How many SCSA Mathematics Specialist questions cover Vectors in three dimensions?

AusGrader has 281 SCSA Mathematics Specialist questions on Vectors in three dimensions, all with instant AI grading and detailed marking feedback.

WACE Mathematics Specialist — Vectors in three dimensions Questions & Answers

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 $\underset{\sim}{\mathrm{r}}$ .

Q7

2024

QCAA

Paper 1

1 mark

Q7

1 mark

$A, B$ and $C$ are points in three-dimensional space.
If $2\vec{AB} = \vec{BC}$ , then

A

$|\vec{AB}|$ is twice the value of $|\vec{BC}|$ .

B

$\vec{AB}$ and $\vec{BC}$ are perpendicular.

C

only one plane contains $A, B$ and $C$ .

D

a straight line passes through $A, B$ and $C$ .

Reveal Answer

A

$|\vec{AB}|$ is twice the value of $|\vec{BC}|$ .

Taking the magnitude of both sides of $2\vec{AB} = \vec{BC}$ gives $2|\vec{AB}| = |\vec{BC}|$ , which means $|\vec{BC}|$ is twice $|\vec{AB}|$ , not the reverse.

B

$\vec{AB}$ and $\vec{BC}$ are perpendicular.

Since $\vec{BC}$ is a scalar multiple of $\vec{AB}$ , the vectors are parallel (collinear), not perpendicular.

C

only one plane contains $A, B$ and $C$ .

Since the points are collinear, there is an infinite number of planes (a pencil of planes) that can pass through the line containing $A, B,$ and $C$ .

D

a straight line passes through $A, B$ and $C$ .

Correct Answer

The equation implies $\vec{AB}$ and $\vec{BC}$ are parallel vectors sharing a common point $B$ , which proves that $A, B,$ and $C$ are collinear.

Q5

2024

VCAA

Paper 2

10 marks

Q5

Consider the points $A(1, -2, 3)$ and $B(2, -5, -1)$ .

Q5d

Another plane has the Cartesian equation $2x - 3y + 4z = 12$ .
This plane intersects the coordinate axes at three points, which form the vertices of a triangle.

Q5a

1 mark

Find a vector equation, in terms of the components $\underset{\sim}{\mathrm{i}}$ , $\underset{\sim}{\mathrm{j}}$ and $\underset{\sim}{\mathrm{k}}$ , for the line passing through these points.

Reveal Answer

Many correct answers exist, such as:

$\underset{\sim}{r}(t) = \underset{\sim}{i} - 2\underset{\sim}{j} + 3\underset{\sim}{k} + t(\underset{\sim}{i} - 3\underset{\sim}{j} - 4\underset{\sim}{k}), (t \in R)$

Marking Criteria

Descriptor	Marks
Provides the correct vector equation using the first point	1
None of the above	0

Q5b

3 marks

Consider the different line $L_1: \underset{\sim}{\mathrm{r_1}}(t) = 2\underset{\sim}{\mathrm{i}} + \underset{\sim}{\mathrm{j}} - 3\underset{\sim}{\mathrm{k}} + t(-\underset{\sim}{\mathrm{i}} + 2\underset{\sim}{\mathrm{j}} + \underset{\sim}{\mathrm{k}}), t \in R$ .

Find the shortest distance from $L_1$ to point $A$ .

Give your answer in the form $\frac{a\sqrt{b}}{c}$ where $a$ , $b$ and $c$ are positive integers.

Reveal Answer

$\frac{5\sqrt{66}}{6}$

Marking Criteria

Descriptor	Marks
Correct solution, giving the final answer $\frac{5\sqrt{66}}{6}$ .	3
Substantial progress (e.g., applies a correct method to find the shortest distance but makes a minor calculation error).	2
Partial solution (e.g., identifies a valid method or finds a relevant vector).	1
No valid response.	0

Q5c

3 marks

Let $C$ be the point $(0, 2, -5)$ .

Find the Cartesian equation of the plane that contains the points $A$ , $B$ and $C$ .

Reveal Answer

$40x + 12y + z = 19$

Marking Criteria

Descriptor	Marks
Finds two valid vectors in the plane (e.g., $\vec{AB}$ and $\vec{AC}$ ).	1
Calculates a normal vector to the plane (e.g., using the cross product).	1
Correct Cartesian equation: $40x + 12y + z = 19$ .	1

Q5d (i)

1 mark

Find the coordinates of these three points.

