z = 2.
i (b)
j
v= 1 3 −2 =
2 1
(c)
k 3
3 −2 1
3
i−
1 −2 2
3
j+
1 3 2 1
k = 11i – 7j – 5k.
u = m(i + 3j – 2k) + n(2i + j + 3k) = (m + 2n)i + (3m + n)j + (–2m + 3n)k Therefore, v . u = 11(m + 2n) – 7(3m + n) – 5(–2m + 3n) = 11m + 22n – 21m – 7n + 10m – 15n = 0, for all m and n. That is, v is perpendicular to u for all values of m and n.
(G3)
3
(M1)(C2)
3
(C1) (M1)(C1) (AG)
OR v is perpendicular to both a and b [from part (b)]. Therefore, v . a = v . b = 0, so v . u = m(v . a) + n(v . b) = 0, and hence v is perpendicular to u for all values of m and n.
(M1) (M1)(R1)(AG)
3
16
(d)
The normal to the plane, 3i – j + k, and v are both perpendicular to the required line, l. Therefore, the direction of l is given by i j k − 7 − 5 11 − 5 11 − 7 v × (3i – j + k) = 11 − 7 − 5 = − 1 1 i − 3 1 j + 3 − 1 k 3 −1 1 = –12i – 26j + 10k Thus, an equation for l is r = i – j + 2k + λ(6i + 13j – 5k), where λ is a scalar. [Any form of the correct answer is quite acceptable.]
(M1)(C2)
(C1)
4 [13]
25.
v . w = 2 + 3 +2 = 7 ⏐v⏐ = ⏐w⏐ =
(A1)
6 14
(A1) (A1)
⎛ v ⋅ w ⎞ ⎟ θ = arccos ⎜⎜ ⎟ v w ⎝ ⎠
⎛
(M1)
7
⎞ ⎟ ⎝ 6 14 ⎠
= arccos ⎜
(A1)
= 0.702 radians.
(A1) [6]
26.
Solve 1 + λ = 1 + 2µ, 1 + 2λ = 4 + µ, 1 + 3λ = 5 + 2µ Solving, λ = 2, (or µ = 1). P has position vector 3i + 5j + 7k.
(M1)(A1) (A1) (A1)(A1)(A1) (C6) [6]
27.
(a)
(i)
AB = –j + 3k and AC = i + 5k i
j
k
AB × AC = 0 – 1 3 1 0 5 = –5i + 3j + k
(A1)(A1)
(M1) (A1)
17
1 × ⏐–5i + 3j + k⏐ 2 35 = 2 (accept 2.96)
(ii)
(b)
Area =
(M1) (A1)
(i)
The equation of the plane Π is (x, y, z).(–5, 3, 1) = c, that is, – 5x + 3y + z = c (M1)(A1) where c = –5 + 9 + 1 = 5, that is, – 5x + 3y + z = 5 (A1) Note: Award (M1)(A1)(A0) if answer not given in Cartesian form.
(ii)
Equations of L are
x–5 y+2 = =z–1 –5 3 Note: form. (c)
(d)
(M1)(A1)
5
Award (M1)(A0) if answer not given in Cartesian
L meets Π where –5(5 – 5λ) + 3(3λ – 2) + λ + 1 = 5 ⇒λ=1 Point of intersection is (0, 1, 2). (Accept j + 2k.)
Perpendicular distance is =
6
(M1) (A1) (A1)
52 + 32 + 12
3
(M1)
35 (Accept ± 35 or ± 5.92)
(A1)
2 [16]
28.
METHOD 1 R 1:
1
2
–3
2
R 2:
2
3
–5
3
R 1:
1
2
–3
2
R2 – 2R1:
0
–1
1
–1 (M2)(A1)
Let z = t, then y = t + 1 and x = t. Therefore the line of intersection is x = t, y = t + 1, z = t (or equivalent). (A1)(A1)(A1)(C6)
18
METHOD 2 Let z = 0 => x + 2y = 2 2x + 3y = 3 =>x = 0, y = 1 The direction vector of the line of intersection is i
j
k
1
2
–3
2
3
–5
(M1) (A1)
(M1) (A1)
=–i–j–k
⎛ 0 ⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ Therefore the line of intersection is r = ⎜ 1 ⎟ + t ⎜1⎟ (or equivalent) ⎜ 0 ⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠
(A2) (C6) [6]
29.
METHOD 1 Let a + b and a – b be diagonals of a parallelogram ABCD with sides AB and AD equal to a and b respectively. (M1)
B
C a–b
a
A
b
a+b
D
If ⏐a + b⏐ = ⏐a – b⏐ then the diagonals AC and BD are equal in length. (M1)(A1) Therefore ABCD is a rectangle and a . b = 0. (R1)(A1) (C6) METHOD 2 ⏐a + b⏐ = ⏐a – b⏐ => ⏐a + b⏐2 = ⏐a – b⏐2 => (a + b) . (a + b) = (a – b) . (a – b) => ⏐a⏐2 + 2a . b + ⏐b⏐2 = ⏐a⏐2 – 2a . b + ⏐b⏐2 => 4a . b = 0 => a.b=0
(M1) (M1)(A1) (A1) (A1) (A1) (C6) [6]
30.
(a)
⎛ 5 ⎞ ⎜ ⎟ AB = ⎜ – 10 ⎟ . Direction vector of line is 1: –2: 5. ⎜ 25 ⎟ ⎝ ⎠ (Accept any multiple of 1: –2: 5) Therefore the equation of l in parametric form is
(M1)
19
x = λ + l, y = –2λ + 3, z = 5λ – 17 (A1)(A1)(A1) (or x = λ + 6, y = –2λ – 7, z = 5λ + 8, or any equivalent parametric form) (b)
4
P on l => P can be written as (p + 1, –2p + 3, 5p – 17).
⎛ p + 1 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ OP ⊥ l => ⎜ – 2 p + 3 ⎟ ⋅ ⎜ – 2 ⎟ = 0 ⎜ 5 p – 17 ⎟ ⎜ 5 ⎟ ⎝ ⎠ ⎝ ⎠ p + 1 + 4p – 6 + 25p – 85 = 0 30p = 90 => p = 3 Therefore P is (4, –3, –2)
(M1) (A1) (A1)
3 [7]
31.
METHOD 1
a×b=
i
j
1
2 –1
–3 2
k (M1)(A1)
2
= i(4 + 2) + j(3 – 2) + k(2 + 6) = 6i + j + 8k (6i + j + 8k) ⋅ (2i – 3j + 4k) = 41
(M1) (A1) (M1)(A1) (C6)
METHOD 2
1 (a × b) ⋅ c = – 3
2
2
–1
2
2
–3
4
= 1(8 + 6) – 2(–12 – 4) –1(9 – 4) = 14 + 32 – 5 = 41
(M1)(A1) (A1)(A1)(A1) (A1) (C6) [6]
20
32.
⎛ x ⎞ ⎛ 1 ⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ Equation of line is ⎜ y ⎟ = ⎜ 1 ⎟ + λ ⎜ 1 ⎟ ⎜ z ⎟ ⎜ 9 ⎟ ⎜ – 1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Coordinates of foot satisfy 2(1 + 2λ) + (1 + λ) – (9 – λ) = 6 ⇒ 6λ = 12 λ=2 Foot of perpendicular is (5, 3, 7)
(M1)(A1)
(M1)(A1) (A1) (A1) (C2)(C2)(C2) [6]
33.
(a)
(i)
Using row reduction, x + 2y + z = k –3y + 2z = 6 – 2k 6y – 4z = k – 9
(M1) (A1) (A1)
x + 2y + z = k –3y + 2z = 6 – 2k 0z = 3 – 3k Not a unique solution because the coefficient of z in the third equation is zero. (ii)
In order for the system to have a solution, 3 – 3k = 0, consistent for k = 1.
