How many QCAA Specialist Mathematics questions cover Vectors in two and three dimensions?

AusGrader has 167 QCAA Specialist Mathematics questions on Vectors in two and three dimensions, all with instant AI grading and detailed marking feedback.

QCAA Specialist Mathematics — Vectors in two and three dimensions Questions & Answers

Q6

2024

QCAA

Paper 2

1 mark

Q6

1 mark

Two concurrent forces represented in the polar form of $F_1 = (1.21 \text{ N}, 120^\circ)$ and $F_2 = (1.30 \text{ N}, -160^\circ)$ act on an object.
Determine the magnitude of the resultant force.

A

0.50 N

B

1.92 N

C

2.51 N

D

3.70 N

Reveal Answer

A

0.50 N

This value is incorrect. It is close to the result of $\sqrt{F_2^2 - F_1^2}$ (approx. $0.48$ N), which implies an incorrect application of the Pythagorean theorem rather than vector addition.

B

1.92 N

Correct Answer

The resultant is found by vector addition using the Law of Cosines or component method. The angle between the forces is $|120^\circ - (-160^\circ)| = 280^\circ$ , which is equivalent to $80^\circ$ . Calculating $R = \sqrt{1.21^2 + 1.30^2 + 2(1.21)(1.30)\cos(80^\circ)}$ yields approximately $1.92$ N.

C

2.51 N

This represents the scalar sum of the magnitudes ( $1.21 + 1.30 = 2.51$ N). This would only be correct if the two forces were acting in exactly the same direction.

D

3.70 N

This value corresponds to the square of the resultant force ( $R^2 \approx 3.70$ N $^2$ ). The final step of taking the square root to find the magnitude was omitted.

Q14

2024

VCAA

Paper 2

1 mark

Q14

1 mark

Consider the vectors $\underset{\sim}{\mathrm{r}}$ and $\underset{\sim}{\mathrm{s}}$ where $|\underset{\sim}{\mathrm{r}}| = 9$ and $\underset{\sim}{\mathrm{s}} = 2\underset{\sim}{\mathrm{i}} - 2\underset{\sim}{\mathrm{j}} + \underset{\sim}{\mathrm{k}}$ .
If the vector resolute of $\underset{\sim}{\mathrm{r}}$ in the direction of $\underset{\sim}{\mathrm{s}}$ is equal to $-4\underset{\sim}{\mathrm{i}} + 4\underset{\sim}{\mathrm{j}} - 2\underset{\sim}{\mathrm{k}}$ , then the scalar resolute of $\underset{\sim}{\mathrm{s}}$ in the direction of $\underset{\sim}{\mathrm{r}}$ is equal to

A

$-18$

B

$-2$

C

$2$

D

$3$

Reveal Answer

A

$-18$

This is the dot product $\underset{\sim}{\mathrm{r}} \cdot \underset{\sim}{\mathrm{s}}$ , not the scalar resolute. The dot product is found by multiplying the scalar resolute of $\underset{\sim}{\mathrm{r}}$ on $\underset{\sim}{\mathrm{s}}$ ( $-6$ ) by $|\underset{\sim}{\mathrm{s}}|$ ( $3$ ).

B

$-2$

Correct Answer

The vector resolute of $\underset{\sim}{\mathrm{r}}$ on $\underset{\sim}{\mathrm{s}}$ gives a scalar resolute of $-6$ , meaning $\underset{\sim}{\mathrm{r}} \cdot \underset{\sim}{\mathrm{s}} = -6 \times |\underset{\sim}{\mathrm{s}}| = -18$ . The scalar resolute of $\underset{\sim}{\mathrm{s}}$ on $\underset{\sim}{\mathrm{r}}$ is then $\frac{\underset{\sim}{\mathrm{r}} \cdot \underset{\sim}{\mathrm{s}}}{|\underset{\sim}{\mathrm{r}}|} = \frac{-18}{9} = -2$ .

C

$2$

This is the absolute value of the scalar resolute. However, because the dot product $\underset{\sim}{\mathrm{r}} \cdot \underset{\sim}{\mathrm{s}}$ is negative, the scalar resolute must also be negative.