Reveal Answer

$(6,0,0), \ (0,-4,0), \ (0,0,3)$

Marking Criteria

Descriptor	Marks
Correctly identifies all three coordinates: $(6,0,0), \ (0,-4,0), \ (0,0,3)$ .	1

Q5d (ii)

2 marks

Find the area of the triangle.

Give your answer in the form $m\sqrt{n}$ where $m$ and $n$ are integers.

Reveal Answer

$3\sqrt{29}$

Marking Criteria

Descriptor	Marks
Correctly determines the spanning vectors and sets up the area calculation (e.g., half the magnitude of the cross product).	1
Correct final answer: $3\sqrt{29}$ .	1

Q1

2023

QCAA

Paper 1

1 mark

Q1

1 mark

The position of a particle is given by $r = (t+2)\hat{i} + t^2\hat{j}$ for $t \ge 0$ .
Determine the corresponding Cartesian equation.

A

$y=x^2-4$

B

$y=x^2+4$

C

$y=x^2-4x+4$

D

$y=x^2+4x+4$

Reveal Answer

A

$y=x^2-4$

This option is incorrect. Substituting $x = t+2$ into this equation yields $y = (t+2)^2 - 4 = t^2 + 4t$ , which does not match the given component $y = t^2$ .

B

$y=x^2+4$

This option is incorrect. Substituting $x = t+2$ into this equation yields $y = (t+2)^2 + 4 = t^2 + 4t + 8$ , which does not match the given component $y = t^2$ .

C

$y=x^2-4x+4$

Correct Answer

From the vector $\vec{r}$ , we have $x = t+2$ and $y = t^2$ . Solving for $t$ gives $t = x-2$ . Substituting this into the $y$ equation yields $y = (x-2)^2 = x^2 - 4x + 4$ .

D

$y=x^2+4x+4$

This option corresponds to $y = (x+2)^2$ . This implies an incorrect substitution where $t = x+2$ instead of the correct relationship $t = x-2$ .

Q5

2023

SCSA

Paper 1

5 marks

Q5

Consider two planes given by their Cartesian equations:
$x - 3y + 3z = 9$
$2x + y - z = 4$

Q5a

1 mark

Explain why these planes are not parallel.

Reveal Answer

The normal vectors for each plane $\begin{pmatrix} 1 \\ -3 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ are not scalar multiples of each other. Hence the planes cannot be parallel to each other.

Marking Criteria

Descriptor	Marks
explains that the normal vectors are not multiples of each other	1

Q5b

1 mark

State the geometric interpretation of the solution in the above simultaneous equations.

Reveal Answer

The two planes will intersect in a line in space

Marking Criteria

Descriptor	Marks
states that the planes intersect in a line	1

Q5c

3 marks

Determine the vector equation for the intersection of these two planes.

Reveal Answer

$x-3y+3z = 9 \quad ... (1)$
$2x+y-z = 4 \quad ... (2)$

Consider $(1) + 3 \times (2)$ :

\begin{align*} 7x+0y+0z &= 9+12\\ \text{i.e.\quad} 7x &= 21\\ \therefore x &= 3 \end{align*}

Substituting $x = 3$ into $(1)$ :

\begin{align*} 3-3y+3z &= 9\\ \text{i.e.\quad} z &= y+2 \text{\quad where \quad} y \in \mathbb{R} \end{align*}

i.e. there are infinitely many ordered triples for $x, y, z$ .
Hence the intersection of the two planes is a line in space.

Vector equation for this line:
$\underset{\sim}{r} = \begin{pmatrix} 3 \\ \lambda \\ \lambda+2 \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$

Marking Criteria

Descriptor	Marks
eliminates a variable correctly from the pair of equations	1
obtains the relationship $z = y+2$	1
forms the vector equation of the line using a parameter correctly	1

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

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 $+2$ .

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 $(\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0$ .

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

Q16

2022

QCAA

Paper 1

7 marks

Q16

Consider this system of equations that corresponds to three planes.

$x + 5y = 1 + 2z$
$x + z = 3y + 3$
$8y - \lambda = 3z$

Q16a

4 marks

Use a Gaussian technique to determine the value of $\lambda$ for which this system of equations has infinitely many solutions.

