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O Level Elementary Mathematics Vectors Matrices Quiz
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Questions
O-Level Elementary Mathematics Quiz - Vectors Matrices
Name: ________________________
Class: ________________________
Date: ________________________
Score: ______ / 40
Duration: 45 minutes
Total Marks: 40
Instructions:
- Answer ALL questions.
- Show all working clearly. Marks are awarded for method.
- Give non-exact answers to 3 significant figures unless otherwise stated.
- Diagrams are not drawn to scale unless indicated.
Section A: Short Answer (10 marks)
Answer all questions in the spaces provided.
1. Given that a = (\begin{pmatrix} 3 \ -2 \end{pmatrix}) and b = (\begin{pmatrix} -1 \ 5 \end{pmatrix}), find a + b.
[2 marks]
2. Given that p = (\begin{pmatrix} 4 \ 1 \end{pmatrix}), find the magnitude of p, giving your answer correct to 3 significant figures.
[2 marks]
3. Given that u = (\begin{pmatrix} 2 \ -3 \end{pmatrix}) and v = (\begin{pmatrix} -4 \ 6 \end{pmatrix}), express v in terms of u.
[2 marks]
4. The position vectors of points A and B are (\begin{pmatrix} 5 \ 2 \end{pmatrix}) and (\begin{pmatrix} -1 \ 4 \end{pmatrix}) respectively. Find the vector (\overrightarrow{AB}).
[2 marks]
5. Given that c = (\begin{pmatrix} 1 \ -2 \end{pmatrix}) and d = (\begin{pmatrix} 3 \ 4 \end{pmatrix}), find 2c − d.
[2 marks]
Section B: Structured Questions (10 marks)
Answer all questions in the spaces provided. Show all working.
6. In the diagram, O is the origin. Points A, B, and C have position vectors a, b, and c respectively.
It is given that a = (\begin{pmatrix} 2 \ 1 \end{pmatrix}), b = (\begin{pmatrix} 6 \ 5 \end{pmatrix}), and C lies on AB such that AC : CB = 1 : 3.
(a) Express (\overrightarrow{AB}) in terms of a and b, and find (\overrightarrow{AB}) as a column vector. [2 marks]
(b) Find the position vector of C. [2 marks]
(c) Find the magnitude of (\overrightarrow{OC}), giving your answer correct to 3 significant figures. [2 marks]
7. The diagram shows a parallelogram OABC, where O is the origin.
(\overrightarrow{OA}) = a and (\overrightarrow{OC}) = c.
(a) Express, in terms of a and c, the vector (\overrightarrow{OB}). [1 mark]
(b) The point D lies on AB such that AD : DB = 2 : 1.
Express (\overrightarrow{OD}) in terms of a and c. [3 marks]
Section C: Structured Questions (10 marks)
Answer all questions in the spaces provided. Show all working.
8. The point E lies on BC such that BE : EC = 1 : 1.
Express (\overrightarrow{OE}) in terms of a and c. [2 marks]
9. Given that p = (\begin{pmatrix} 3 \ -1 \end{pmatrix}) and q = (\begin{pmatrix} -2 \ 4 \end{pmatrix}),
(a) Find the vector r such that 3p + 2r = q. [3 marks]
(b) Find the magnitude of r, giving your answer correct to 3 significant figures. [1 mark]
10. The diagram shows triangle OAB with O as the origin.
(\overrightarrow{OA}) = a and (\overrightarrow{OB}) = b.
M is the midpoint of OA and N is the midpoint of OB.
(a) Express (\overrightarrow{MN}) in terms of a and b. [2 marks]
(b) Hence, state the relationship between MN and AB. Justify your answer. [2 marks]
Section D: Application & Reasoning (10 marks)
Answer all questions in the spaces provided. Show all working clearly.
11. Points P, Q, and R have position vectors p = (\begin{pmatrix} 1 \ 3 \end{pmatrix}), q = (\begin{pmatrix} 5 \ 7 \end{pmatrix}), and r = (\begin{pmatrix} -3 \ -1 \end{pmatrix}) respectively.
(a) Show that P, Q, and R are collinear. [3 marks]
(b) Find the ratio PQ : QR. [1 mark]
12. In the diagram, O is the origin. Points A and B have position vectors a = (\begin{pmatrix} 2 \ 4 \end{pmatrix}) and b = (\begin{pmatrix} 8 \ 1 \end{pmatrix}) respectively.
(a) Find the vector (\overrightarrow{AB}). [1 mark]
(b) Point C lies on AB extended such that AB : BC = 2 : 1.