(i)
Consider
(R1) (M1) (A1)
OR
⎛ 1 2 1 ⎞ ⎜ ⎟ ⎜ 2 1 4 ⎟ = 1 × 21 – 2 × 6 + 1 × –9 ⎜ 1 – 4 5 ⎟ ⎝ ⎠
(M1)
=0 (A1) The zero value confirms that the equations do not have a unique solution. (R1) (ii)
Consider
⎛ 1 2 k ⎞ ⎜ ⎟ ⎜ 2 1 4 ⎟ = 9 – 9k ⎜ 1 – 4 5 ⎟ ⎝ ⎠ Consistent when this determinant is zero, ie k = 1.
(M1)(A1) (R1)
6
21
(b)
The general solution is z = λ, y =
(2λ – 4) , x = (11 – 7λ ) . 3
3
(M1)(A1)(A1)
3 [9]
34.
METHOD 1
⎛ 0 ⎞ ⎛ 5 ⎞ ⎜ ⎟ ⎜ ⎟ (A1)(A1) BA = ⎜ − 3 ⎟, BC = ⎜ 0 ⎟ ⎜ − 4 ⎟ ⎜ − 12 ⎟ ⎝ ⎠ ⎝ ⎠ Note: Award (A1), (A1) for any two correct vectors used to find area. j k ⎞ ⎛ i ⎜ ⎟ BA × BC = ⎜ 0 − 3 − 4 ⎟ ⎜ 5 0 − 12 ⎟ ⎝ ⎠ = 36i – 20j + 15k 1 1 BA × BC = 36 2 + 20 2 + 15 2 Area = 2 2 1 1921 = 2 = 21.9
(M1) (A1)
(M1) (A1) (C6)
METHOD 2
⎛ 0 ⎞ ⎛ 5 ⎞ ⎜ ⎟ ⎜ ⎟ BA = ⎜ − 3 ⎟, BC = ⎜ 0 ⎟ ⎜ − 4 ⎟ ⎜ − 12 ⎟ ⎝ ⎠ ⎝ ⎠
(A1)(A1)
BA = 5
(A1)(A1)
BC = 13
1 ⎛ ⎛ 48 ⎞ ⎞ × 5 × 13 sin ⎜⎜ cos −1 ⎜ ⎟ ⎟⎟ 2 ⎝ 65 ⎠ ⎠ ⎝ = 21. 9
Area =
(M1) (A1) (C6) [6]
22
35.
⎛ 0 ⎞ ⎜ ⎟ z-axis has direction vector ⎜ 0 ⎟ ⎜ 1 ⎟ ⎝ ⎠ Let θ equal the angle between the line and the normal to the plane. ⎛ 0 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 0 ⎟ . ⎜ − 2 ⎟ ⎜ 1 ⎟ ⎜ 4 ⎟ ⎝ ⎠ ⎝ ⎠ cos θ = 1 32 + 2 2 + 4 2 4 cos θ = 29 θ = 42° The angle between the line and the plane is (90° – θ). The angle is 48°.
(A1)
(M1) (A1) (A1) (M1) (A1) (C6) [6]
36.
(a)
⎛1⎞ ⎜ ⎟ n = ⎜1⎟ , hence equation of L through A(2, 5, –1) is given ⎜1⎟ ⎝ ⎠ by
(b)
x − 2 y − 5 z +1 = = . 1 1 1
A general point on L is (2 + λ, 5 + λ, –1 + λ). At intersection of line L and the plane (2 + λ) + (5 + λ) + (–1 + λ) –1 = 0 ⇒ 3λ = –5 5 ⇒λ=– 3 ⎛ 1 10 8 ⎞ ⇒ point of intersection ⎜ , , − ⎟ 3 ⎠ ⎝ 3 3
(M1)(A1)
2
(A1) (M1) (A1) (A1)
4
23
(c)
=0
A
Plane – –5 3 – 10 — A' 3 Let Aʹ′(x, y, z) be the reflection of A. Note: Diagram does not have to be given. EITHER
10 3 ⎛ 4 5 13 ⎞ ⇒ A′ = ⎜ − , , − ⎟ 3 ⎠ ⎝ 3 3 At A′ λ = –
(M1) (A1)
OR Since point of intersection of L and the plane is midpoint of AA′ ⎛ 1 ⎞ ⎜ ⎟ ⎛ 2 ⎞ ⎛ x ⎞ ⎜ 3 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 10 ⎟ ⎜ 5 ⎟ + ⎜ y ⎟ = 2⎜ ⎟ ⎜ − 1⎟ ⎜ z ⎟ ⎜ 3 ⎟ 8 ⎝ ⎠ ⎝ ⎠ ⎜ − ⎟ ⎝ 3 ⎠ ⎛ 4 5 13 ⎞ ⇒ A′ = ⎜ − , ,− ⎟ ⎝ 3 3 3 ⎠
(M1)
(A1)
2
24
(d)
A X
B(2, 0, 6)
Plane
Let X be foot of perpendicular from B to L ⇒ d = BX
BX = OX − OB
⎛ 2 + λ ⎞ ⎛ 2 ⎞ ⎛ λ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎜ 5 + λ ⎟ − ⎜ 0 ⎟ = ⎜ 5 + λ ⎟ ⎜ − 1 + λ ⎟ ⎜ 6 ⎟ ⎜ − 7 + λ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(A1)
Now BX ⋅ n = 0
(M1)
⇒ λ + (5 + λ ) + (−7 + λ ) = 0
⇒ 3λ = 2 ⎛ 2 ⎞ ⎜ ⎟ ⎜ 3 ⎟ 2 17 ⎟ ⇒ λ = ⇒ BX = ⎜ ⎜ 3 3 ⎟ ⎜ 19 ⎟ ⎜ − ⎟ ⎝ 3 ⎠
Hence d =
(A1)
4 289 361 + + 9 9 9 =
654 = 8.52 (units) 3
(A1)
25
OR
AB × n d= n i
⎛ 0 ⎞ ⎜ ⎟ where AB = ⎜ − 5 ⎟ ⎜ 7 ⎟ ⎝ ⎠ j
k
AB × n = 0 − 5 7 = − 12i + 7 j + 5k 1
⇒d =
1
(M1)(A1)
(A1)
1
144 + 49 + 25 3
= 8.52 (units)
(A1)
4 [12]
37.
(a)
(b)
38.
3x + y + z = 1 I 2x + y – z = 4 II 5x + y + bz = 1 III Solving for z III – II also II – I 3 × V + IV
⇒ 3x + bz + z = –3 IV ⇒ –x – 2z = 3 V ⇒ bz – 5z = 6 6 ⇒z= b−5
If b = 5, z is undefined. Hence equation has unique solution if b ≠ 5.
x = 1 + µ, y = –µ, z = 1 + 2µ 2(1 + µ) + µ + 1 + 2µ + 2 = 0 µ = –1 P is (0, 1, –1) (Accept any form of notation, including vectors.)
(M1) (M1) (A1) (A1)
4
(R1) (A1)
2 [6]
(M1)(A1) (M1)(A1) (A1) (A1) (C6) [6]
39.