D

$3$

This is the magnitude of vector $\underset{\sim}{\mathrm{s}}$ ( $|\underset{\sim}{\mathrm{s}}| = \sqrt{2^2 + (-2)^2 + 1^2} = 3$ ), not its scalar resolute in the direction of $\underset{\sim}{\mathrm{r}}$ .

Q4

2024

QCAA

Paper 1

1 mark

Q4

1 mark

A plane contains the point $(1, 3, 1)$ and is normal to the vector $\hat{i} + \hat{j} + 2\hat{k}$ .

The vector equation of the plane is

A

$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}$

B

$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$

C

$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \times \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}$

D

$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$

Reveal Answer

A

$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}$

This option incorrectly uses the position vector of the point $\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}$ as the normal vector. The dot product must be taken with the normal vector $\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ .

B

$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$

Correct Answer

The vector equation of a plane is given by $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$ , where $\mathbf{n}$ is the normal vector and $\mathbf{a}$ is a point on the plane. Substituting $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$ and $\mathbf{a} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}$ yields this result.

C

$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \times \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \times \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}$

The equation of a plane is defined using the scalar (dot) product to establish orthogonality, not the vector (cross) product.

D

$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$

While this involves the correct vectors, the cross product is typically used for the equation of a line. The plane equation requires the condition $(\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0$ .

Q19

2021

QCAA

Paper 1

7 marks

Q19

7 marks

The velocity vectors of two objects A and B (in m s $^{-1}$ ) at time $t$ (in s) are given respectively by

$\boldsymbol{v}_{\mathrm{A}} = 6\sin(3t)\hat{\boldsymbol{i}} + 6\cos(3t)\hat{\boldsymbol{j}}$
$\boldsymbol{v}_{\mathrm{B}} = \cos(t)\hat{\boldsymbol{i}} - \sin(t)\hat{\boldsymbol{j}}$

Objects A and B are initially at $(-2, 0, 2)$ and $(0, 1, -1)$ respectively. Determine the position of Object A when it is 4 metres away from Object B for the first time.

Reveal Answer

$v_A = 6\sin(3t)i + 6\cos(3t)j$
$v_B = \cos(t)i - \sin(t)j$

$r_A = \int v_A dt = -2\cos(3t)i + 2\sin(3t)j + c_A$
When $t = 0$
$-2i + 2k = -2\cos(0)i + 2\sin(0)j + c_A \Rightarrow c_A = 2k$
$\therefore r_A = -2\cos(3t)i + 2\sin(3t)j + 2k$

$r_B = \int v_B dt = \sin(t)i + \cos(t)j + c_B$
When $t = 0$
$j - k = \sin(0)i + \cos(0)j + c_B \Rightarrow c_B = -k$
$\therefore r_B = \sin(t)i + \cos(t)j - k$

$r_B - r_A$
$= (\sin(t)i + \cos(t)j - k) \dots$
$\dots - (-2\cos(3t)i + 2\sin(3t)j + 2k)$
$= (\sin(t) + 2\cos(3t))i + (\cos(t) - 2\sin(3t))j - 3k$

$|r_B - r_A|$
$= \sqrt{\sin^2(t) + 4\sin(t)\cos(3t) + 4\cos^2(3t) + \dots}$
$\overline{\dots \cos^2(t) - 4\cos(t)\sin(3t) + 4\sin^2(3t) + 9}$
$= \sqrt{14 - 4(\sin(3t)\cos(t) - \cos(3t)\sin(t))}$
$= \sqrt{14 - 4\sin(2t)}$

Given $|r_B - r_A| = 4$
$\sqrt{14 - 4\sin(2t)} = 4$
$\sin(2t) = -\frac{1}{2}$
$2t = \frac{7\pi}{6}$
$t = \frac{7\pi}{12}$ s (first positive solution)

Position of A
$r_A = -2\cos(3t)i + 2\sin(3t)j + 2k$
$= -2\cos(\frac{7\pi}{4})i + 2\sin(\frac{7\pi}{4})j + 2k$
$= -\sqrt{2}i - \sqrt{2}j + 2k$ (m)