Find the position vector of C. [3 marks]
13. Given that u = (\begin{pmatrix} 5 \ -2 \end{pmatrix}) and v = (\begin{pmatrix} -1 \ 3 \end{pmatrix}), find the vector w such that 2u − w = v.
[2 marks]
14. The position vectors of points X and Y are (\begin{pmatrix} 3 \ -1 \end{pmatrix}) and (\begin{pmatrix} -2 \ 5 \end{pmatrix}) respectively. Find the unit vector in the direction of (\overrightarrow{XY}).
[2 marks]
15. Given that a = (\begin{pmatrix} 4 \ 0 \end{pmatrix}) and b = (\begin{pmatrix} 1 \ 2 \end{pmatrix}), find the value of (k) such that (|\textbf{a} + k\textbf{b}| = 5).
[2 marks]
16. In triangle ABC, (\overrightarrow{AB} = \begin{pmatrix} 3 \ 1 \end{pmatrix}) and (\overrightarrow{AC} = \begin{pmatrix} -2 \ 4 \end{pmatrix}). Find (\overrightarrow{BC}).
[2 marks]
17. The points D, E, and F have position vectors d = (\begin{pmatrix} 2 \ 5 \end{pmatrix}), e = (\begin{pmatrix} -1 \ 3 \end{pmatrix}), and f = (\begin{pmatrix} 4 \ 1 \end{pmatrix}). Find the vector (\overrightarrow{DE} + \overrightarrow{EF}).
[2 marks]
18. Given that p = (\begin{pmatrix} 2 \ -1 \end{pmatrix}) and q = (\begin{pmatrix} -3 \ 4 \end{pmatrix}), find the vector projection of p onto q.
[2 marks]
19. The position vectors of points G and H are g and h respectively. Given that (\overrightarrow{GH} = \begin{pmatrix} 5 \ -2 \end{pmatrix}) and g = (\begin{pmatrix} 1 \ 3 \end{pmatrix}), find h.
[2 marks]
20. In the diagram, O is the origin. Points J and K have position vectors j = (\begin{pmatrix} 3 \ 2 \end{pmatrix}) and k = (\begin{pmatrix} -1 \ 4 \end{pmatrix}). Point L lies on JK such that JL : LK = 3 : 2. Find the position vector of L.
[2 marks]
END OF QUIZ
Check your work carefully.
Answers
O-Level Elementary Mathematics Quiz - Vectors Matrices
ANSWER KEY
Total Marks: 40
Section A: Short Answer (10 marks)
1. a + b = (\begin{pmatrix} 3 \ -2 \end{pmatrix} + \begin{pmatrix} -1 \ 5 \end{pmatrix} = \begin{pmatrix} 2 \ 3 \end{pmatrix})
[M1] Correct addition of components
[A1] (\begin{pmatrix} 2 \ 3 \end{pmatrix})
(2 marks)
2. |p| = (\sqrt{4^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17})
= 4.12 (3 s.f.)
[M1] Correct use of magnitude formula
[A1] 4.12
(2 marks)
3. v = (\begin{pmatrix} -4 \ 6 \end{pmatrix} = -2\begin{pmatrix} 2 \ -3 \end{pmatrix} = -2\textbf{u})
[M1] Identifying scalar multiple relationship
[A1] v = −2u
(2 marks)
4. (\overrightarrow{AB} = \begin{pmatrix} -1 \ 4 \end{pmatrix} - \begin{pmatrix} 5 \ 2 \end{pmatrix} = \begin{pmatrix} -6 \ 2 \end{pmatrix})
[M1] Subtracting position vectors (b − a)
[A1] (\begin{pmatrix} -6 \ 2 \end{pmatrix})
(2 marks)
5. 2c − d = (2\begin{pmatrix} 1 \ -2 \end{pmatrix} - \begin{pmatrix} 3 \ 4 \end{pmatrix} = \begin{pmatrix} 2 \ -4 \end{pmatrix} - \begin{pmatrix} 3 \ 4 \end{pmatrix} = \begin{pmatrix} -1 \ -8 \end{pmatrix})
[M1] Scalar multiplication and subtraction
[A1] (\begin{pmatrix} -1 \ -8 \end{pmatrix})
(2 marks)
Section B: Structured Questions (10 marks)
6.
(a) (\overrightarrow{AB} = \textbf{b} - \textbf{a} = \begin{pmatrix} 6 \ 5 \end{pmatrix} - \begin{pmatrix} 2 \ 1 \end{pmatrix} = \begin{pmatrix} 4 \ 4 \end{pmatrix})
[M1] Correct expression or column subtraction
[A1] (\begin{pmatrix} 4 \ 4 \end{pmatrix})
(2 marks)
(b) AC : CB = 1 : 3, so C divides AB in ratio 1 : 3 from A.