(a)
(b)
⎛ x ⎞ ⎛ 3 ⎞ ⎛ 1 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ y ⎟ • ⎜ − 4 ⎟ = ⎜ 2 ⎟ • ⎜ − 4 ⎟ ⎜ z ⎟ ⎜ 1 ⎟ ⎜11⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 3x – 4y + z = 6 (i)
1 + 3 × 2 – 11 = –4 ⇒ P lies in π2
(M1) (A1)
2
(A1) (AG)
26
(ii)
⎛ 3 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ − 4 ⎟ × ⎜ 3 ⎟ = ⎜ 4 ⎟ ⎜ 1 ⎟ ⎜ − 1⎟ ⎜13⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ 2 + λ ⎜ 4 ⎟ r = ⎜ ⎟ ⎜11⎟ ⎜13⎟ ⎝ ⎠ ⎝ ⎠
(M1)(A1)
(M1)(A1)
5
Note: Award (M1)(A0) if equation given in incorrect form. (c)
METHOD 1 ⎛ 1 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ • ⎜ − 4 ⎟ = ⎜ − 1⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠
⎛ 1 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ ⎜ − 4 ⎟ cos θ ⎜ − 1⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠
⎛ 1 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ • ⎜ − 4 ⎟ = – 10 ⎜ − 1⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 1 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ = 11 ⎜ − 4 ⎟ = 26 ⎜ − 1⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ − 10 cos θ = (= –0.5913) 11 26 θ = 2.2035 radians (or θ = 126.3°) The angle between the planes is π – 2.2035 = 0.938 radians (or 180° – 126.3° = 53.7°)
(A1)
(A1)
(M1) (A1) (A1) (N2)
METHOD 2
⎛ 1 ⎞ ⎛ 3 ⎞ ⎛ 1 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ × ⎜ − 4 ⎟ = ⎜ 3 ⎟ ⎜ − 4 ⎟ sin θ ⎜ − 1⎟ ⎜ 1 ⎟ ⎜ − 1⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(M1)
⎛ 1 ⎞ ⎛ 3 ⎞ 1 ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ × ⎜ − 4 ⎟ = 4 = 186 ⎜ − 1⎟ ⎜ 1 ⎟ 13 ⎝ ⎠ ⎝ ⎠
(A1)
⎛ 1 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ = 11, ⎜ − 4 ⎟ = 26 ⎜ − 1⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠
(A1)
sin θ =
186
(= 0.8064 ) 11 26 θ = 0.938 radians (or θ = 53.7)
(M1) (A1) (N2)
5 [12]
27
40.
⎛ x ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ Equation of (AB) is ⎜ y ⎟ = ⎜ 4 ⎟ + λ ⎜ 1 ⎟ ⎜ z ⎟ ⎜ – 1⎟ ⎜ – 1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(A1)
⎛ x ⎞ ⎛ 5 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ and of (CD) is ⎜ y ⎟ = ⎜ 6 ⎟ + µ ⎜ 2 ⎟ ⎜ z ⎟ ⎜ 3 ⎟ ⎜1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(A1)
at point of intersection of two lines 1 + λ = 5 + 3µ 4 + λ = 6 + 2µ –1 – λ = 3 + µ
(M1)
solving simultaneously any two of these three equations gives λ = –2 and µ = –2 (only one value required).
(A2)
⇒ point of intersection (–1, 2, 1) (A1) (C6) Note: Since question states that lines intersect, there is no need to check the solution in the third equation. [6]
41.
(a) (b)
M
(3µ – 2, µ, 9 – 2µ)
(A1)
(i)
⎛ 4 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ x – 4 y z +3 = = or r = ⎜ 0 ⎟ + λ ⎜ 1 ⎟ 3 1 –2 ⎜ – 3⎟ ⎜ – 2 ⎟ ⎝ ⎠ ⎝ ⎠
(ii)
⎛ 3µ – 2 – 4 ⎞ ⎜ ⎟ PM = ⎜ µ ⎟ ⎜ 9 – 2µ + 3⎟ ⎝ ⎠
(M1)
⎛ 3µ – 6 ⎞ ⎜ ⎟ = ⎜ µ ⎟ ⎜12 – 2µ ⎟ ⎝ ⎠
(A1)
1
(M1)(A1)
4
28
(c)
⎛ 3µ – 6 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ µ ⎟ ⋅ ⎜ 1 ⎟ = 0 ⎜12 – 2µ ⎟ ⎜ – 2 ⎟ ⎝ ⎠ ⎝ ⎠
(i)
(M1)
9µ – 18 + µ – 24 + 4µ = 0 µ=3 (ii)
⎛ 3 ⎞ ⎜ ⎟ PM = ⎜ 3 ⎟ ⎜ 6 ⎟ ⎝ ⎠
(A1)
PM = 32 + 32 + 6 2
(M1)
= 3 6 (accept i (d)
(A1)
j
n= 3 3
54 or 7.35)
(A1)
5
k 6 = –12i + 24j – 6k
(M1)(A1)
3 1 −2 = –6(2i – 4j + k)
⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ 4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ – 4 ⎟ • r = ⎜ – 4 ⎟ • ⎜ 0 ⎟ ⎜ 1⎟ ⎜ 1 ⎟ ⎜ –3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(M1)
2x – 4y + z = 5 (e)
(A1)
4
EITHER l1is on π1 from part (d). Testing l1 on π2 gives (3µ – 2) –5(µ) – (9 – 2µ) = –11. Therefore l1 is also on π2 and is therefore the line of intersection.
(M1) (R1)
OR 2x – 4y + z = 5 x – 5 y – z = – 11 3x – 9 y = – 6 x – 3y = –2 If y = λ, x = –2 + 3λ, z = –2λ + 9
or
(M1)
x+2 y z –9 = = which is l1. 3 1 –2
(A1)
2 [16]
42.
(a)
⎛ 4 ⎞ ⎜ ⎟ Finding correct vectors AB = ⎜ 3 ⎟ ⎜ − 1⎟ ⎝ ⎠
⎛ − 3 ⎞ ⎜ ⎟ AC = ⎜ 1 ⎟ ⎜ 1 ⎟ ⎝ ⎠
Substituting correctly in scalar product AB ⋅ AC = 4(–3) + 3(1) – 1(1) = –10
A1A1
A1 AG
N0
3 29
(b)
AB = 26
AC = 11
⌢ Attempting to use scalar product formula, cos BAC =
(A1)(A1)
− 10 26 11
= –0.591 (to 3 sf)
M1 AG
3 [6]
43.
(a)
⎛ 0.1 0.4 0.1⎞ ⎜ ⎟ A = ⎜ − 0.7 0.2 0.3 ⎟ ⎜ − 1.2 0.2 0.8 ⎟ ⎝ ⎠
A2
(b)
⎛ x ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ −1 For attempting to calculate ⎜ y ⎟ = A ⎜ 2 ⎟ ⎜ z ⎟ ⎜ 3 ⎟ ⎝ ⎠ ⎝ ⎠
M1
x = 1.2, y = 0.6, z = 1.6 (So the point is (1.2, 0.6, 1.6)) (1.2, 0.6, 1.6) lies on x + y + z = d ∴ d = 3.4.
(c)
44.
(a)
(b)
–1
L1 : x = 2 + λ; y = 2 + 3λ; z = 3 + λ L2 : x = 2 + µ; y = 3 + 4µ; z = 4 + 2µ At the point of intersection 2+λ=2+µ (1) 2 + 3λ = 3 + 4µ (2) 3 + λ = 4 + 2µ (3) From (1), λ = µ. Substituting in (2), 2 + 3λ = 3 + 4λ ⇒ λ = µ = –1. We need to show that these values satisfy (3). They do because LHS = RHS = 2; therefore the lines intersect. So P is (l, –1, 2). The normal to Π is normal to both lines. It is therefore given by the vector product of the two direction vectors. ⎛ i j k ⎞ ⎜ ⎟ Therefore, normal vector is given by ⎜1 3 1 ⎟ ⎜1 4 2 ⎟ ⎝ ⎠ = 2i – j + k The Cartesian equation of ∏ is 2x – y + z = 2 + 1 + 2 ie 2x – y + z = 5
N2
2
A2
N2
3
A1
N1
1 [6]
(A1) (A1) (M1)
A1 A1 M1 R1 A1
N3
8
N2
6
M1A1 A2 (M1) A1
30
(c)
The midpoint M of [PQ] is (2, 3/2, 5/2).