Reveal Answer

Rearranging equations:
$x + 5y - 2z = 1$
$x - 3y + z = 3$
$8y - 3z = \lambda$
Expressing in matrix form:
$\begin{bmatrix} 1 & 5 & -2 & 1 \\ 1 & -3 & 1 & 3 \\ 0 & 8 & -3 & \lambda \end{bmatrix}$
$R_2' = R_2 - R_1$
$\begin{bmatrix} 1 & 5 & -2 & 1 \\ 0 & -8 & 3 & 2 \\ 0 & 8 & -3 & \lambda \end{bmatrix}$
$R_3' = R_3 + R_2$
$\begin{bmatrix} 1 & 5 & -2 & 1 \\ 0 & -8 & 3 & 2 \\ 0 & 0 & 0 & \lambda + 2 \end{bmatrix}$
For $\lambda = -2$ there are infinitely many solutions.

Marking Criteria

Descriptor	Marks
correctly rearranges the three equations	1
establishes an augmented matrix	1
establishes a row of zeros in the row containing $\lambda$	1
determines a value of $\lambda$	1

Q16b

3 marks

Use the result from Question 16a) to determine the infinitely many solutions. Express your answer in the form of a vector equation of a line.

Reveal Answer

Using $\begin{bmatrix} 1 & 5 & -2 & 1 \\ 0 & -8 & 3 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix}$
Letting $z = k (k \in \mathbb{R})$
Row 2: $-8y + 3z = 2$
$8y - 3k = -2 \Rightarrow y = \frac{3k-2}{8}$
Row 1: $x + 5y - 2z = 1$
$x + 5(\frac{3k-2}{8}) - 2k = 1$
$x + \frac{15k-10}{8} - \frac{16k}{8} = 1 \Rightarrow x = \frac{k}{8} + \frac{9}{4}$
The solutions in vector form are:
$\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} \frac{1}{8} \\ \frac{3}{8} \\ 1 \end{bmatrix} k + \begin{bmatrix} \frac{9}{4} \\ -\frac{1}{4} \\ 0 \end{bmatrix}$

Marking Criteria

Descriptor	Marks
Expresses y in terms of a parameter	1
Expresses x in terms of a parameter	1
Determines the infinite solutions expressed in the form of a vector equation of a line	1

Q16

2025

VCAA

Paper 2

1 mark

Q16

1 mark

The position vector of a particle at time $t$ is given by $\underset{\sim}{\text{r}}(t) = ne^{-2t}\underset{\sim}{\text{i}} - t^2\underset{\sim}{\text{j}}$ , where $n$ is a positive constant.

For what value of $n$ is the particle's acceleration perpendicular to its velocity when $t = \frac{1}{2}$ ?

A

$2e$

B

$\frac{e^{0.5}}{2}$

C

$\frac{e}{2}$

D

$\frac{e}{2\sqrt{2}}$

Reveal Answer

A

$2e$

Incorrect. This value results from an algebraic error when solving the dot product equation $-8n^2e^{-2} + 2 = 0$ , perhaps by incorrectly isolating $n$ .

B

$\frac{e^{0.5}}{2}$

Incorrect. This value likely comes from incorrectly evaluating the derivatives at a different time or making an exponent error during differentiation.

C

$\frac{e}{2}$

Correct Answer

Correct. The velocity is $\mathbf{v}(t) = -2ne^{-2t}\mathbf{i} - 2t\mathbf{j}$ and acceleration is $\mathbf{a}(t) = 4ne^{-2t}\mathbf{i} - 2\mathbf{j}$ . Setting their dot product at $t=1/2$ to zero gives $-8n^2e^{-2} + 2 = 0$ , which yields $n = \frac{e}{2}$ .

D

$\frac{e}{2\sqrt{2}}$

Incorrect. This answer stems from a miscalculation in the dot product, such as setting $8n^2e^{-2} = 1$ instead of $2$ , leading to $n = \frac{e}{2\sqrt{2}}$ .

Q15

2020

VCAA

Paper 2

1 mark

Q15

1 mark

Two forces, $\mathbf{F}_A=4\mathbf{i}-2\mathbf{j}$ and $\mathbf{F}_B=2\mathbf{i}+5\mathbf{j}$ , act on a particle of mass 3 kg. The particle is initially at rest at position $\mathbf{i}+\mathbf{j}$ . All force components are measured in newtons and displacements are measured in metres.