(\overrightarrow{OC} = \overrightarrow{OA} + \frac{1}{4}\overrightarrow{AB} = \begin{pmatrix} 2 \ 1 \end{pmatrix} + \frac{1}{4}\begin{pmatrix} 4 \ 4 \end{pmatrix} = \begin{pmatrix} 2 \ 1 \end{pmatrix} + \begin{pmatrix} 1 \ 1 \end{pmatrix} = \begin{pmatrix} 3 \ 2 \end{pmatrix})
[M1] Using section formula or vector addition with correct ratio
[A1] (\begin{pmatrix} 3 \ 2 \end{pmatrix})
(2 marks)
(c) |(\overrightarrow{OC})| = (\sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13})
= 3.61 (3 s.f.)
[M1] Magnitude formula
[A1] 3.61
(2 marks)
7.
(a) (\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{OC} = \textbf{a} + \textbf{c})
[A1] a + c
(1 mark)
(b) D lies on AB with AD : DB = 2 : 1.
(\overrightarrow{OD} = \overrightarrow{OA} + \frac{2}{3}\overrightarrow{AB})
(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (\textbf{a} + \textbf{c}) - \textbf{a} = \textbf{c})
(\overrightarrow{OD} = \textbf{a} + \frac{2}{3}\textbf{c})
[M1] Identifying (\overrightarrow{AB} = \textbf{c})
[M1] Using correct ratio for section formula
[A1] (\textbf{a} + \frac{2}{3}\textbf{c})
(3 marks)
Section C: Structured Questions (10 marks)
8. E lies on BC with BE : EC = 1 : 1, so E is the midpoint of BC.
(\overrightarrow{OE} = \overrightarrow{OB} + \frac{1}{2}\overrightarrow{BC})
(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \textbf{c} - (\textbf{a} + \textbf{c}) = -\textbf{a})
(\overrightarrow{OE} = (\textbf{a} + \textbf{c}) + \frac{1}{2}(-\textbf{a}) = \textbf{a} + \textbf{c} - \frac{1}{2}\textbf{a} = \frac{1}{2}\textbf{a} + \textbf{c})
[M1] Finding (\overrightarrow{BC}) or using midpoint formula
[A1] (\frac{1}{2}\textbf{a} + \textbf{c})
(2 marks)
9.
(a) 3p + 2r = q
2r = q − 3p
r = (\frac{1}{2}(\textbf{q} - 3\textbf{p}))
3p = (3\begin{pmatrix} 3 \ -1 \end{pmatrix} = \begin{pmatrix} 9 \ -3 \end{pmatrix})
q − 3p = (\begin{pmatrix} -2 \ 4 \end{pmatrix} - \begin{pmatrix} 9 \ -3 \end{pmatrix} = \begin{pmatrix} -11 \ 7 \end{pmatrix})
r = (\frac{1}{2}\begin{pmatrix} -11 \ 7 \end{pmatrix} = \begin{pmatrix} -5.5 \ 3.5 \end{pmatrix})
[M1] Rearranging equation
[M1] Correct substitution and scalar multiplication
[A1] (\begin{pmatrix} -5.5 \ 3.5 \end{pmatrix}) or (\begin{pmatrix} -\frac{11}{2} \ \frac{7}{2} \end{pmatrix})
(3 marks)
(b) |r| = (\sqrt{(-5.5)^2 + 3.5^2} = \sqrt{30.25 + 12.25} = \sqrt{42.5})
= 6.52 (3 s.f.)
[A1] 6.52
(1 mark)
10.
(a) M is midpoint of OA: (\overrightarrow{OM} = \frac{1}{2}\textbf{a})
N is midpoint of OB: (\overrightarrow{ON} = \frac{1}{2}\textbf{b})
(\overrightarrow{MN} = \overrightarrow{ON} - \overrightarrow{OM} = \frac{1}{2}\textbf{b} - \frac{1}{2}\textbf{a} = \frac{1}{2}(\textbf{b} - \textbf{a}))
[M1] Finding position vectors of M and N
[A1] (\frac{1}{2}(\textbf{b} - \textbf{a}))
(2 marks)
(b) (\overrightarrow{AB} = \textbf{b} - \textbf{a})
(\overrightarrow{MN} = \frac{1}{2}(\textbf{b} - \textbf{a}) = \frac{1}{2}\overrightarrow{AB})
Therefore, MN is parallel to AB and MN = (\frac{1}{2})AB.
