Vectors: Additional Mathematics Tingkatan 4 questions

Section A: objective questions

12 marks

Choose the correct answer. Each question carries 1 mark.

Which of the following is a vector quantity?

The mass of a box, 5 kg
The temperature of a room, 30°C
The velocity of a car, 60 km h⁻¹ due north
The area of a tile, 120 cm²

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A vector has magnitude and direction. Velocity with a direction (due north) is a vector; mass, temperature and area have magnitude only.

For the teacher: Remembering, low difficulty, standard 8.1.1

Textbook p. 212

Given |a| = 4 and →CD = −2a. Which statement about →CD is correct?

Its magnitude is 2 and it points the same way as a
Its magnitude is 8 and it points opposite to a
Its magnitude is −8 and it points opposite to a
Its magnitude is 6 and it points opposite to a

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|k a| = |k| × |a| = 2 × 4 = 8. k = −2 is negative, so the direction is opposite to a. A magnitude can never be negative.

For the teacher: Understanding, low difficulty, standard 8.1.3

Textbook p. 217

Given that a and b are non-zero and not parallel, and (h − 3)a = (k + 4)b, find h + k.

−1
7
1
−7

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Since a and b are not parallel, both coefficients are zero: h − 3 = 0 and k + 4 = 0. So h = 3, k = −4 and h + k = −1.

For the teacher: Applying, low difficulty, standard 8.1.4

Textbook p. 219

PQRS is a parallelogram with →PQ = a and →PS = b. Find →QS.

a + b
a − b
−a − b
b − a

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Triangle law: QS = QP + PS = −a + b = b − a. The option a − b is the vector SQ, the negative of the correct one.

For the teacher: Applying, low difficulty, standard 8.2.1

Textbook p. 223

Find the magnitude of the vector −5i + 12j.

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√((−5)² + 12²) = √169 = 13. Adding the components gives 7, and 169 is the value before taking the square root.

For the teacher: Applying, low difficulty, standard 8.3.1

Textbook p. 228

What is the unit vector in the direction of 3i − 4j?

3i − 4j
15(3i − 4j)
37i − 47j
125(3i − 4j)

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Magnitude = √(9 + 16) = 5. Divide the vector by 5. Dividing by 25 uses the square of the magnitude.

For the teacher: Applying, low difficulty, standard 8.3.2

Textbook p. 230

Given p = 3−1 and q = 14, find 2p − q.

5−6
7−2
52
−56

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2p = (6, −2). Then (6 − 1, −2 − 4) = (5, −6).

For the teacher: Applying, low difficulty, standard 8.3.3

Textbook p. 233

Vector 4i − 6j is parallel to 6i + kj. Find k.

9
−4
−9
4

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Parallel means 6 = m × 4 for some m, so m = 3⁄2 and k = m × (−6) = −9. Equivalent: 4 ÷ 6 = −6 ÷ k.

For the teacher: Analysing, medium difficulty, standard 8.3.3

Textbook p. 229

Given →PQ = 2a + 3b and →QR = 6a + hb. If P, Q and R are collinear, find h.

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Collinear with common point Q: QR = k PQ. a: 6 = 2k so k = 3. b: h = 3k = 9.

For the teacher: Analysing, high difficulty, standard 8.2.2

Textbook p. 224

A boat heads north at 5 m s⁻¹ in a stream flowing east at 12 m s⁻¹. What is the speed of the boat relative to the ground?

7 m s⁻¹
17 m s⁻¹
10 m s⁻¹
13 m s⁻¹

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The two velocities are at right angles, so speed = √(5² + 12²) = √169 = 13 m s⁻¹. Adding the speeds (17) only works for the same direction.

For the teacher: Applying, low difficulty, standard 8.2.1

Textbook p. 223

Which of the following is a unit vector?

11
0.60.6
1323
−0.60.8

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A unit vector has magnitude 1. √((−0.6)² + 0.8²) = √1 = 1. The others have magnitudes √2, about 0.85 and about 0.75.

For the teacher: Understanding, low difficulty, standard 8.3.2

Textbook p. 230

Given |a| = 7, find |−3a|.

21
−21
4
10

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|−3a| = 3 × 7 = 21. The negative sign changes the direction only, not the magnitude.

For the teacher: Understanding, low difficulty, standard 8.1.3

Textbook p. 217

Section B: structured questions

31 marks

Answer all questions. Show your working.

Vectors a and b are non-zero and not parallel.

[2 marks] (a)Given →AB = 4a + 6b and →BC = 10a + 15b, show that A, B and C are collinear.

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BC = 2.5 × AB, so the vectors are parallel [1]1
B is the common point, so A, B and C are collinear [1]1

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[2 marks] (b)Given (2h − 1)a = (k + 3)b, find h and k.