Marking Criteria

Descriptor	Marks
correctly determines the expression for the position of Object A	1
correctly determines the expression for the position of Object B	1
determines an expression to represent the relative position of Objects A and B	1
determines an expression to represent the distance (or square of the distance) between the objects	1
uses a trigonometric identity to determine an expression in terms of a single trigonometric function that represents the distance (or square of the distance) between the objects	1
determines the first time that Object A is 4 metres away from Object B	1
determines position of Object A	1

Q14

2020

VCAA

Paper 2

1 mark

Q14

1 mark

The magnitude of the component of the force $\mathbf{F}=\mathbf{i}+6\mathbf{j}-18\mathbf{k}$ that acts in the direction $\mathbf{d}=2\mathbf{i}-3\mathbf{j}-6\mathbf{k}$ is

A

$\frac{92}{19}$

B

$\frac{92}{7}$

C

$\frac{124}{7}$

D

$\frac{92}{11}$

E

$\frac{18}{7}$

Reveal Answer

A

$\frac{92}{19}$

Incorrect. This option uses an incorrect magnitude for the direction vector $\mathbf{d}$ . The correct magnitude is $\sqrt{2^2 + (-3)^2 + (-6)^2} = 7$ .

B

$\frac{92}{7}$

Correct Answer

Correct. The component of $\mathbf{F}$ in the direction of $\mathbf{d}$ is found using $\frac{\mathbf{F} \cdot \mathbf{d}}{|\mathbf{d}|}$ . Calculating this gives $\frac{(1)(2) + (6)(-3) + (-18)(-6)}{\sqrt{2^2 + (-3)^2 + (-6)^2}} = \frac{92}{7}$ .

C

$\frac{124}{7}$

Incorrect. This answer likely stems from an arithmetic error when calculating the dot product $\mathbf{F} \cdot \mathbf{d}$ , which should be $92$ .

D

$\frac{92}{11}$

Incorrect. While the dot product is correctly calculated as $92$ , the magnitude of the direction vector $|\mathbf{d}|$ is $7$ , not $11$ .

E

$\frac{18}{7}$

Incorrect. This answer comes from incorrectly calculating the dot product $\mathbf{F} \cdot \mathbf{d}$ . The correct dot product is $92$ .

Q3

2024

QCAA

Paper 2

1 mark

Q3

1 mark

Given $a = \hat{j} + \hat{k}$ and $b = 2\hat{i} + \hat{k}$ , determine $a \times b$ .

A

$\hat{i} - 2\hat{j} - 2\hat{k}$

B

$\hat{i} - 2\hat{j} + 2\hat{k}$

C

$\hat{i} + 2\hat{j} - 2\hat{k}$

D

$\hat{i} + 2\hat{j} + 2\hat{k}$

Reveal Answer

A

$\hat{i} - 2\hat{j} - 2\hat{k}$

This option has the wrong sign for the $\hat{j}$ component. The calculation for the $\hat{j}$ term involves a negative sign: $-[(0)(1) - (1)(2)] = -(-2) = +2$ .

B

$\hat{i} - 2\hat{j} + 2\hat{k}$

This option has incorrect signs for both the $\hat{j}$ and $\hat{k}$ components. The $\hat{k}$ component is calculated as $(0)(0) - (1)(2) = -2$ .

C

$\hat{i} + 2\hat{j} - 2\hat{k}$

Correct Answer

Using the determinant method with rows $\hat{i}, \hat{j}, \hat{k}$ ; $0, 1, 1$ ; and $2, 0, 1$ , the result is $\hat{i}(1-0) - \hat{j}(0-2) + \hat{k}(0-2) = \hat{i} + 2\hat{j} - 2\hat{k}$ .

D

$\hat{i} + 2\hat{j} + 2\hat{k}$

This option has the wrong sign for the $\hat{k}$ component. It is calculated as $a_x b_y - a_y b_x = (0)(0) - (1)(2) = -2$ , not $+2$ .

Q11

2021

VCAA

Paper 2

1 mark

Q11

1 mark

Let $\underset{\sim}{i}$ be a unit vector pointing east and let $\underset{\sim}{j}$ be a unit vector pointing north.

A group of hikers travels 5 km in the direction south $30^\circ$ west and then north for 10 km.