[M1] Finding (\overrightarrow{AB}) and comparing
[A1] Correct relationship with justification (parallel and half the length)
(2 marks)
Section D: Application & Reasoning (10 marks)
11.
(a) (\overrightarrow{PQ} = \textbf{q} - \textbf{p} = \begin{pmatrix} 5 \ 7 \end{pmatrix} - \begin{pmatrix} 1 \ 3 \end{pmatrix} = \begin{pmatrix} 4 \ 4 \end{pmatrix})
(\overrightarrow{PR} = \textbf{r} - \textbf{p} = \begin{pmatrix} -3 \ -1 \end{pmatrix} - \begin{pmatrix} 1 \ 3 \end{pmatrix} = \begin{pmatrix} -4 \ -4 \end{pmatrix})
(\overrightarrow{PR} = -\overrightarrow{PQ})
Since (\overrightarrow{PR}) is a scalar multiple of (\overrightarrow{PQ}), points P, Q, and R are collinear.
[M1] Finding (\overrightarrow{PQ})
[M1] Finding (\overrightarrow{PR}) and showing scalar multiple relationship
[A1] Conclusion with justification
(3 marks)
(b) (\overrightarrow{PQ} = \begin{pmatrix} 4 \ 4 \end{pmatrix}), (\overrightarrow{QR} = \textbf{r} - \textbf{q} = \begin{pmatrix} -3 \ -1 \end{pmatrix} - \begin{pmatrix} 5 \ 7 \end{pmatrix} = \begin{pmatrix} -8 \ -8 \end{pmatrix} = -2\begin{pmatrix} 4 \ 4 \end{pmatrix})
So PQ : QR = 1 : 2
[A1] 1 : 2
(1 mark)
12.
(a) (\overrightarrow{AB} = \textbf{b} - \textbf{a} = \begin{pmatrix} 8 \ 1 \end{pmatrix} - \begin{pmatrix} 2 \ 4 \end{pmatrix} = \begin{pmatrix} 6 \ -3 \end{pmatrix})
[A1] (\begin{pmatrix} 6 \ -3 \end{pmatrix})
(1 mark)
(b) AB : BC = 2 : 1, so C lies beyond B.
(\overrightarrow{AC} = \frac{3}{2}\overrightarrow{AB}) (since AB : AC = 2 : 3)
(\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = \textbf{a} + \frac{3}{2}\overrightarrow{AB})
= (\begin{pmatrix} 2 \ 4 \end{pmatrix} + \frac{3}{2}\begin{pmatrix} 6 \ -3 \end{pmatrix} = \begin{pmatrix} 2 \ 4 \end{pmatrix} + \begin{pmatrix} 9 \ -4.5 \end{pmatrix} = \begin{pmatrix} 11 \ -0.5 \end{pmatrix})
[M1] Correct interpretation of ratio (C beyond B)
[M1] Using section formula or vector addition with correct scalar
[A1] (\begin{pmatrix} 11 \ -0.5 \end{pmatrix}) or (\begin{pmatrix} 11 \ -\frac{1}{2} \end{pmatrix})
(3 marks)
13. 2u − w = v ⇒ w = 2u − v
2u = (2\begin{pmatrix} 5 \ -2 \end{pmatrix} = \begin{pmatrix} 10 \ -4 \end{pmatrix})
w = (\begin{pmatrix} 10 \ -4 \end{pmatrix} - \begin{pmatrix} -1 \ 3 \end{pmatrix} = \begin{pmatrix} 11 \ -7 \end{pmatrix})
[M1] Rearranging and scalar multiplication
[A1] (\begin{pmatrix} 11 \ -7 \end{pmatrix})
(2 marks)
14. (\overrightarrow{XY} = \begin{pmatrix} -2 \ 5 \end{pmatrix} - \begin{pmatrix} 3 \ -1 \end{pmatrix} = \begin{pmatrix} -5 \ 6 \end{pmatrix})
|(\overrightarrow{XY})| = (\sqrt{(-5)^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61})
Unit vector = (\frac{1}{\sqrt{61}}\begin{pmatrix} -5 \ 6 \end{pmatrix})
[M1] Finding (\overrightarrow{XY}) and its magnitude
[A1] (\frac{1}{\sqrt{61}}\begin{pmatrix} -5 \ 6 \end{pmatrix}) or equivalent
(2 marks)