M1A1
The direction of MS is the same as the normal to ∏, ie 2i – j + k
(R1)
The coordinates of a general point R on MS are therefore ⎛⎜ 2 + 2λ , 3 − λ , 5 + λ ⎞⎟ 2 2 ⎝ ⎠ It follows that PR = (1 + 2λ)i + ⎛⎜ 5 − λ ⎞⎟ j + ⎛⎜ 1 + λ ⎞⎟ k ⎝ 2 ⎠ ⎝ 2 ⎠
(M1)
A1A1A1
At S, length of PR is 3, ie (1 + 2λ)2 + (5/2 – λ)2 + (1/2 + λ)2 = 9 1 + 4λ + 4λ2 + 25 / 4 – 5λ + λ2 + 1 / 4 + λ + λ2 = 9 6 6λ2 = 4 1 λ=± 2 Substituting these values,
(M1) A1 (A1)
the possible positions of S are (3, 1, 3) and (1, 2, 2)
A1A1
A1 A1 (M1) N2
15 [29]
→
45.
(a)
→
ˆ = OP ⋅ OQ cos POQ → → ⎜OP ⎟⎜ OQ⎟
=
(M1)
− 6 ⎛ 6 ⎞ (A1)(A1) ⎜ = − ⎟ 14 24 ⎝ 336 ⎠ Note: Award (A1) for scalar product, and (A1) for correct values of magnitudes.
ˆ = 109o (1.90 radians) POQ
(A1) (C4)
31
(b)
METHOD 1 area ΔPOQ =
1 → → ˆ ⎜PO⎟ ⎜OQ⎟ sin POQ 2
(M1)
= 8.66 (Accept 8.67)
(A1) (C2)
METHOD 2 area ΔPOQ =
=
→ 1 → ⎜OP × OQ⎟ 2
(M1)
1 10i + 10 j + 10k = 75 2
(= 5 3 )
(A1) (C2) [6]
46.
(a)
(i)
For points which lie in π1 and π2 (1)
2 − 2λ + µ = 2 + s + t
(2)
1 + λ − 3µ = 2s + t
(3)
1 + 8λ − 9µ = 1 + s + t
(M1)(A1)
subtracting (3) from (1)
(ii)
1 − 10λ + 10µ = 1
(M1)
⇒λ = µ
(AG)
On the line of intersection λ = µ
⎛ 2 ⎞ ⎛ − 2 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⇒ an equation of the line is r = ⎜ 1 ⎟ + λ ⎜ 1 ⎟ + λ ⎜ − 3 ⎟ ⎜ 1 ⎟ ⎜ 8 ⎟ ⎜ − 9 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 2 ⎞ ⎛ − 1 ⎞ ⎜ ⎟ ⎜ ⎟ = ⎜ 1 ⎟ + λ ⎜ − 2 ⎟ ⎜ 1 ⎟ ⎜ − 1 ⎟ ⎝ ⎠ ⎝ ⎠ (b)
The plane π3 contains, eg the point (2, 0, –1).
⎛ x ⎞ ⎛ 3 ⎞ ⎛ 2 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ The equation of the plane is ⎜ y ⎟ • ⎜ – 2 ⎟ = ⎜ 0 ⎟ • ⎜ – 2 ⎟ = 5. ⎜ z ⎟ ⎜ 1 ⎟ ⎜ – 1⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ The cartesian equation of the plane is 3x − 2 y + z = 5 .
(c)
N0
(M1)
(A1)
N1
5
N1
4
(A1)
(M1)(A1)
(A1)
⎛ 2 ⎞ ⎛ −1 ⎞ ⎜ ⎟ Intersection between line r = 1 + λ ⎜ −2 ⎟ and π3 . ⎜ ⎟ ⎜ ⎟ ⎜ 1 ⎟ ⎜ −1 ⎟ ⎝ ⎠ ⎝ ⎠
3x − 2 y + z = 5 ⇒ 3(2 − λ) − 2(1 − 2λ) + 1 − λ = 5
(M1)(A1) 32
This equation is satisfied by any real value of λ ⇒ the 3 planes
⎛ 2 ⎞ ⎛ − 1 ⎞ ⎜ ⎟ ⎜ ⎟ intersect at the line r = ⎜ 1 ⎟ + λ ⎜ − 2 ⎟ . ⎜ 1 ⎟ ⎜ − 1 ⎟ ⎝ ⎠ ⎝ ⎠
47.
(R1)
N1
3 [12]
A(3, 2, 0), B(7, − 1, − 1), C(10, − 3, 0), D(6, 0, 1)
⎛ 7 ⎞ ⎛ 3 ⎞ ⎛ 4 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ Two bounding vectors are AB = ⎜ − 1⎟ − ⎜ 2 ⎟ = ⎜ − 3 ⎟ and ⎜ − 1⎟ ⎜ 0 ⎟ ⎜ − 1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(A1)
⎛ 6 ⎞ ⎛ 3 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AD = ⎜ 0 ⎟ − ⎜ 2 ⎟ = ⎜ − 2 ⎟ ⎜ 1 ⎟ ⎜ 0 ⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(A1)
j k ⎞ ⎛ i ⎜ ⎟ Area of parallelogram = ⎜ 4 − 3 − 1⎟ ⎜ 3 − 2 1 ⎟ ⎝ ⎠
(M1)
→
→
= =
(− 3 − 2)i − (4 + 3) j − (− 8 + 9)k = (− 5)2 + (− 7 )2 + (− 1)2 75 (accept 5 3 or 8.66)
(M1)(A1) (A1) (C6) [6]
33
48.
⎛ 1 ⎞ ⎜ ⎟ Line direction is ⎜ 2 ⎟ and plane normal is ⎜ 2 ⎟ ⎝ ⎠
⎛1⎞ ⎜ ⎟ ⎜1⎟ ⎜1⎟ ⎝ ⎠
(M1)
Angle θ between these two given by
cos θ = =
⎛ 1 ⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 2 ⎟ • ⎜1⎟ ⎜ 2 ⎟ ⎜1⎟ ⎝ ⎠ ⎝ ⎠ 12 + 2 2 + 2 2 12 + 12 + 12
(M1)(A1)
5 3 3
θ = arccos
5 3 3
θ = 0.276 radians Angle between line and plane is
(A1)
π − 0.276 =1.295 radians 2
Angle between line and its reflection is 1.295 × 2 = 2.59 radians Note: Do not award the final (A1) if the answer is given in degrees, 148° .
(A1) (A1) (C6)
[6]
34
49.
(a)
⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ A vector in the plane is ⎜ 1 ⎟ − ⎜ − 2 ⎟ = ⎜ 3 ⎟ ⎜ 5 ⎟ ⎜ 3 ⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(M1)(A1) (N1)
⎛ 0 ⎞ ⎛ 2 ⎞ ⎛ 12 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ Normal vector to plane is ⎜ 3 ⎟ × ⎜ 3 ⎟ = ⎜ 4 ⎟ ⎜ 2 ⎟ ⎜ 6 ⎟ ⎜ − 6 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(M1)(A1) (N1)
⎛ 6 ⎞ ⎛ 1 ⎞ ⎛ 6 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ Equation of plane is r • ⎜ 2 ⎟ = ⎜ − 2 ⎟ • ⎜ 2 ⎟ ⎜ − 3 ⎟ ⎜ 3 ⎟ ⎜ − 3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 6 ⎞ ⎜ ⎟ r • ⎜ 2 ⎟ = 6 − 4 − 9 ⎜ − 3 ⎟ ⎝ ⎠ ⎛ 6 ⎞ ⎜ ⎟ r • ⎜ 2 ⎟ = − 7 ⎜ − 3 ⎟ ⎝ ⎠ ⇒ 6x + 2y − 3z = − 7
(M1)(A1)
(A1)
(AG) (N0)
35
(b)
METHOD 1 Any point P on normal from origin O to plane is (6k, 2k, − 3k) 2
Distance OP = k 6 2 + 2 2 + (− 3) = 7k
(M1) (A1)
P lies on plane 6(6k) + 2(2k) − 3(− 3k) = − 7 36k + 4k + 9k = − 7 k= − Distance = 7 × −
1 7
(A1)
1 =1 7
(A1) (N2)
METHOD 2
ax 0 + by 0 + cz 0 + d
Using distance =
a2 + b2 + c2
(M1)
(x0 , y0, z0) is (0, 0, 0)
−7
distance =
2
6 2 + 2 2 + (− 3)
Note:
Award (A1) for the numerator, (A1) for the denominator.