The position vector $\underset{\sim}{a}$ of the group of hikers with respect to the starting point is

A

$\underset{\sim}{a} = -\frac{5}{2}\underset{\sim}{i} - \frac{5\sqrt{3}}{2}\underset{\sim}{j}$

B

$\underset{\sim}{a} = -\frac{5}{2}\underset{\sim}{i} + \left(10 - \frac{5\sqrt{3}}{2}\right)\underset{\sim}{j}$

C

$\underset{\sim}{a} = -\frac{5}{2}\underset{\sim}{i} + 10\underset{\sim}{j}$

D

$\underset{\sim}{a} = -\frac{5\sqrt{3}}{2}\underset{\sim}{i} + \frac{15}{2}\underset{\sim}{j}$

E

$\underset{\sim}{a} = \frac{5}{2}\underset{\sim}{i} + \left(10 + \frac{5\sqrt{3}}{2}\right)\underset{\sim}{j}$

Reveal Answer

A

$\underset{\sim}{a} = -\frac{5}{2}\underset{\sim}{i} - \frac{5\sqrt{3}}{2}\underset{\sim}{j}$

This represents only the first leg of the journey (5 km south $30^\circ$ west) and fails to add the second displacement of 10 km north.

B

$\underset{\sim}{a} = -\frac{5}{2}\underset{\sim}{i} + \left(10 - \frac{5\sqrt{3}}{2}\right)\underset{\sim}{j}$

Correct Answer

The first displacement is $-5\sin(30^\circ)\underset{\sim}{i} - 5\cos(30^\circ)\underset{\sim}{j}$ and the second is $10\underset{\sim}{j}$ . Adding these vectors yields the correct final position.

C

$\underset{\sim}{a} = -\frac{5}{2}\underset{\sim}{i} + 10\underset{\sim}{j}$

This incorrectly ignores the south component of the first displacement, only combining the west component with the 10 km north displacement.

D

$\underset{\sim}{a} = -\frac{5\sqrt{3}}{2}\underset{\sim}{i} + \frac{15}{2}\underset{\sim}{j}$

This incorrectly interprets "south $30^\circ$ west" as $30^\circ$ south of west, which swaps the sine and cosine terms for the first displacement.

E

$\underset{\sim}{a} = \frac{5}{2}\underset{\sim}{i} + \left(10 + \frac{5\sqrt{3}}{2}\right)\underset{\sim}{j}$

This incorrectly assumes the first displacement is north $30^\circ$ east, making both the $\underset{\sim}{i}$ and $\underset{\sim}{j}$ components positive instead of negative.

Q12

2022

VCAA

Paper 2

1 mark

Q12

1 mark

Consider the vectors $\underset{\sim}{u}(x) = -\text{cosec}(x)\underset{\sim}{i} + \sqrt{3}\underset{\sim}{j}$ and $\underset{\sim}{v}(x) = \cos(x)\underset{\sim}{i} + \underset{\sim}{j}$ .

If $\underset{\sim}{u}(x)$ is perpendicular to $\underset{\sim}{v}(x)$ , then possible values for $x$ are

A

$\frac{\pi}{6}$ and $\frac{7\pi}{6}$

B

$\frac{\pi}{3}$ and $\frac{4\pi}{3}$

C

$\frac{5\pi}{6}$ and $\frac{11\pi}{6}$

D

$\frac{2\pi}{3}$ and $\frac{5\pi}{3}$

E

$\frac{\pi}{6}$ and $\frac{5\pi}{6}$

Reveal Answer

A

$\frac{\pi}{6}$ and $\frac{7\pi}{6}$

Correct Answer

Setting the dot product to zero gives $-\cot(x) + \sqrt{3} = 0$ , which simplifies to $\tan(x) = \frac{1}{\sqrt{3}}$ . The solutions for this equation are $x = \frac{\pi}{6}$ and $\frac{7\pi}{6}$ .

B

$\frac{\pi}{3}$ and $\frac{4\pi}{3}$

These values correspond to $\tan(x) = \sqrt{3}$ , which is incorrect. This mistake likely comes from confusing $\cot(x)$ with $\tan(x)$ when solving the dot product equation.

C

$\frac{5\pi}{6}$ and $\frac{11\pi}{6}$

These values correspond to $\tan(x) = -\frac{1}{\sqrt{3}}$ . This would be the solution if the dot product equation was $\cot(x) + \sqrt{3} = 0$ , missing the negative sign.

