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

AusGrader has 15 QCAA Specialist Mathematics Alternative Sequence questions on Vectors in two and three dimensions, all with instant AI grading and detailed marking feedback.

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

Q19

2021

QCAA

Paper 1

7 marks

Q19

7 marks

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

$v_A=6\sin(3t)\,\hat{\mathbf{i}}+6\cos(3t)\,\hat{\mathbf{j}}$

$v_B=\cos(t)\,\hat{\mathbf{i}}-\sin(t)\,\hat{\mathbf{j}}$

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

$\boldsymbol{v}_A = 6\sin(3t)\hat{\boldsymbol{\imath}} + 6\cos(3t)\hat{\boldsymbol{\jmath}}$
$\boldsymbol{v}_B = \cos(t)\hat{\boldsymbol{\imath}} - \sin(t)\hat{\boldsymbol{\jmath}}$

$\boldsymbol{r}_A = \int \boldsymbol{v}_A \, dt = -2\cos(3t)\hat{\boldsymbol{\imath}} + 2\sin(3t)\hat{\boldsymbol{\jmath}} + \boldsymbol{c}_A$
When $t = 0$
$-2\hat{\boldsymbol{\imath}} + 2\hat{\boldsymbol{k}} = -2\cos(0)\hat{\boldsymbol{\imath}} + 2\sin(0)\hat{\boldsymbol{\jmath}} + \boldsymbol{c}_A \Rightarrow \boldsymbol{c}_A = 2\hat{\boldsymbol{k}}$
$\therefore \boldsymbol{r}_A = -2\cos(3t)\hat{\boldsymbol{\imath}} + 2\sin(3t)\hat{\boldsymbol{\jmath}} + 2\hat{\boldsymbol{k}}$

$\boldsymbol{r}_B = \int \boldsymbol{v}_B \, dt = \sin(t)\hat{\boldsymbol{\imath}} + \cos(t)\hat{\boldsymbol{\jmath}} + \boldsymbol{c}_B$
When $t = 0$
$\hat{\boldsymbol{\jmath}} - \hat{\boldsymbol{k}} = \sin(0)\hat{\boldsymbol{\imath}} + \cos(0)\hat{\boldsymbol{\jmath}} + \boldsymbol{c}_B \Rightarrow \boldsymbol{c}_B = -\hat{\boldsymbol{k}}$
$\therefore \boldsymbol{r}_B = \sin(t)\hat{\boldsymbol{\imath}} + \cos(t)\hat{\boldsymbol{\jmath}} - \hat{\boldsymbol{k}}$

$\boldsymbol{r}_B - \boldsymbol{r}_A$
$= (\sin(t)\hat{\boldsymbol{\imath}} + \cos(t)\hat{\boldsymbol{\jmath}} - \hat{\boldsymbol{k}}) \dots$
$\dots - (-2\cos(3t)\hat{\boldsymbol{\imath}} + 2\sin(3t)\hat{\boldsymbol{\jmath}} + 2\hat{\boldsymbol{k}})$
$= (\sin(t) + 2\cos(3t))\hat{\boldsymbol{\imath}} + (\cos(t) - 2\sin(3t))\hat{\boldsymbol{\jmath}} - 3\hat{\boldsymbol{k}}$

$|\boldsymbol{r}_B - \boldsymbol{r}_A| = \sqrt{\sin^2(t) + 4\sin(t)\cos(3t) + 4\cos^2(3t) + \dots}$
$\overline{\dots \cos^2(t) - 4\cos(t)\sin(3t) + 4\sin^2(3t) + 9}$
$= \sqrt{14 - 4(\sin(3t)\cos(t) - \cos(3t)\sin(t))}$
$= \sqrt{14 - 4(\sin(3t - t))}$
$= \sqrt{14 - 4\sin(2t)}$