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2h − 1 = 0, so h = 1/2 [1]1
k + 3 = 0, so k = −3 [1]1

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For the teacher: Applying, low difficulty, standard 8.1.4

Textbook p. 219

Diagram 1 shows a parallelogram PQRS with →PQ = 3x and →PS = 2y. M is the midpoint of PR.

Parallelogram PQRS with PQ along the bottom labelled 3x and PS up the left side labelled 2y; diagonal PR drawn with M at its midpoint, and diagonal QS — Diagram 1

[2 marks] (a)Express →PR and →QS in terms of x and y.

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PR = PQ + PS = 3x + 2y [1]1
QS = QP + PS = −3x + 2y = 2y − 3x [1]1

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[2 marks] (b)Express →QM in terms of x and y.

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PM = ½PR = (3/2)x + y [1]1
QM = QP + PM = −3x + (3/2)x + y = y − (3/2)x [1]1

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[1 mark] (c)Show that Q, M and S are collinear.

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QM = ½(2y − 3x) = ½QS, so QM is parallel to QS with Q common [1]1

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For the teacher: Applying, medium difficulty, standard 8.2.1

Textbook p. 223

Given →OA = 2a + b, →OB = 4a − b and →OC = ha + 5b, where h is a constant. A, B and C are collinear.

[1 mark] (a)Express →AB in terms of a and b.

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AB = −OA + OB = 2a − 2b [1]1

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[1 mark] (b)Express →AC in terms of h, a and b.

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AC = −OA + OC = (h − 2)a + 4b [1]1

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[3 marks] (c)Find the value of h.

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AC = kAB: (h − 2)a + 4b = 2ka − 2kb [1]1
4 = −2k so k = −2 [1]1
h − 2 = 2k = −4, so h = −2 [1]1

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For the teacher: Analysing, high difficulty, standard 8.2.2

Textbook p. 224

A swimmer heads straight across a river (north) at 1.5 m s⁻¹. The stream flows east at 2.0 m s⁻¹. The river is 60 m wide.

[2 marks] (a)Find the speed of the swimmer relative to the river bank.

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√(1.5² + 2.0²) [1]1
= 2.5 m s⁻¹ [1]1

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[2 marks] (b)Find the bearing of the swimmer's path.

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tan θ = 2.0 ÷ 1.5 [1]1
θ = 53.13°, bearing 053.13° [1]1

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[1 mark] (c)How long does the swimmer take to cross?

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60 ÷ 1.5 = 40 s (only the northward speed counts) [1]1

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For the teacher: Applying, low difficulty, standard 8.2.1

Textbook p. 223

Given p = 3−4 and q = −25.

[1 mark] (a)Find 2p + q as a column vector.

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(6 − 2, −8 + 5) = (4, −3) [1]1

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[1 mark] (b)Find |2p + q|.

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√(16 + 9) = 5 [1]1

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[1 mark] (c)Find the unit vector in the direction of 2p + q, in the form xi + yj.

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(4i − 3j) ÷ 5 = 0.8i − 0.6j [1]1

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[1 mark] (d)Express p − 3q in the form xi + yj.

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(3 + 6)i + (−4 − 15)j = 9i − 19j [1]1

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[2 marks] (e)Find k if p + kq is parallel to the x-axis.

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p + kq = (3 − 2k, −4 + 5k) [1]1
Parallel to the x-axis means y-component 0: k = 4/5 [1]1

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For the teacher: Applying, low difficulty, standard 8.3.3

Textbook p. 233

A particle leaves A(2, 9) with constant velocity (3i − 2j) m s⁻¹. After t seconds its position vector is →OS = 2 + 3t9 − 2t. A second particle leaves B(20, 0) with velocity (−3i + j) m s⁻¹.

[1 mark] (a)Find the speed of the first particle.

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√(3² + (−2)²) = √13 = 3.61 m s⁻¹ [1]1

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[1 mark] (b)Find the position of the first particle after 3 seconds.

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(2 + 9, 9 − 6) = (11, 3) [1]1

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[1 mark] (c)When is the first particle on the x-axis?

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9 − 2t = 0, t = 4.5 s [1]1

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[3 marks] (d)Write the position vector of the second particle after t seconds and show that the two particles meet. State when and where.

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Second: (20 − 3t, t) [1]1
Equate: 2 + 3t = 20 − 3t gives t = 3; 9 − 2t = t gives t = 3 (same t) [1]1
They meet after 3 s at (11, 3) [1]1

Tick each point you got.

For the teacher: Evaluating, high difficulty, standard 8.3.4

Textbook p. 234

Section A: objective questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Section B: structured questions

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18