7
distance =
(A1)(A1)
49
=1
(A1) (N2) [11]
50.
(a)
a×b=
i
j
k
2
1
0
(M1)(A1)
−1 p 6 = 6i − 12j + (2p + 1)k
A1
N3
36
(b)
METHOD 1 a × b parallel to c ⇒
⎛ 6 ⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ − 12 ⎟ = k ⎜ − 4 ⎟ ⎜ 2 p + 1⎟ ⎜ 3 ⎟ ⎝ ⎠ ⎝ ⎠ ⇒
M1
k=3
(A1)
p=4
A1
N2
METHOD 2 b perpendicular to c ⇒
⎛ − 1⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ p ⎟ • ⎜ − 4 ⎟ = 0 ⎜ 6 ⎟ ⎜ 3 ⎟ ⎝ ⎠ ⎝ ⎠
M1
− 2 − 4p + 18 = 0
(A1)
p=4
A1
⇒
51.
(a)
(i)
[6]
⎛ − 1⎞ ⎜ ⎟ BA = ⎜ − 1⎟ ⎜ − 1⎟ ⎝ ⎠
⎛ − 2 ⎞ ⎜ ⎟ BC = ⎜ m ⎟ ⎜ 1 ⎟ ⎝ ⎠
→
→
→
→
→
→
(A1)
A1
BA• BC =1 − m (ii)
N2
→
→
ˆC BA• BC = BA BC cos AB 1− m = 3 5 + m 2 ×
N2
(M1)
2 3
A1
EITHER
(1 − m)2 = 2 (5 + m 2 ) 3
(⇒ m
2
− 6m − 7 = 0
)
M1
Solving (gives m = −1 or m = 7)
A1
⇒m=−1
AG
N0
M1 A1 AG
N0
OR For a correct GDC sketch Showing m = − 1 is the point of intersection ⇒m=−1
37
(b)
Using vector product
i
j
(M1)
k
n = −1 −1 −1 = − 2 i + 3 j − k − 2 −1
A1
1
Substituting coordinates of a point (eg A(2, −1, 0)) ⇒ − 2(x − 2) + 3(y + 1) − z = 0 (−2x + 3y − z = −7) (c)
(M1) A1
N2
METHOD 1
2 3
ˆC = cos AB
7 ˆ ( B = 61.87° ..., 1.0799 radians) 3
ˆC = ⇒ sin AB area ABC =
=
→ 1 → ˆC BA BC sin AB 2
1 7 14 3 6 = 2 3 2
A1 (M1)
(=1.87)
A1
N2
METHOD 2 Area ABC =
1 → → BA × BC 2
(M1)
⎛ − 2 ⎞ 1 ⎜ ⎟ = ⎜ 3 ⎟ 2 ⎜ ⎟ ⎝ − 1 ⎠
A1
14 2
A1
=
(=1.87)
N2
38
(d)
(i)
line perpendicular to plane ABC ⇒ line parallel to n
⎛ 2 ⎞ ⎛ − 2 ⎞ ⎜ ⎟ ⎜ ⎟ equation of line is r = ⎜ − 1⎟ + λ ⎜ 3 ⎟ ⎜ 0 ⎟ ⎜ − 1 ⎟ ⎝ ⎠ ⎝ ⎠ (ii)
(M1) M1A1A1
⎛ 4 ⎞ ⎜ ⎟ AD = ⎜ − 6 ⎟ ⎜ 2 ⎟ ⎝ ⎠ →
(A1)
volume of pyramid =
=
1 area ABC 3
→
AD
(M1)
1 14 56 (FT from any 3 2 →
=
numerical error in AD )
A1
14 ( = 4.67, accept 4.66) 3
A1
N2 [21]
52.
The direction vectors of L1 and L2 are
⎛ 2 ⎞ ⎛ − 1⎞ ⎜ ⎟ ⎜ ⎟ L1 = ⎜ 3 ⎟ , L2 = ⎜ 4 ⎟ ⎜ − 1⎟ ⎜ 2 ⎟ ⎝ ⎠ ⎝ ⎠ Note: L i• L 2 = 8
Allow FT on incorrect vectors. A1
L1 = 14 , L2 = 21 cos θ =
A1A1
8 14 21
θ = 1.09 radians (62.2° )
A1 (A1) A1 [6]
39
53.
General point on line is (1 + λ, λ, −2 −2λ)
(M1)(A1)
Intersection of line and plane 1 + λ + λ − 2(−2 −2λ) + 3 = 0 ⇒ λ=
−4 3
M1A1 Note: Allow FT on incorrect λ.
⎛ − 4 ⎞ ⎟ ⎝ 3 ⎠
⇒ at P ʹ′ λ = 2⎜
M1
⎛ − 5 − 8 10 ⎞ , , ⎟ ⎝ 3 3 3 ⎠
⇒ P ʹ′ is ⎜
A1
N3 [6]
54.
(a)
⎛ 1 ⎞ ⎜ ⎟ Direction vector of l1 = ⎜ − 4 ⎟ ⎜ − 3 ⎟ ⎝ ⎠ ⇒ Equations of line through point A are:
(M1)
x y −1 z − 2 = = 1 −4 −3
A1
N2
Note: Accept any correct cartesian form.
(b)
⎛ 1 ⎞ ⎜ ⎟ Direction vector of l2 = ⎜ − 4 ⎟ ⎜ − 3 ⎟ ⎝ ⎠ ⎛ 3 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ ⇒ vector equation of l2 is r = ⎜ − 8 ⎟ + µ ⎜ − 4 ⎟ ⎜ − 11⎟ ⎜ − 3 ⎟ ⎝ ⎠ ⎝ ⎠ Note:
(M1)
A1
N2
⎛ x ⎞ ⎜ ⎟ Accept only this form but allow ⎜ y ⎟ ⎜ z ⎟ ⎝ ⎠ in place of r.
40
(c)
(i)
General point on l1 is (λ, 1 − 4λ, 2 − 3λ) →
→
(A1)
→
PQ = OQ − OP ⎛ λ ⎞ ⎛ 3 ⎞ ⎛ λ − 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎜ 1 − 4 λ ⎟ − ⎜ − 8 ⎟ = ⎜ 9 − 4 λ ⎟ ⎜ 2 − 3λ ⎟ ⎜ − 11⎟ ⎜13 − 3λ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ →
M1A1
→
⇒ PQ • l1 = 0
(M1)
⎛ λ − 3 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ ⇒ ⎜ 9 − 4 λ ⎟ • ⎜ − 4 ⎟ = 0 ⎜13 − 3 λ ⎟ ⎜ − 3 ⎟ ⎝ ⎠ ⎝ ⎠
A1
⇒ λ − 3 − 36 + 16λ − 39 + 9λ = 0
(ii)
⇒ 26λ = 78 ⇒ λ = 3
A1
⇒ Q = (3, −11, −7)
A1
→
d = PQ =
N0
(M1)
0 + 9 + 16
M1
=5
A1
N1 [14]
1 2 k 55.