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

Position of A
$\boldsymbol{r}_A = -2\cos(3t)\hat{\boldsymbol{\imath}} + 2\sin(3t)\hat{\boldsymbol{\jmath}} + 2\hat{\boldsymbol{k}}$
$= -2\cos\left(\frac{7\pi}{4}\right)\hat{\boldsymbol{\imath}} + 2\sin\left(\frac{7\pi}{4}\right)\hat{\boldsymbol{\jmath}} + 2\hat{\boldsymbol{k}}$
$= -\sqrt{2}\hat{\boldsymbol{\imath}} - \sqrt{2}\hat{\boldsymbol{\jmath}} + 2\hat{\boldsymbol{k}}$ (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

Q8

2021

QCAA

Paper 2

1 mark

Q8

1 mark

The vectors $3a\hat{i} - a\hat{j} + 2\hat{k}$ and $a\hat{i} - \hat{j} - 2\hat{k}$ (where $a \in R$ ) are perpendicular vectors.

Determine all possible values of $a$ .

A

$-1$ and $-\frac{4}{3}$

B

$-1$ and $\frac{4}{3}$

C

$1$ and $-\frac{4}{3}$

D

$1$ and $\frac{4}{3}$

Reveal Answer

A

$-1$ and $-\frac{4}{3}$

Incorrect. This results from a sign error when factoring the quadratic equation $3a^2 + a - 4 = 0$ , incorrectly yielding $a = -1$ .

B

$-1$ and $\frac{4}{3}$

Incorrect. This results from incorrectly solving the dot product equation $3a^2 + a - 4 = 0$ , likely by making sign errors on both roots.

C

$1$ and $-\frac{4}{3}$

Correct Answer

Correct. Two vectors are perpendicular when their dot product is zero. Setting $(3a)(a) + (-a)(-1) + (2)(-2) = 3a^2 + a - 4 = 0$ and factoring yields $(a - 1)(3a + 4) = 0$ , giving $a = 1$ and $a = -\frac{4}{3}$ .

D

$1$ and $\frac{4}{3}$

Incorrect. This results from a sign error when solving $3a + 4 = 0$ , incorrectly yielding $a = \frac{4}{3}$ instead of $-\frac{4}{3}$ .

Q2

2021

QCAA

Paper 1

1 mark

Q2

1 mark

Which line is parallel to the vector $\begin{pmatrix}2\\3\\4\end{pmatrix}$ ?

A

$\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}$

B

$2x=3y=4z$

C

$x-2=y-3=z-4$

D

$x+2=y+3=z+4$

Reveal Answer

A

$\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}$

Correct Answer

In the symmetric equations of a line $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$ , the denominators $a, b, c$ represent the direction vector. Here, the denominators match the given vector $\begin{pmatrix}2\\3\\4\end{pmatrix}$ .

B

$2x=3y=4z$

This equation can be rewritten as $\frac{x}{1/2} = \frac{y}{1/3} = \frac{z}{1/4}$ , meaning its direction vector is $\begin{pmatrix}1/2\\1/3\\1/4\end{pmatrix}$ , which is not parallel to $\begin{pmatrix}2\\3\\4\end{pmatrix}$ .

C

$x-2=y-3=z-4$

The denominators in this symmetric equation are all implicitly $1$ , meaning its direction vector is $\begin{pmatrix}1\\1\\1\end{pmatrix}$ . The numbers $2, 3, 4$ represent a point $(2, 3, 4)$ the line passes through, not its direction.

D

$x+2=y+3=z+4$

Like option C, the denominators are all implicitly $1$ , giving a direction vector of $\begin{pmatrix}1\\1\\1\end{pmatrix}$ . The numbers $2, 3, 4$ indicate the line passes through the point $(-2, -3, -4)$ .

Q12

2021

QCAA

Paper 1

8 marks

Q12

Consider the plane $x-y-2z=15$ .

Q12d

The line $l$ and the plane intersect at point $S$ .

Q12a

1 mark

Determine a vector $\mathbf{n}$ that is perpendicular to the plane.

Reveal Answer

A vector perpendicular to $x - y - 2z = 15$ is
$\boldsymbol{n} = \begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix}$

Marking Criteria

Descriptor	Marks
correctly determines a suitable vector $\boldsymbol{n}$	1

Q12b

1 mark

Determine the vector equation of the line $l$ that is perpendicular to the plane and contains the point $A(-2,1,3)$ .