The cartesian equation of the path of the particle is

A

$y=\frac{x}{2}$

B

$y=\frac{x}{2}-\frac{1}{2}$

C

$y=\frac{(x+1)^2}{2}+1$

D

$y=\frac{(x-1)^2}{2}+1$

E

$y=\frac{x}{2}+\frac{1}{2}$

Reveal Answer

A

$y=\frac{x}{2}$

This equation represents a path starting from the origin, but the particle initially starts at position $\mathbf{i}+\mathbf{j}$ .

B

$y=\frac{x}{2}-\frac{1}{2}$

This equation has the correct gradient but an incorrect y-intercept, likely resulting from a sign error when substituting the initial position coordinates.

C

$y=\frac{(x+1)^2}{2}+1$

This equation incorrectly assumes a quadratic relationship between $x$ and $y$ . Since the particle starts from rest and experiences constant acceleration, its path is a straight line.

D

$y=\frac{(x-1)^2}{2}+1$

This incorrectly assumes $t = x-1$ instead of $t^2 = x-1$ , which would imply the $x$ -component of velocity is constant rather than accelerating from rest.

E

$y=\frac{x}{2}+\frac{1}{2}$

Correct Answer

The total force gives an acceleration of $2\mathbf{i}+\mathbf{j}$ . Since the particle starts from rest at $(1,1)$ , its path is a straight line with a gradient of $1/2$ passing through $(1,1)$ , yielding $y = \frac{x}{2} + \frac{1}{2}$ .

Q17

2025

VCAA

Paper 2

1 mark

Q17

1 mark

The acceleration vector of a particle that starts from rest is given by
$\underset{\sim}{\text{a}}(t) = 4\cos(2t)\underset{\sim}{\text{i}} + 10\sin(2t)\underset{\sim}{\text{j}} - 6e^{-2t}\underset{\sim}{\text{k}}$ , where $t \geq 0$ .

The velocity vector of the particle, $\underset{\sim}{\text{v}}(t)$ , is given by

A

$\underset{\sim}{\text{v}}(t) = 2\sin(2t)\underset{\sim}{\text{i}} - 5\cos(2t)\underset{\sim}{\text{j}} + 3e^{-2t}\underset{\sim}{\text{k}}$

B

$\underset{\sim}{\text{v}}(t) = 2\sin(2t)\underset{\sim}{\text{i}} - 5(\cos(2t) + 1)\underset{\sim}{\text{j}} + 3(e^{-2t} + 1)\underset{\sim}{\text{k}}$

C

$\underset{\sim}{\text{v}}(t) = 2\sin(2t)\underset{\sim}{\text{i}} - 5(\cos(2t) - 1)\underset{\sim}{\text{j}} + 3(e^{-2t} - 1)\underset{\sim}{\text{k}}$

D

$\underset{\sim}{\text{v}}(t) = -8\sin(2t)\underset{\sim}{\text{i}} + 20\cos(2t)\underset{\sim}{\text{j}} + 12e^{-2t}\underset{\sim}{\text{k}}$

Reveal Answer

A

$\underset{\sim}{\text{v}}(t) = 2\sin(2t)\underset{\sim}{\text{i}} - 5\cos(2t)\underset{\sim}{\text{j}} + 3e^{-2t}\underset{\sim}{\text{k}}$

This option represents the indefinite integral without accounting for the initial condition $\underset{\sim}{\text{v}}(0) = 0$ . It fails to include the necessary constants of integration.

B

$\underset{\sim}{\text{v}}(t) = 2\sin(2t)\underset{\sim}{\text{i}} - 5(\cos(2t) + 1)\underset{\sim}{\text{j}} + 3(e^{-2t} + 1)\underset{\sim}{\text{k}}$

This option has the wrong signs for the constants of integration. Evaluating this vector at $t=0$ gives $-10\underset{\sim}{\text{j}} + 6\underset{\sim}{\text{k}}$ instead of the required zero vector.

C

$\underset{\sim}{\text{v}}(t) = 2\sin(2t)\underset{\sim}{\text{i}} - 5(\cos(2t) - 1)\underset{\sim}{\text{j}} + 3(e^{-2t} - 1)\underset{\sim}{\text{k}}$

Correct Answer

Integrating the acceleration vector gives $\underset{\sim}{\text{v}}(t) = (2\sin(2t) + C_1)\underset{\sim}{\text{i}} + (-5\cos(2t) + C_2)\underset{\sim}{\text{j}} + (3e^{-2t} + C_3)\underset{\sim}{\text{k}}$ . Applying the initial condition $\underset{\sim}{\text{v}}(0) = 0$ yields the correct constants: $C_1=0$ , $C_2=5$ , and $C_3=-3$ .