(a)
det = 1 3 1 =1(15 − 8) − 2 (5 − k ) + k (8 − 3k )
M1A1
k 8 5 det = −3k2 + 10k − 3
A1
Now det = 0 ⇒ 3k2 − 10k + 3 = 0 ⇒ (3k − 1) (k − 3) = 0 ⇒ k=
(A1)
1 or k = 3 3
(A1)
For unique solution det ≠ 0 ⇒ k ∈ Note:
,k ≠
1 or k ≠ 3 3
R1
N0
Allow FT from previous line for R1.
41
(b)
k=
1 ⇒ 3
x + 2y +
1 z =0 3
(1)
x + 3y + z = 3
(2)
1 x + 8 y + 5z = 6 3
(3)
Attempting to eliminate a variable
M1
3 × equation (1) − equation (2) ⇒ 2x + 3y = −3
A1
15 × equation (2) − 3 × equation (3) ⇒ 14x + 21y = 27 ⇒ 2x + 3y = 9
A1
which is a contradiction so no solution.
A1
N0
k=3⇒ x + 2y + 3z = 0
(1)
x + 3y + z = 3
(2)
3x + 8y + 5z = 6
(3)
Attempting to eliminate a variable equation (2) − equation (1) ⇒ y − 2z = 3 ⇒ z =
M1
y −3 2
4 × equation (1) − equation (3) ⇒ x + 7z = −6 ⇒ z =
A1
x+6 −7
A1
EITHER Hence there is an infinite number of solutions in the line
x +6 y −3 = =z −7 2
A1
N0
A1
N0
OR General solution is (−6 −7λ, 3 + 2λ, λ) Note: Other correct forms are possible.
[14]
42
56.
⎛ 2 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ a • b = ⎜ 3 ⎟ • ⎜ 2 ⎟ ⎜ − 1⎟ ⎜ 5 ⎟ ⎝ ⎠ ⎝ ⎠ =3
(A1)
⎛ 3 ⎞ ⎛ µ ⎞ ⎜ ⎟ ⎜ ⎟ s = 3 ⎜ 1 ⎟ + ⎜ − 2 ⎟ ⎜ λ ⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ Note:
M1 Allow FT on a • b provided a • b is scalar.
⎛ 9 + µ ⎞ ⎜ ⎟ s = ⎜ 1 ⎟ ⎜ 3 λ + 1⎟ ⎝ ⎠
A1
⎛ 9 + µ ⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ s • a = 0 ⇒ ⎜ 1 ⎟ • ⎜ 3 ⎟ = 0 ⎜ 3 λ + 1⎟ ⎜ −1⎟ ⎝ ⎠ ⎝ ⎠
(M1)
Note: Allow FT for s. 18 + 2µ + 3 − 3λ − 1 = 0 (20 + 2µ = 3λ)
λ=
20 + 2 µ 3
A1 A1
N3 [6]
57.
n1 = −2i + j − k and n2 = i + 2j − k
n1 = 6 and n2 = 6
(A1)
n1 • n2 n1 n2
cos θ =
cos θ =
(A1)(A1)
(− 2 i + j − k )• (i + 2 j − k )
cos θ =
6× 6 1 6
( 0.167 to 3 sf )
M1(A1)
A1
N4 [6]
43
58.
METHOD 1 →
→
AB = 2 j − k and AC = − 3 i + 2 j →
i
→
j
AB× AC = 0
k
2 −1
−3 2
M1
0
= 2i + 3j + 6k
A1
→ 1 → AB × AC 2
Area Δ ABC =
(M1)
7 2
=
(A1)(A1)
A1
N4
METHOD 2 →
→
AB = 2 j − k and AC = − 3 i + 2 j
(A1)(A1)
Attempting to use the scalar product to find θ ie →
→
→
→
AB • AC = AB AC cos θ
cos θ =
M1
4
⎛ ⎞ 4 7 ⎜⎜ θ = arc cos = arc sin = 60.255...° ⎟⎟ 65 ⎝ 65 65 ⎠ 1 → → AB AC sin θ 2
Area Δ ABC =
⎛ 1 7 ⎞ ⎜⎜ = × 5 × 13 × ⎟⎟ 65 ⎠ ⎝ 2
7 2
=
A1
(M1)
A1
N4 [6]
i 59.
(a)
(i)
− 2 2 3 = − 3i + 6 j − 6 k 2
(ii)
j k M1A1
N2
1 0
n = i − 2 j + 2 k so equation of π1 is x − 2y + 2z = D (or r • n = a • n)
M1
substituting (3, 1, 5) ⇒ 3 − 2 + 10 = D
M1
so 11 = D Therefore the equation of plane π1 is x − 2y + 2z = 11
AG
N0
44
(b)
Using scalar product of normal vectors
⎛ 1 ⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ − 2 ⎟ • ⎜ − 1 ⎟ = 2 + 2 − 4 = 0 ⎜ 2 ⎟ ⎜ − 2 ⎟ ⎝ ⎠ ⎝ ⎠ The normals are perpendicular, so the planes are perpendicular. (c)
(M1) (M1)
A1
R1 AG
METHOD 1 Elimination of one variable. Choosing a parameter.
(M1) (M1)
METHOD 2 Finding direction of the line Finding a point on the line
(M1) (M1)
x = 2λ − 1; y = 2λ − 6; z = λ
A1A1A1
N5
1 ⎛ ⎞ OR ⎜ x = µ + 5, y = µ, z = µ + 3 ⎟ 2 ⎝ ⎠ 1 1 ⎞ ⎛ OR ⎜ x = t , y = t − 5, z = t + ⎟ 2 2 ⎠ ⎝
⎛ ⎛ − 1 ⎞ ⎛ 2 ⎞ ⎞ ⎜ ⎜ ⎟ ⎜ ⎟ ⎟ OR ⎜ r = ⎜ − 6 ⎟ + λ ⎜ 2 ⎟ ⎟ ⎜ ⎜ 0 ⎟ ⎜ 1 ⎟ ⎟ ⎝ ⎝ ⎠ ⎝ ⎠ ⎠ (d)
Direction vector of l2 = 2i + 2j + k equation of l 2 :
x − 3 y + 5 z +1 = = 2 2 1
(A1) A1A1
N3
45
(e)
(i)
(ii)
Recognizing that the equation of line (PQ) is needed
(M1)
⎛ 3 ⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ equation of line (PQ) is r = ⎜ − 5 ⎟ + t ⎜ − 1 ⎟ ⎜ − 1 ⎟ ⎜ − 2 ⎟ ⎝ ⎠ ⎝ ⎠
(A1)
Q on π2 ⇒ 2(3 + 2t) − (−5 −t) −2 (−1 −2t) = 4
M1
t = −1
A1
⇒ Q = (1, −4, 1)
A1
(3 −1)2 + (− 5 + 4)2 + (− 1− 1)2
PQ = =3
N0
(M1) A1
N2 [23]
60.
(a)
Substituting for a, b and c into c = ma + nb
(M1)
Forming any 2 of the following equations
A1A1
m+n=2
Eq(1)
−m + 2n = −5
Eq(2)
m + 4n = −1
Eq(3) Accept equations in vector form.
Note: Solving for m and n
(M1)
m = 3 and n = −1 (b)
A1
N3
METHOD 1
i
j
k
a × b = 1 −1 1
1
2
(M1)(A1)
4
= −6i − 3j + 3k
A1
Attempting to find | a × b | u=
(=
1 (−6i − 3j + 3k) 54 Note:
54 = 3 6
)
⎛ 1 ⎜⎜ = (− 2 i − j + k )⎞⎟⎟ 6 ⎝ ⎠
(M1) A1
N3
Award as above for b × a = 6i + 3j − 3k 1 and u = (−6i − 3j + 3k). 54
46
METHOD 2 Stating 2 equations derived from a • u and b • u where u = xi + yj + zk. x−y+z=0
Eq(1)
x + 2y + 4z = 0
Eq(2)
Attempting to solve the above system of equations
A1
(M1)
Solution sets include x = −2z and y = −z
A1
OR y = x and z = − x
2
A1
2
OR z = − y and x = 2y Note: Accept any correct numerical solution such as x = 2, y = 1, z = −1. Using x2 + y2 + z2 = 1 (ie| u | = 1) to find values for x, y and z.