Reveal Answer

Vector equation of line $l$ is $\boldsymbol{r} = \boldsymbol{a} + k\boldsymbol{d}$
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix} + k \begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix} \quad k \in R$

Marking Criteria

Descriptor	Marks
determines vector equation of the line	1

Q12c

1 mark

Use the result from Question 12b) to express the equation of the line $l$ in parametric form.

Reveal Answer

Equation of line $l$ in parametric form
$x = -2 + k$
$y = 1 - k$
$z = 3 - 2k \quad k \in R$

Marking Criteria

Descriptor	Marks
expresses equation of the line in parametric form	1

Q12d

3 marks

Show that the coordinates of $S$ are $(2,-3,-5)$ .

Reveal Answer

Method 1
Given S lies on the plane $x - y - 2z = 15$
$(-2 + k) - (1 - k) - 2(3 - 2k) = 15$
$6k - 9 = 15 \Rightarrow 6k = 24$
$k = 4$
The coordinates of S are
$(-2 + 4, 1 - 4, 3 - 8) = (2, -3, -5)$

Marking Criteria

Descriptor	Marks
substitutes result from 12c) into the equation of the plane	1
determines value of the parameter	1
determines coordinates of S	1

Q12e

1 mark

Determine $\overrightarrow{AS}$ .

Reveal Answer

$\overrightarrow{AS} = s - a = \begin{pmatrix} 2 \\ -3 \\ -5 \end{pmatrix} - \begin{pmatrix} -2 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ -4 \\ -8 \end{pmatrix}$

Marking Criteria

Descriptor	Marks
determines $\overrightarrow{AS}$	1

Q12f

1 mark

Use a property of parallel vectors to verify that $\overrightarrow{AS}$ and $\mathbf{n}$ are parallel.

Reveal Answer

$\begin{pmatrix} 4 \\ -4 \\ -8 \end{pmatrix} = 4 \begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix} \Rightarrow \text{result is verified}$

Marking Criteria

Descriptor	Marks
shows that $\overrightarrow{AS}$ is a scalar multiple of $\boldsymbol{n}$	1

Q6

2021

QCAA

Paper 2

1 mark

Q6

1 mark

The Cartesian equation of a particular sphere is given by $x^2 + y^2 + z^2 + 2x - 2y = 7$ .

The centre and radius of the sphere are

A

$(-1, 1, 0)$ and 3 respectively.

B

$(-1, 1, 0)$ and 9 respectively.

C

$(1, -1, 0)$ and 3 respectively.

D

$(1, -1, 0)$ and 9 respectively.

Reveal Answer

A

$(-1, 1, 0)$ and 3 respectively.

Correct Answer

Completing the square gives $(x + 1)^2 + (y - 1)^2 + z^2 = 9$ . This matches the standard sphere equation $(x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 = r^2$ , revealing a centre of $(-1, 1, 0)$ and a radius of $\sqrt{9} = 3$ .

B

$(-1, 1, 0)$ and 9 respectively.

While the centre is correct, the radius is incorrect. The constant term after completing the square is $r^2 = 9$ , so the radius $r$ must be $\sqrt{9} = 3$ , not 9.

C

$(1, -1, 0)$ and 3 respectively.

The signs for the centre coordinates are reversed. Completing the square yields $(x + 1)^2$ and $(y - 1)^2$ , which means the centre is at $x = -1$ and $y = 1$ .

D

$(1, -1, 0)$ and 9 respectively.

Both the centre and radius are incorrect. The signs for the centre coordinates are reversed, and the radius should be the square root of the constant term ( $\sqrt{9} = 3$ ), not 9.

Q5

2021

QCAA

Paper 2

1 mark

Q5

1 mark

A vector normal to the plane that contains the vectors $\begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$ is

A

$6\hat{i} + 2\hat{j} + 3\hat{k}$

B

$6\hat{i} + 2\hat{j} - 3\hat{k}$

C

$6\hat{i} - 2\hat{j} + 3\hat{k}$

D

$6\hat{i} - 2\hat{j} - 3\hat{k}$

Reveal Answer

A

$6\hat{i} + 2\hat{j} + 3\hat{k}$

This vector is not orthogonal to the plane. Taking its dot product with the first vector $\begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix}$ yields $12$ , not $0$ .

