Page 71 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Answers

5. Let z = 3 + i (Normal vector n = Parallel vector of the line L )
1
1
2
|z| = ( 3) + (1) = 2 Vector equation of L : r = (3i – 6j + 2k) + l(4i – 9j – 7k)
2
1
–1 1 π Parametric equations of L :
Arg z = tan = 1
3 6 x = 3 + 4l, y = –6 – 9l, z = 2 – 7l
π
π
z = 21 cos + i sin 2 2 5 –3
1 2 1 2 1 2
6 6 →
(b) (i) EF = 2 – 1 = 1
π
π
12
12
( 3 + i) = 3 21 cos + i sin 24 ; 3 –3 6
6 6 Vector equation of L :
π
π
2
12
( 3 – i) = 3 21 cos 1 – 2 + i sin 1 – 224 r = (5i + j – 3k) + µ(–3i + j + 6k)
12
6 6
Parametric equations of L : 1
( 3 + i) + ( 3 – i) x = 3 + 4l, y = –6 – 9l, z = 2 – 7l
12
12
π
π
π
π
12
= 3 21 cos + i sin 24 + 3 21 cos 1 – 2 + i sin 1 – 224 12 Parametric equations of L : 2
6 6 6 6 x = 5 – 3µ, y = 1 + µ, z = –3 + 6µ
12
= 2 3 cos 12π + i sin 12π 4 + 2 3 cos 1 – 12π 2 + i sin 1 – 12π 24
12
6 6 6 6 Equating the two lines:
12
12
= 2 [cos 2π + i sin 2π] + 2 [cos(–2π) + i sin (–2π)] 3 + 4l = 5 – 3µ
= 2 [1 + i(0)] + 2 [1 + i(0)] 3µ + 4l = 2 ......................a
12
12
= 2 (shown) –6 – 9l = 1 + µ
13
µ = –7 – 9l .....................b
2 – 7l = –3 + 6µ .....................c
12
3
12
2z – [( 3 + i) + ( 3 – i) ]i = 0
2z – 2 i = 0 Substitute b into a:
13
3
3
12
z = 2 i 3(–7 – 9l) + 4l = 2
z = 2 (0 + i) –21 – 27l + 4l = 2
12
3
23l = –23
π
π
z = 2 1 cos + i sin 2 l = –1
12
3
1 2 2 1
— π π — When l = –1, from b: µ = –7 – 9(–1) = 2,
12 3
3
z = (2 ) 3 cos 1 + 2kπ2 + i sin 1 + 2kπ24
2 2 checking from equation c:
π
π
z = 16 3 cos 1 + 2kπ 2 + i sin 1 + 2kπ 24 , k = 0, 1, 2 l = –1 , LHS = 2 – 7l = 2 – 7(–1) = 9
6 3 6 3 µ = 2 , RHS = –3 + 6µ = –3 + 6(2) = 9 = LHS

π
π
k = 0, z = 161 cos + i sin 2 The equations are consistent with l = – 1 and µ = 2,
6 6 RHS = LHS, therefore the two lines L and L
1
= 161 3 + i 2 intersect. (shown) 1 2
2 2
= 8( 3 + i) Position vector of point of intersection = –i + 3j + 9k
Coordinates of point of intersection = (–1, 3, 9)
5π
k = 1, z = 161 cos 5π + i sin 2 n . n
6 6 (ii) cos q = 1 2
1
2
1
= 161 – 3 + i 2 |n ||n |
.
2 2 (4i – 9j – 7k) (–3i + j + 6k)
=
= 8(– 3 + i) 146 . 46
3π
k = 2, z = 161 cos 3π + i sin 2 = –63
2 2 6 716
= 16(0 – i) q = 140.2°
= –16i
Acute angle = 180° – 140.2° = 39.8°
The roots are 8( 3 + i), 8(– 3 + i) and –16i.
7. (a) p(x) has a factor (x – 3),
2
3
→ 5 1 4 p(3) = a(3) – (1 + 6a)(3) + 2(8a – b)(3) – 27 = 0
5 –
4 = 1
6. (a) AB = 1 2 1 2 1 2 27a – 9 – 54a + 48a – 6b – 27 = 0
–1 –2 1 21a – 6b = 36
1 2 1 2 1 2
→ 4 1 3 7a – 2b = 12
AC = 3 – 4 = –1 (shown)
1 –2 3
i j k (b) From 7a – 2b = 12,

substitute –2b = 12 – 7a into
→
→
n = AB × AC = 4 1 1
3
2
1 p(x) = ax – (1 + 6a)x + (16a – 2b)x – 27,
3 –1 3 p(x) = ax – (1 + 6a)x + (16a + 12 – 7a)x – 27
2
3
= (3 + 1)i – (12 – 3)j + (–4 – 3)k p(x) = ax – (1 + 6a)x + 3(3a + 4)x – 27
3
2
= 4i – 9j – 7k
290
Answers STPM Math T S1.indd 290 3/28/18 4:25 PM