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

AusGrader has 150 SCSA Mathematics Specialist questions on Vectors in three dimensions, all with instant AI grading and detailed marking feedback.

SCSA (WACE) Mathematics Specialist — Vectors in three dimensions Questions & Answers | AusGrader

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

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

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

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

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

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

D

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

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.

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.

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.

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

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

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.

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.

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

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 $\Pi$ .

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

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 $\Pi$ .

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

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

Marking Criteria

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

SCSA Mathematics Specialist Vectors in three dimensions

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