D

$\underset{\sim}{\text{v}}(t) = -8\sin(2t)\underset{\sim}{\text{i}} + 20\cos(2t)\underset{\sim}{\text{j}} + 12e^{-2t}\underset{\sim}{\text{k}}$

This option is the derivative of the acceleration vector (known as jerk), rather than the integral. Velocity is found by integrating acceleration, not differentiating it.

Q13

2025

VCAA

Paper 2

1 mark

Q13

1 mark

From an open window, a person projects a ball vertically up using an outstretched arm so the ball does not strike any part of the building. The point of projection of the ball is $50 \text{ m}$ above the ground and its velocity of projection is $20 \text{ m s}^{-1}$ .

The time, in seconds, it takes for the ball to reach the tray of a truck that is $1 \text{ m}$ above the ground directly below the point of projection is closest to

A

$1.72$

B

$5.80$

C

$5.83$

D

$1.75$

Reveal Answer

A

$1.72$

This is the magnitude of the negative root of the kinematic equation, which represents the time if the trajectory was extended backwards before projection.

B

$5.80$

Correct Answer

Using the kinematic equation $\Delta y = ut + \frac{1}{2}at^2$ with a displacement of $\Delta y = -49 \text{ m}$ (since the tray is $1 \text{ m}$ above the ground), $u = 20 \text{ m s}^{-1}$ , and $a = -9.8 \text{ m s}^{-2}$ yields $t \approx 5.80 \text{ s}$ .

C

$5.83$

This is the time it would take for the ball to reach the ground ( $\Delta y = -50 \text{ m}$ ), failing to account for the truck tray being $1 \text{ m}$ above the ground.

D

$1.75$

This is the magnitude of the negative root if the displacement was incorrectly set to $-50 \text{ m}$ (reaching the ground instead of the truck tray).

Q17

2023

QCAA

Paper 2

6 marks

Q17

6 marks

An object is projected upwards from ground level with an initial velocity of $15 \text{ m s}^{-1}$ at an angle of $54^\circ$ to the horizontal.

The object just passes over a drone hovering in the air. An observer is positioned directly below the drone and at a horizontal distance of $20 \text{ m}$ from where the object is projected.

The observer commented that:

it took the object around 2 to 2.5 seconds after its projection to reach the drone
the object was still moving in an upwards direction as it passed the drone.

Assuming that air resistance is negligible, use a vector calculus approach to evaluate the reasonableness of the observer's comments.

Reveal Answer

Let $\hat{i}$ and $\hat{j}$ be the horizontal and vertical unit vectors respectively. Let $t$ represent the time in seconds after the projection of the object.
$a(t) = -9.8\hat{j}$
$v(t) = \int a(t) dt = -9.8t\hat{j} + c$
Given $v(0) = 15\cos(54^\circ)\hat{i} + 15\sin(54^\circ)\hat{j}$
$v(t) = 15\cos(54^\circ)\hat{i} + (15\sin(54^\circ) - 9.8t)\hat{j}$
$r(t) = \int v(t) dt$
$= 15\cos(54^\circ)t\hat{i} + (15\sin(54^\circ)t - 4.9t^2)\hat{j} + c$
Let origin be at the release point: $r(0) = 0\hat{i} + 0\hat{j}$
$r(t) = 15\cos(54^\circ)t\hat{i} + (15\sin(54^\circ)t - 4.9t^2)\hat{j}$
When $r_x = 20 \Rightarrow 15\cos(54^\circ)t = 20$
Time object just passes drone: $t = 2.27\text{s}$
Finding maximum value of $r_y: 15\sin(54^\circ)t - 4.9t^2$
Using GDC
Time object reaches maximum height: $t = 1.24\text{s}$

While the estimation of the time taken for the object to reach the drone is reasonable, the comment regarding the direction of the object as it passed the drone is not reasonable as it would have been moving in a downward direction at that time.

Marking Criteria

Descriptor	Marks
correctly determines the velocity function of the object using vector calculus	1
determines displacement function of the object	1
determines time when the object just passes drone	1
determines time when the object reaches maximum height	1
uses mathematical justification to evaluate the reasonableness of both comments based on prior mathematical reasoning	1
shows logical organisation, communicating key steps	1

SCSA Mathematics Specialist Vectors in three dimensions

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