1
Either u =
6
(2i + j − k )
Note:
(c)
(i)
or u =
1 6
(− 2i − j + k )
A1
(M1) A1
N3
(M1) M1 A1
N3
(M1) M1 A1
N3
Ignore any additional answers, even if incorrect.
METHOD 1 Equation of π1 is of the form x + 2y + 4z = d Substituting (1, −1, 1) (⇒ d = 3) ⇒ ( x + 2y + 4z = 3 METHOD 2 r • (i + 2j + 4k) = d Evaluate the scalar product a • (i + 2j + 4k) ⇒ x + 2y + 4z = 3
(ii)
(d)
(i)
3 ⎞ ⎛ 3 ⎞ ⎛ L(3, 0, 0), M ⎜ 0, , 0 ⎟ and N ⎜ 0, 0, ⎟ 4 ⎠ ⎝ 2 ⎠ ⎝
(M1)A1
P has coordinates (x, y, z) = ( λ, 2λ, 4λ) Substituting the coordinates of P into the equation of π1 λ + 4λ + 16λ = 3
(A1) (M1) A1
λ= 1
(A1)
7
⎛ 1 2 4 ⎞ P⎜ , , ⎟ ⎝ 7 7 7 ⎠
(ii)
(= 3)
Distance =
A1
1 2 1 + 22 + 42 7
N3
M1
47
21 or equivalent (= 0.655) 7 Note: Award M0A0 for any other method. =
(e)
A1
N2
(Given θ is the angle between π2 and a line and α is the angle
⎛ π ⎞ between the normal and a line) cos α = cos ⎜ − θ ⎟ = sin θ ⎝ 2 ⎠
(R1)
a •b a •b or cosα = a b a b
M1
Using the scalar product eg sin θ =
(sin θ )= (i − j + k ) • (i + 2 j + 4k ) i − j + k i + 2 j + 4k
=
1 (or equivalent) 7
θ = 0.388 (= 22.2°)
⎛ ⎞ ⎜⎜ = arcsin 1 ⎟⎟ 7 ⎠ ⎝
(A1)
(A1)
A1
N2 [27]
48
61.
Eliminating any one of the variables Using the first two equations this could be y − 2z = 1 Let z = α ⇒ y = 2α + 1
M1 A1 M1 A1
From the first equation x = 3 + 2α + 1 + α
M1
⇒ x = 3α + 4
A1
Hence x = 3α + 4, y = 2α + 1, z = α
⎛ x ⎞ ⎛ 4 ⎞ ⎛ 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ OR ⎜ y ⎟ = ⎜ 1 ⎟ + α ⎜ 2 ⎟ ⎜ z ⎟ ⎜ 0 ⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ OR
x − 4 y −1 z = = 3 2 1
OR x =
3α + 5 α −1 , y= α, z = 2 2
OR x = α , y =
2α − 5 α−4 ,z= 3 3 [6]
62.
(a)
⎛ 4 + λ ⎞ ⎜ ⎟ l1 r = ⎜ 3 + 5 λ ⎟ ⎜ − 2 λ ⎟ ⎝ ⎠ for l1 for x = 2, λ = − 2
A1
⇒ y = −7 ⇒z=4 Therefore point fits on line.
R1
49
(b)
4+λ=2 3 + 5λ = − 1 + 2µ −2λ = 3 − 3µ
Eq (1) Eq (2) Eq (3)
From Eq (1), λ = −2
(M1) A1
From Eq (2), 3 − 10 = −1 + 2µ −7 = −1 + 2µ
µ = −3
A1
Substituting in Eq (3) ⇒4=3+9 ⇒ lines do not intersect
63.
R1
N0 [6]
EITHER The parametric equations of the line are x=4−λ y = −2 + 2λ z = 6 + 4λ
A1 A1 A1
Substituting into the left handside of the equation of the plane 2(4 − λ) − (−2 + 2λ) + (6 + 4λ) = 8 − 2λ + 2 − 2λ + 6 + 4λ
M1A1
= 16 This equals the right handside. Hence the line is contained in the plane.
R1
OR We first need to prove that that the line and the plane are parallel. If true, the scalar product is zero.
⎛ − 1⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 2 ⎟ • ⎜ − 1⎟ = − 2 − 2 + 4 = 0 ⎜ 4 ⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ Now we need to show that a point on the line lies in the plane. A point on the line is (4, −2, 6)
⎛ 4 ⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ − 2 ⎟ • ⎜ − 1⎟ = 8 + 2 + 6 =16 ⎜ 6 ⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ Hence this is true. Therefore the line is contained in the plane.
M1A1
A1 M1A1
R1 [6]
64.
(a)
(i)
METHOD 1
50
⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AB = b − a = ⎜ 2 ⎟ − ⎜ 1 ⎟ = ⎜ 1 ⎟ ⎜ 3 ⎟ ⎜ 2 ⎟ ⎜ 1 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(A1)
⎛ 3 ⎞ ⎛ 1 ⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ AC = c − a = ⎜ 0 ⎟ − ⎜ 1 ⎟ = ⎜ − 1⎟ ⎜ 1 ⎟ ⎜ 2 ⎟ ⎜ − 1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(A1)
i j k 1 1 AB × AC = 0 2 −1 −1
M1
→
→
→
→
= i (−1 + 1) − j(0 − 2) + k (0 − 2)
(A1)
= 2j − 2k Area of triangle ABC = Note:
A1
1 1 2 j − 2k = 8 = 2 sq. units M1A1 2 2
(
)
Allow FT on final A1.
51
METHOD 2
AB = 2 , BC = 12 , AC = 6 A1A1A1 Using cosine rule, eg on Cˆ cos C =
6 + 12 − 2 2 72
=
M1
2 2 3
A1
1 ∴Area Δ ABC = absin C 2 =
1 12 6 sin 2
M1
⎛ ⎞ ⎜ arccos 2 2 ⎟ ⎜ 3 ⎟⎠ ⎝
⎛ 2 2 ⎞⎟ = 2 = 3 2 sin ⎜⎜ arccos 3 ⎟⎠ ⎝
(
)
A1
Note: Allow FT on final A1. (ii)
AB =
2 =
2
A1
1 1 AB × h = 2 × h , h equals the shortest distance 2 2
⇒h=2 (iii)
(M1) A1
METHOD 1
⎛ 0 ⎞ ⎜ ⎟ π has form r• ⎜ 2 ⎟ = d ⎜ − 2 ⎟ ⎝ ⎠
(M1)
Since (1, 1, 2) is on the plane
⎛ 1 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ d = ⎜ 1 ⎟ • ⎜ 2 ⎟ = 2 − 4 = − 2 ⎜ 2 ⎟ ⎜ − 2 ⎟ ⎝ ⎠ ⎝ ⎠
M1A1
⎛ 0 ⎞ ⎜ ⎟ Hence r• ⎜ 2 ⎟ = −2 ⎜ − 2 ⎟ ⎝ ⎠ 2y − 2z = −2 (or y − z = −1)
A1
METHOD 2
⎛ 1 ⎞ ⎛ 0 ⎞ ⎛ 2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ r = ⎜ 1 ⎟ + λ ⎜ 1 ⎟ + µ ⎜ − 1⎟ ⎜ 2 ⎟ ⎜ 1 ⎟ ⎜ − 1⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ x = 1 + 2µ
(i)
y=1+λ−µ
(ii)
z = 2 + λ − µ (iii) Note: Award A1 for all three correct, A0
(M1)
A1
52
otherwise. From (i) µ =
x −1 2 ⎛ x −1 ⎞ ⎟ ⎝ 2 ⎠
substitute in (ii) y = 1 + λ − ⎜
⎛ x −1 ⎞ ⎟ ⎝ 2 ⎠
⇒ λ = y − 1 + ⎜
substitute λ and µ in (iii)
M1
⎛ x − 1 ⎞ ⎛ x −1 ⎞ ⎟ − ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠
⇒ z = 2 + y − 1 + ⎜ ⇒ y − z = −1
A1
53
(b)
(i)
The equation of OD is
⎛ 0 ⎞ ⎜ ⎟ r = λ ⎜ 2 ⎟ , ⎜ − 2 ⎟ ⎝ ⎠
⎛ ⎛ 0 ⎞ ⎞ ⎜ ⎜ ⎟ ⎟ ⎜ or r = λ ⎜ 1 ⎟ ⎟ ⎜ ⎜ − 1⎟ ⎟ ⎝ ⎠ ⎠ ⎝
M1
This meets π where 2λ + 2λ = −1
λ= −
(M1)
1 4
A1
1 1 ⎞ ⎛ Coordinates of D are ⎜ 0 , − , ⎟ 2 2 ⎠ ⎝ 2
(ii)
A1
2
→ 1 ⎛ 1 ⎞ ⎛ 1 ⎞ OD = 0 + ⎜ − ⎟ + ⎜ ⎟ = 2 ⎝ 2 ⎠ ⎝ 2 ⎠
(M1)A1 [20]
54
65.