D

$\frac{2\pi}{3}$ and $\frac{5\pi}{3}$

These values correspond to $\tan(x) = -\sqrt{3}$ , which does not satisfy the dot product equation $-\cot(x) + \sqrt{3} = 0$ .

E

$\frac{\pi}{6}$ and $\frac{5\pi}{6}$

While $\frac{\pi}{6}$ is a valid solution, $\frac{5\pi}{6}$ yields $\tan(x) = -\frac{1}{\sqrt{3}}$ , which does not satisfy the condition for perpendicular vectors.

Q14

2025

VCAA

Paper 2

1 mark

Q14

1 mark

For non-zero vectors $\underset{\sim}{a}$ and $\underset{\sim}{b}$ , if $\underset{\sim}{a} \cdot \underset{\sim}{b} = |\underset{\sim}{a} \times \underset{\sim}{b}|$ , then the angle between $\underset{\sim}{a}$ and $\underset{\sim}{b}$ is

A

$0$

B

$\frac{\pi}{4}$

C

$\frac{\pi}{2}$

D

$\frac{3\pi}{4}$

Reveal Answer

A

$0$

If the angle is $0$ , the dot product is maximized ( $\cos(0)=1$ ) and the cross product is zero ( $\sin(0)=0$ ), so they are not equal.

B

$\frac{\pi}{4}$

Correct Answer

Using the formulas $\underset{\sim}{a} \cdot \underset{\sim}{b} = |\underset{\sim}{a}||\underset{\sim}{b}|\cos\theta$ and $|\underset{\sim}{a} \times \underset{\sim}{b}| = |\underset{\sim}{a}||\underset{\sim}{b}|\sin\theta$ , equating them gives $\cos\theta = \sin\theta$ . This simplifies to $\tan\theta = 1$ , which means the angle $\theta$ is $\frac{\pi}{4}$ .

C

$\frac{\pi}{2}$

If the angle is $\frac{\pi}{2}$ , the dot product is zero ( $\cos(\frac{\pi}{2})=0$ ) while the magnitude of the cross product is maximized ( $\sin(\frac{\pi}{2})=1$ ).

D

$\frac{3\pi}{4}$

If the angle is $\frac{3\pi}{4}$ , the dot product is negative ( $\cos(\frac{3\pi}{4}) < 0$ ) while the magnitude of the cross product is positive, so they cannot be equal.

Q8

2025

QCAA

Paper 2

1 mark

Q8

1 mark

Consider the plane that contains both the $x$ -axis and the $y$ -axis. A sphere centred at (3, 4, 6) touches this plane.

The length of the radius of the sphere is

A

3 units.

B

4 units.

C

5 units.

D

6 units.

Reveal Answer

A

3 units.

This is the distance from the center to the $yz$ -plane ( $x=0$ ), not the $xy$ -plane.

B

4 units.

This is the distance from the center to the $xz$ -plane ( $y=0$ ), not the $xy$ -plane.

C

5 units.

This is the distance from the center to the $z$ -axis ( $\sqrt{3^2+4^2} = 5$ ), not the distance to the $xy$ -plane.

D

6 units.

Correct Answer

The plane containing the $x$ -axis and $y$ -axis is the $xy$ -plane ( $z=0$ ). The distance from the center $(3, 4, 6)$ to this plane is the absolute value of the $z$ -coordinate, which is $6$ .

Q13

2024

VCAA

Paper 2

1 mark

Q13

1 mark

If the angle between the vectors $2\underset{\sim}{\mathrm{i}} - \underset{\sim}{\mathrm{j}} + 2\underset{\sim}{\mathrm{k}}$ and $2\underset{\sim}{\mathrm{i}} + m\underset{\sim}{\mathrm{j}} + 6\underset{\sim}{\mathrm{k}}$ is $\cos^{-1}\left(\frac{13}{21}\right)$ , then the value of $m$ , where $m \in R^+$ , is

A

$2$

B

$3$

C

$4$

D

$5$

Reveal Answer

A

$2$

Substituting $m=2$ into the dot product formula yields $\cos \theta = \frac{14}{3\sqrt{44}}$ , which does not simplify to $\frac{13}{21}$ .

