Page 73 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Answers

1 2 1 2 1 2
1
→ → → 2 2 0 (d) n = 6i + 4j + 5k From the answer in (b)
TM = OM – OT = 5 – 0 = 5 = a parallel vector of the normal vector
0 4 –4 Choose a point of plane RTM, point R(0, 3, 4)
.

1
→ → i j k Equation of the plane: r n = D
TR × TM = –2 3 0  6 0 6
1 2 1 2 1 2
.
.
0 5 –4 r 4 = 3 4
= (–12 – 0)i – (8 – 0)j + (–10 – 0)k 5 4 5
= –12i – 8j – 10k 6
1 2
.
= –2(6i + 4j + 5k) r 4 = 0 + 12 + 20
→
1 →
Area of triangle RTM = TR × TM  5
2 r (6i + 4j + 5k) = 32
.
2
2
= 1 × 2 × 6 + 4 + 5 2 Cartesian equation: 6x + 4y + 5z = 32
2
= 77 (e) Normal vector of the plane RTM:
n = 6i + 4j + 5k From the answer in (b)
1
(c) Let the acute angle between TR and TM = q Normal vector of the plane OTM:
Method 1 : Cross Product i j k
→ → → → → →  
TR × TM  = TR TM  sin q OT × OM = 2 0 4
→ → 2 5 0
TR × TM  = 2 × 77 From the answer in (b) = (0 – 20)i – (0 – 8)j + (10 – 0)k
2 77 = –20i + 8j + 10k
sin q = = –2(10i – 4j – 5k)
2 + 3 2 . 5 + 4 2 n = 10i – 4j – 5k
2
2
2
2 77
= Let the acute angle between the plane RTM and the
13 . 41 plane OTM = a
q = 49.5° n . n 2
1
cos a =
Method 2 : Scalar Product n n  2
1
10
6
→ →
→
→
TR TM = TR TM  cos q 1 2 1 2
.
.
4
–4

–2
0
–5
5
1 2 1 2 = 6 + 4 + 5 2 . 10 + 4 + 5 2
.
5
3

2
2
2
2

cos q = 0 –4 60 – 16 – 25
2 + 3 2 . 5 + 4 2 =
2
2
= 15 77 . 141
13 . 41 = 19
77 . 141
q = 49.5°
a = 79.5°
292
Answers STPM Math T S1.indd 292 3/28/18 4:25 PM