METHOD 1 Use of | a × b | = | a | | b | sinθ
(M1)
| a × b |2 = | a |2 | b |2 sin2θ Note: Only one of the first two marks can be implied.
(A1)
= | a |2 | b |2 (1 − cos2θ)
A1
= | a |2 | b |2 − | a |2 | b |2 cos2θ
(A1)
= | a |2 | b |2 − (| a | | b | cosθ)2 Note: Only one of the above two A1 marks can be implied.
(A1)
= | a |2 | b |2 − (a • b)2 Hence LHS = RHS
A1 AG
N0
METHOD 2 Use of a • b = | a | | b | cosθ
(M1)
| a |2 | b |2 − (a • b)2 = | a |2 | b |2 − (| a | | b | cosθ)2
(A1)
= | a |2 | b |2 − | a |2 | b |2 cos2θ Note: Only one of the above two A1 marks can be implied.
(A1)
= | a |2 | b |2 (1 − cos2θ)
A1
= | a |2 | b |2 sin2θ
A1
= | a × b |2
A1
Hence LHS = RHS
AG Notes:
N0
Candidates who independently correctly simplify both sides and show that LHS = RHS should be awarded full marks. If the candidate starts off with expression that they are trying to prove and concludes that sin2θ = (1 − cos2θ) award M1A1A1A1A0A0. If the candidate uses two general 3D vectors and explicitly finds the expressions correctly award full marks. Use of 2D vectors gains a maximum of 2 marks. If two specific vectors are used no marks are gained. [6]
66.
(a)
Use of cosθ =
→
→
→
→
OA • AB
(M1)
OA AB
55
→
AB = i − j + k →
→
3 and OA = 3 2
AB = →
A1
→
OA • AB = 6
A1
substituting gives cosθ =
(b)
→
→
L1: r = OA + s AB
2 ⎛⎜ 6 ⎞⎟ or equivalent = ⎜ 6 ⎝ 3 ⎟⎠
or equivalent
L1: r = i − j + 4k + s(i − j + k) Note:
(c)
A1
or equivalent
M1
N1
(M1) A1
Award (M1)A0 for omitting “r =” in the final answer.
Equating components and forming equations involving s and t 1 + s = 2 + 2t, −1 − s = 4 + t, 4 + s = 7 + 3t Having two of the above three equations Attempting to solve for s or t Finding either s = −3 or t = −2 Explicitly showing that these values satisfy the third equation Point of intersection is (− 2, 2,1) Note: Position vector is not acceptable for final A1.
(M1) A1A1 (M1) A1 R1 A1
N1
56
(d)
METHOD 1
⎛ 1 ⎞ ⎛ 2 ⎞ ⎛ − 3 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ r = ⎜ − 1⎟ + λ ⎜ 1 ⎟ + µ ⎜ 3 ⎟ ⎜ 4 ⎟ ⎜ 3 ⎟ ⎜ − 3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
(A1)
x = 1 + 2λ − 3µ, y = −1 + λ + 3µ and z = 4 + 3λ − 3µ Elimination of the parameters x + y = 3λ so 4(x + y) = 12λ and y + z = 4λ + 3 so 3(y + z) = 12λ + 9
M1A1 M1
3(y + z) = 4(x + y) + 9
A1
Cartesian equation of plane is 4x + y − 3z = −9 (or equivalent)
A1
N1
METHOD 2 EITHER The point (2, 4, 7) lies on the plane. The vector joining (2, 4, 7) and (1, − 1, 4) and 2i + j + 3k are parallel to the plane. So they are perpendicular to the normal to the plane. (i − j + 4k) − (2i + 4j + 7k) = − i − 5j − 3k
i
j
k
n = −1 − 5 − 3 2
(A1)
1
M1
3
= − 12i − 3j + 9k
or equivalent parallel vector
A1
OR L1 and L2 intersect at D (−2, 2,1) →
AD = (−2i + 2j + k) − (i − j + 4k) = −3i + 3j − 3k i
j
k
n= 2
1
3
(A1)
M1
−3 3 −3 = −12i − 3j + 9k
or equivalent parallel vector
A1
THEN r • n = (i − j + 4k) • (−12i − 3j + 9k) = 27 Cartesian equation of plane is 4x + y − 3z = −9 (or equivalent)
67.
⎛ 1 ⎞ ⎜ ⎟ The normal vector to the plane is ⎜ 3 ⎟ . ⎜ 2 ⎟ ⎝ ⎠
M1 A1 A1
N1 [20]
(A1)
EITHER 57
θ is the angle between the line and the normal to the plane. ⎛ 4 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 1 ⎟ • ⎜ 3 ⎟ ⎜ − 2 ⎟ ⎜ 2 ⎟ ⎛ 3 3 ⎞ ⎜⎜ = ⎟⎟ cos θ = ⎝ ⎠ ⎝ ⎠ = 14 21 14 21 ⎝ 7 6 ⎠ ⇒θ
= 79.9° (= 1.394 ...)
The required angle is 10.1° (= 0.176)
(M1)A1A1 A1 A1
OR
φ is the angle between the line and the plane. ⎛ 4 ⎞ ⎛ 1 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ 1 ⎟ • ⎜ 3 ⎟ ⎜ − 2 ⎟ ⎜ 2 ⎟ 3 sin φ = ⎝ ⎠ ⎝ ⎠ = 14 21 14 21
φ = 10.1°
(= 0.176)
(M1)A1A1 A2 [6]
58
68.
METHOD 1 (from GDC)
1 ⎛ ⎜ 1 0 6 ⎜ ⎜ 0 1 − 2 ⎜ 3 ⎜ 0 0 0 ⎜ ⎝
1 ⎞ ⎟ 12 ⎟ 1 − ⎟ 6 ⎟ 0 ⎟ ⎟ ⎠
−
(M1)
1 1 x+ λ=− 6 12
A1
2 1 y− λ=− 3 6
A1
1 ⎞ ⎛ 1 2 ⎛ 1 ⎞ i − j ⎟ + λ ⎜ − i + j + k ⎟ ⎝ 12 6 ⎠ ⎝ 6 3 ⎠
r = ⎜ −
A1A1A1
N3
M1A1 (M1)A1 M1 A1
N3
METHOD 2 (Elimination method either for equations or row reduction of matrix) Eliminating one of the variables Finding a point on the line Finding the direction of the line The vector equation of the line
[6]
59
