A Level H2 Mathematics Vectors Matrices Quiz

A-Level Maths H2 Quiz - Vectors Matrices (Answer Key)

Total Marks: 60
Topic: Vectors & Matrices

Section A: Vector Basics

Q1. [2 marks]
$\mathbf{a} + 2\mathbf{b} = \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} + 2\begin{pmatrix} -1 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} + \begin{pmatrix} -2 \\ 8 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 6 \\ 5 \end{pmatrix}$
Teaching note: Scalar multiplication multiplies each component; then add component-wise.
Common mistake: Forgetting to multiply all components by 2.

Q2. [2 marks]
$|\mathbf{v}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$
Teaching note: Magnitude = square root of sum of squared components.

Q3. [3 marks]
$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 4-1 \\ 3-0 \\ -1-2 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ -3 \end{pmatrix}$
$|\overrightarrow{AB}| = \sqrt{9+9+9} = \sqrt{27} = 3\sqrt{3}$
Unit vector = $\frac{1}{3\sqrt{3}}\begin{pmatrix} 3 \\ 3 \\ -3 \end{pmatrix} = \begin{pmatrix} 1/\sqrt{3} \\ 1/\sqrt{3} \\ -1/\sqrt{3} \end{pmatrix}$
Marks: 1 for $\overrightarrow{AB}$, 1 for magnitude, 1 for unit vector.

Q4. [3 marks]
$\overrightarrow{PQ} = (2,4,4)$, $\overrightarrow{QR} = (2,4,4)$
Since $\overrightarrow{PQ} = \overrightarrow{QR}$, $Q$ is midpoint; ratio $PQ:QR = 1:1$ → collinear.
Teaching note: Collinear if vectors between successive points are parallel (scalar multiples).

Q5. [2 marks]
$|\mathbf{u}| = \sqrt{1+4+4} = 3$; unit vector = $\frac{1}{3}\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 1/3 \\ 2/3 \\ 2/3 \end{pmatrix}$

Section B: Scalar and Vector Products

Q6. [2 marks]
$\mathbf{p}\cdot\mathbf{q} = 1(4) + 3(-1) + (-2)(2) = 4 - 3 - 4 = -3$

Q7. [3 marks]
$\mathbf{m}\cdot\mathbf{n} = 2(1)+1(-1)+0(2)=1$
$|\mathbf{m}|=\sqrt{5}, |\mathbf{n}|=\sqrt{6}$
$\cos\theta = \frac{1}{\sqrt{30}} \Rightarrow \theta \approx 79^\circ$
Marks: 1 dot, 1 magnitudes, 1 angle.

Q8. [3 marks]
$\mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 2 \\ 0 & 3 & 1 \end{vmatrix} = \mathbf{i}(0-6) - \mathbf{j}(1-0) + \mathbf{k}(3-0) = \begin{pmatrix} -6 \\ -1 \\ 3 \end{pmatrix}$

Q9. [2 marks]
$|\mathbf{a}\cdot\hat{\mathbf{n}}|$ is the length of the projection of $\mathbf{a}$ onto the direction of $\hat{\mathbf{n}}$ (absolute value).
Teaching note: Scalar projection magnitude.

Q10. [3 marks]
Area = $\frac{1}{2}|\mathbf{x}\times\mathbf{y}|$
$\mathbf{x}\times\mathbf{y} = \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}\times\begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ -6 \\ 6 \end{pmatrix}$
$|\mathbf{x}\times\mathbf{y}| = \sqrt{16+36+36}=\sqrt{88}=2\sqrt{22}$
Area = $\sqrt{22}$
Marks: 1 cross, 1 magnitude, 1 half.

Section C: 3D Vector Geometry

Q11. [2 marks]
$\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$

Q12. [3 marks]
$\mathbf{n}\cdot(\mathbf{r}-\mathbf{r}_0)=0 \Rightarrow 1(x-0) -2(y-1) +1(z-2)=0$
$x -2y +2 + z -2 = 0 \Rightarrow x -2y + z = 0$

Q13. [4 marks]
Let foot $F = \begin{pmatrix} 1+2\lambda \\ \lambda \\ 1-\lambda \end{pmatrix}$
$\overrightarrow{PF} = \begin{pmatrix} -2+2\lambda \\ -2+\lambda \\ 1-\lambda \end{pmatrix}$
Perp: $\overrightarrow{PF}\cdot\begin{pmatrix}2\\1\\-1\end{pmatrix}=0$
$2(-2+2\lambda)+1(-2+\lambda)-1(1-\lambda)=0 \Rightarrow -4+4\lambda-2+\lambda-1+\lambda=0 \Rightarrow 6\lambda=7 \Rightarrow \lambda=7/6$
$F = \begin{pmatrix} 1+7/3 \\ 7/6 \\ 1-7/6 \end{pmatrix} = \begin{pmatrix} 10/3 \\ 7/6 \\ -1/6 \end{pmatrix}$
Marks: 2 for setup, 1 solve, 1 coordinates.

Q14. [3 marks]
Direction $L_2 = 2\times$direction $L_1$ → parallel.
Check if same line: pt $(1,0,0)$ in $L_2$? $(0,1,2)+t(2,2,2)=(1,0,0)$ no solution → parallel distinct.

Q15. [3 marks]
Normals: $\mathbf{n}_1=(1,-1,1), \mathbf{n}_2=(2,1,-1)$
$\cos\theta = \frac{|1(2)+(-1)(1)+1(-1)|}{\sqrt{3}\sqrt{6}} = \frac{0}{\sqrt{18}}=0 \Rightarrow \theta=90.0^\circ$

Section D: Matrices and Mixed

Q16. [2 marks]
$\mathbf{MN} = \begin{pmatrix} 1&2\\3&4 \end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix} = \begin{pmatrix} 2&1\\4&3 \end{pmatrix}$

Q17. [3 marks]
$\det = 2(3)-1(5)=1 \neq 0$ → invertible.
Marks: 2 det, 1 statement.

Q18. [3 marks]
$\begin{pmatrix}2&1\\1&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}5\\1\end{pmatrix}$
Inverse = $\frac{1}{-3}\begin{pmatrix}-1&-1\\-1&2\end{pmatrix}$
$\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}2\\1\end{pmatrix}$
Marks: 1 matrix, 1 inverse, 1 solution.

Q19. [4 marks]
$\mathbf{T}$ rotates $90^\circ$ anticlockwise about origin.
Image of $(2,1)$: $\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}2\\1\end{pmatrix}=\begin{pmatrix}-1\\2\end{pmatrix}$
Marks: 2 description, 2 image.

Q20. [4 marks]
$\overrightarrow{AB}=(-1,2,0), \overrightarrow{AC}=(-1,0,3)$
Cross = $\begin{pmatrix}6\\3\\2\end{pmatrix}$, magnitude = 7
Area = $7/2$
Marks: 1 vectors, 2 cross, 1 area.