B

$3$

Correct Answer

Using the dot product formula $\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ , we get $\frac{16 - m}{3\sqrt{m^2 + 40}} = \frac{13}{21}$ . Solving this equation for $m > 0$ yields $m = 3$ .

C

$4$

Substituting $m=4$ into the dot product formula yields $\cos \theta = \frac{12}{3\sqrt{56}}$ , which does not simplify to $\frac{13}{21}$ .

D

$5$

Substituting $m=5$ into the dot product formula yields $\cos \theta = \frac{11}{3\sqrt{65}}$ , which does not simplify to $\frac{13}{21}$ .

Q11

2025

SCSA

Paper 2

6 marks

Q11

Two lines in space are defined by: $\underset{\sim}{r_1} = \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -4 \\ 1 \end{pmatrix}, \underset{\sim}{r_2} = \begin{pmatrix} 4 \\ 8 \\ -16 \end{pmatrix} + \mu \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix}$ .

Determine the ... :

Q11a

3 marks

... position vector of the intersection of these two lines.

Reveal Answer

Intersection is given by $\begin{pmatrix} 2+\lambda \\ 1-4\lambda \\ -3+\lambda \end{pmatrix} = \begin{pmatrix} 4-\mu \\ 8+\mu \\ -16+2\mu \end{pmatrix}$

Hence solving simultaneously:
$2+\lambda = 4-\mu \quad ...(1)$
$1-4\lambda = 8+\mu \quad ...(2)$
yields $\lambda = -3$ , $\mu = 5$

Testing with (3): $-3+(-3) = -16+2(5), \quad$ this is true.
This shows the lines do intersect.

Intersect at $\underset{\sim}{r_2} = \begin{pmatrix} 4 \\ 8 \\ -16 \end{pmatrix} + 5\begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -1 \\ 13 \\ -6 \end{pmatrix}$

Marking Criteria

Descriptor	Marks
forms the equation(s) that produce the intersection of lines	1
solves for the parameters $\lambda$ and $\mu$	1
determines the position vector correctly	1

Q11b

3 marks

... Cartesian equation of the plane that contains both lines.

Reveal Answer

If the plane contains both lines then its normal vector $\underset{\sim}{n} \perp \underset{\sim}{d_1}$ and $\underset{\sim}{n} \perp \underset{\sim}{d_2}$ .

Hence $\underset{\sim}{n} \parallel (\underset{\sim}{d_1}\times \underset{\sim}{d_2})\quad$ i.e. use $\underset{\sim}{n} = \begin{pmatrix} 1 \\ -4 \\ 1 \end{pmatrix} \times \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} -9 \\ -3 \\ -3 \end{pmatrix}$ OR $\underset{\sim}{n} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}$

The plane contains any point on each line i.e. use the point $(2,1,-3)$

Equation of the plane is given by: $\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} = 6+1-3$

i.e. Cartesian equation is $3x+y+z=4$

Marking Criteria

Descriptor	Marks
determines the normal vector for the plane using the cross product correctly	1
uses a point on either line correctly	1
determines the Cartesian equation correctly	1

Q17

2024

QCAA

Paper 2

6 marks

Q17

6 marks

An object moves with a constant speed of $v$ in a circular path.

The position vector of the object is given by

$r = r\cos(\omega t)\hat{i} + r\sin(\omega t)\hat{j}$

where

$r$ is the radius (metres) of the circle
$\omega$ is the angular velocity (radians per second)
$t$ is the time (seconds) of motion for $t \ge 0$ .

Use vector calculus to prove that the magnitude of the acceleration of the object is $|a| = \frac{v^2}{r}$ .