B

$6\hat{i} + 2\hat{j} - 3\hat{k}$

This vector is not orthogonal to the plane. Taking its dot product with the first vector $\begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix}$ yields $12$ , not $0$ .

C

$6\hat{i} - 2\hat{j} + 3\hat{k}$

While this vector is orthogonal to the first vector, its dot product with the second vector $\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$ yields $12$ , not $0$ .

D

$6\hat{i} - 2\hat{j} - 3\hat{k}$

Correct Answer

A normal vector can be found by taking the cross product of the two vectors. Calculating the determinant of $\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & 0 \\ 1 & 0 & 2 \end{vmatrix}$ yields $6\hat{i} - 2\hat{j} - 3\hat{k}$ .

Q16

2021

QCAA

Paper 1

6 marks

Q16

Three planes intersect in the line $l$ .

$\begin{aligned}-x+2y+3z&=2\\x+3y+2z&=-7\\2x-y+mz&=n\end{aligned}$

Q16a

3 marks

Use a Gaussian technique of elimination to determine the values of $m$ and $n$ .

Reveal Answer

Expressing the equations as an augmented matrix:
$\begin{bmatrix} -1 & 2 & 3 & | & 2 \\ 1 & 3 & 2 & | & -7 \\ 2 & -1 & m & | & n \end{bmatrix} \quad \begin{matrix} R_1 \\ R_2 \\ R_3 \end{matrix}$
$\begin{bmatrix} -1 & 2 & 3 & | & 2 \\ 0 & 5 & 5 & | & -5 \\ 0 & 3 & m + 6 & | & n + 4 \end{bmatrix} \quad \begin{matrix} R_1 \\ R_2 \rightarrow R_2 + R_1 \\ R_3 \rightarrow R_3 + 2R_1 \end{matrix}$
$\begin{bmatrix} -1 & 2 & 3 & | & 2 \\ 0 & 1 & 1 & | & -1 \\ 0 & 3 & m + 6 & | & n + 4 \end{bmatrix} \quad \begin{matrix} R_1 \\ R_2 \rightarrow \frac{1}{5}R_2 \\ R_3 \end{matrix}$
$\begin{bmatrix} -1 & 2 & 3 & | & 2 \\ 0 & 1 & 1 & | & -1 \\ 0 & 0 & m + 3 & | & n + 7 \end{bmatrix} \quad \begin{matrix} R_1 \\ R_2 \\ R_3 \rightarrow R_3 - 3R_2 \end{matrix}$
As the planes intersect in a line, there are infinitely many solutions so the values in the last row must all be 0
$\therefore m = -3, n = -7$

Marking Criteria

Descriptor	Marks
correctly expresses the equations as an augmented matrix	1
establishes augmented matrix with two 0s in the third row	1
determines values of $m$ and $n$	1

Q16b

3 marks

Determine the equation of the line $l$ in Cartesian form.

Reveal Answer

Determining the equation of the line, $l$
Let $z = t$
From $R_2$ : $y + z = -1 \Rightarrow y = -1 - z$
$\therefore y = -1 - t \Rightarrow t = -y - 1$
From $R_1$ : $-x + 2y + 3z = 2 \Rightarrow x = 2y + 3z - 2$
$\therefore x = 2(-1 - t) + 3t - 2$
$\therefore x = t - 4 \Rightarrow t = x + 4$
Cartesian equation of the line is
$x + 4 = -y - 1 = z$

Marking Criteria

Descriptor	Marks
expresses $y$ in terms of a parameter	1
expresses $x$ in terms of a parameter	1
determines a Cartesian equation of the line	1

QCAA Specialist Mathematics Alternative Sequence Vectors in two and three dimensions

Frequently Asked Questions

Ready to practise QCAA Specialist Mathematics Alternative Sequence?

QCAA Specialist Mathematics Alternative Sequence Vectors in two and three dimensions

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Frequently Asked Questions

Ready to practise QCAA Specialist Mathematics Alternative Sequence?