15. a + kb = (\begin{pmatrix} 4 \ 0 \end{pmatrix} + k\begin{pmatrix} 1 \ 2 \end{pmatrix} = \begin{pmatrix} 4 + k \ 2k \end{pmatrix})
|a + kb| = (\sqrt{(4+k)^2 + (2k)^2} = \sqrt{16 + 8k + k^2 + 4k^2} = \sqrt{5k^2 + 8k + 16})
Set (\sqrt{5k^2 + 8k + 16} = 5)
(5k^2 + 8k + 16 = 25)
(5k^2 + 8k - 9 = 0)
((5k - ?)(k + ?) = 0) → Using quadratic formula: (k = \frac{-8 \pm \sqrt{64 - 4(5)(-9)}}{10} = \frac{-8 \pm \sqrt{64 + 180}}{10} = \frac{-8 \pm \sqrt{244}}{10} = \frac{-8 \pm 2\sqrt{61}}{10} = \frac{-4 \pm \sqrt{61}}{5})
[M1] Forming magnitude equation and squaring
[A1] (k = \frac{-4 + \sqrt{61}}{5}) or (k = \frac{-4 - \sqrt{61}}{5})
(2 marks)
16. (\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC} = -\overrightarrow{AB} + \overrightarrow{AC} = -\begin{pmatrix} 3 \ 1 \end{pmatrix} + \begin{pmatrix} -2 \ 4 \end{pmatrix} = \begin{pmatrix} -5 \ 3 \end{pmatrix})
[M1] Using vector addition/subtraction correctly
[A1] (\begin{pmatrix} -5 \ 3 \end{pmatrix})
(2 marks)
17. (\overrightarrow{DE} = \textbf{e} - \textbf{d} = \begin{pmatrix} -1 \ 3 \end{pmatrix} - \begin{pmatrix} 2 \ 5 \end{pmatrix} = \begin{pmatrix} -3 \ -2 \end{pmatrix})
(\overrightarrow{EF} = \textbf{f} - \textbf{e} = \begin{pmatrix} 4 \ 1 \end{pmatrix} - \begin{pmatrix} -1 \ 3 \end{pmatrix} = \begin{pmatrix} 5 \ -2 \end{pmatrix})
(\overrightarrow{DE} + \overrightarrow{EF} = \begin{pmatrix} -3 \ -2 \end{pmatrix} + \begin{pmatrix} 5 \ -2 \end{pmatrix} = \begin{pmatrix} 2 \ -4 \end{pmatrix})
[M1] Finding both vectors and adding
[A1] (\begin{pmatrix} 2 \ -4 \end{pmatrix})
(2 marks)
18. Projection of p onto q = (\frac{\textbf{p} \cdot \textbf{q}}{|\textbf{q}|^2} \textbf{q})
p · q = (2)(-3) + (-1)(4) = -6 - 4 = -10
|q|^2 = (-3)^2 + 4^2 = 9 + 16 = 25
Projection = (\frac{-10}{25} \textbf{q} = -\frac{2}{5} \begin{pmatrix} -3 \ 4 \end{pmatrix} = \begin{pmatrix} \frac{6}{5} \ -\frac{8}{5} \end{pmatrix})
[M1] Using projection formula
[A1] (\begin{pmatrix} 1.2 \ -1.6 \end{pmatrix}) or (\begin{pmatrix} \frac{6}{5} \ -\frac{8}{5} \end{pmatrix})
(2 marks)
19. (\overrightarrow{GH} = \textbf{h} - \textbf{g}) ⇒ h = g + (\overrightarrow{GH})
h = (\begin{pmatrix} 1 \ 3 \end{pmatrix} + \begin{pmatrix} 5 \ -2 \end{pmatrix} = \begin{pmatrix} 6 \ 1 \end{pmatrix})
[M1] Using relationship between position vectors and displacement
[A1] (\begin{pmatrix} 6 \ 1 \end{pmatrix})
(2 marks)
20. JL : LK = 3 : 2, so L divides JK in ratio 3 : 2 from J.
(\overrightarrow{OL} = \frac{2\textbf{j} + 3\textbf{k}}{3+2} = \frac{2\begin{pmatrix} 3 \ 2 \end{pmatrix} + 3\begin{pmatrix} -1 \ 4 \end{pmatrix}}{5} = \frac{\begin{pmatrix} 6 \ 4 \end{pmatrix} + \begin{pmatrix} -3 \ 12 \end{pmatrix}}{5} = \frac{\begin{pmatrix} 3 \ 16 \end{pmatrix}}{5} = \begin{pmatrix} 0.6 \ 3.2 \end{pmatrix})
[M1] Using section formula with correct ratio
[A1] (\begin{pmatrix} 0.6 \ 3.2 \end{pmatrix}) or (\begin{pmatrix} \frac{3}{5} \ \frac{16}{5} \end{pmatrix})
(2 marks)
END OF ANSWER KEY