Reveal Answer

Given $r = r\cos(\omega t)\hat{i} + r\sin(\omega t)\hat{j}$

$v = \frac{dr}{dt} = -r\omega\sin(\omega t)\hat{i} + r\omega\cos(\omega t)\hat{j}$

$a = \frac{dv}{dt} = -r\omega^2\cos(\omega t)\hat{i} - r\omega^2\sin(\omega t)\hat{j}$

$v = |v|$

$= \sqrt{(-r\omega\sin(\omega t))^2 + (r\omega\cos(\omega t))^2}$

$= \sqrt{r^2\omega^2(\sin^2(\omega t) + \cos^2(\omega t))}$

$= r\omega$

$|a| = \sqrt{(-r\omega^2\cos(\omega t))^2 + (-r\omega^2\sin(\omega t))^2}$

$|a| = \sqrt{r^2\omega^4(\cos^2(\omega t) + \sin^2(\omega t))}$

$= r\omega^2 = \frac{r^2\omega^2}{r}$

$= \frac{v^2}{r}$

Marking Criteria

Descriptor	Marks
correctly determines the velocity vector	1
determines an acceleration vector	1
determines a simplified expression for the modulus of v	1
determines an expression for the modulus of a	1
correctly completes the proof based on prior evidence	1
shows logical organisation of a fully attempted proof, communicating key steps	1

Q5

2020

VCAA

Paper 1

4 marks

Q5

Let $\underset{\sim}{a} = 2\underset{\sim}{i} - 3\underset{\sim}{j} + \underset{\sim}{k}$ and $\underset{\sim}{b} = \underset{\sim}{i} + m\underset{\sim}{j} - \underset{\sim}{k}$ , where $m$ is an integer.

The vector resolute of $\underset{\sim}{a}$ in the direction of $\underset{\sim}{b}$ is $-\frac{11}{18}(\underset{\sim}{i} + m\underset{\sim}{j} - \underset{\sim}{k})$ .

Q5a

3 marks

Find the value of $m$ .

Reveal Answer

$m = 4$

Marking Criteria

Descriptor	Marks
Equates the scalar resolute formula to the given coefficient, e.g., $\frac{-3m + 1}{m^2 + 2} = -\frac{11}{18}$	1
Rearranges to form a correct quadratic equation, e.g., $11m^2 - 54m + 40 = 0$	1
Solves the quadratic equation and selects the correct integer value, $m = 4$	1

Q5b

1 mark

Find the component of $\underset{\sim}{a}$ that is perpendicular to $\underset{\sim}{b}$ .

Reveal Answer

$\frac{47}{18}\underset{\sim}{i} - \frac{5}{9}\underset{\sim}{j} + \frac{7}{18}\underset{\sim}{k}$

Marking Criteria

Descriptor	Marks
Calculates the correct perpendicular component: $\frac{47}{18}\underset{\sim}{i} - \frac{5}{9}\underset{\sim}{j} + \frac{7}{18}\underset{\sim}{k}$	1

Q2

2020

SCSA

Paper 1

5 marks

Q2

Plane $\Pi$ has vector equation $\underset{\sim}{r}=\begin{pmatrix}0\\0\\4\end{pmatrix}+\lambda\begin{pmatrix}3\\0\\1\end{pmatrix}+\mu\begin{pmatrix}1\\-1\\2\end{pmatrix}.$

Q2a

3 marks

Determine the normal vector $\underset{\sim}{n}$ for plane $\Pi$ .

Reveal Answer

$\text{Normal } \underset{\sim}{n} = \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 0(2) - 1(-1) \\ 1(1) - 3(2) \\ 3(-1) - 0(1) \end{pmatrix} = \begin{pmatrix} 1 \\ -5 \\ -3 \end{pmatrix}$

Marking Criteria

Descriptor	Marks
states that the normal vector can be found using a cross product of the plane direction vectors	1
obtains the correct form for each component of the cross product	1
determines the cross product correctly (or a multiple of this vector)	1

Q2b

2 marks

Determine the Cartesian equation for plane $\Pi$ .

Reveal Answer

$\text{Equation given by } \begin{pmatrix} x \\ y \\ z \end{pmatrix} \bullet \begin{pmatrix} 1 \\ -5 \\ -3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 4 \end{pmatrix} \bullet \begin{pmatrix} 1 \\ -5 \\ -3 \end{pmatrix} = -12$

$\text{i.e. } x - 5y - 3z = -12$

Marking Criteria

Descriptor	Marks
forms the equation using the normal vector $\underset{\sim}{n}$ correctly	1
determines the Cartesian equation correctly	1

QCAA Specialist Mathematics Vectors in two and three dimensions

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