VCAA Specialist Mathematics Space and measurement

15 sample questions with marking guides and sample answers

Q6
2024
QCAA
Paper 2
1 mark
Q6
1 mark

Two concurrent forces represented in the polar form of F1=(1.21 N,120∘)F_1 = (1.21 \text{ N}, 120^\circ) and F2=(1.30 N,−160∘)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 F22−F12\sqrt{F_2^2 - F_1^2} (approx. 0.480.48 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∘−(−160∘)∣=280∘|120^\circ - (-160^\circ)| = 280^\circ, which is equivalent to 80∘80^\circ. Calculating R=1.212+1.302+2(1.21)(1.30)cos⁡(80∘)R = \sqrt{1.21^2 + 1.30^2 + 2(1.21)(1.30)\cos(80^\circ)} yields approximately 1.921.92 N.

C

2.51 N

This represents the scalar sum of the magnitudes (1.21+1.30=2.511.21 + 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 (R2≈3.70R^2 \approx 3.70 N2^2). The final step of taking the square root to find the magnitude was omitted.

Q4
2024
QCAA
Paper 1
1 mark
Q4
1 mark

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

The vector equation of the plane is

A

(xyz)⋅(131)=(112)⋅(131)\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

(xyz)⋅(112)=(131)⋅(112)\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

(xyz)×(131)=(112)×(131)\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

(xyz)×(112)=(131)×(112)\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

(xyz)⋅(131)=(112)⋅(131)\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 (131)\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} as the normal vector. The dot product must be taken with the normal vector (112)\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}.

B

(xyz)⋅(112)=(131)⋅(112)\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 r⋅n=a⋅n\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}, where n\mathbf{n} is the normal vector and a\mathbf{a} is a point on the plane. Substituting n=(112)\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} and a=(131)\mathbf{a} = \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} yields this result.

C

(xyz)×(131)=(112)×(131)\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

(xyz)×(112)=(131)×(112)\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 (r−a)⋅n=0(\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0.

Q17
2025
VCAA
Paper 2
1 mark
Q17
1 mark

The acceleration vector of a particle that starts from rest is given by
a∼(t)=4cos⁡(2t)i∼+10sin⁡(2t)j∼−6e−2tk∼\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≥0t \geq 0.

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

A

v∼(t)=2sin⁡(2t)i∼−5cos⁡(2t)j∼+3e−2tk∼\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

v∼(t)=2sin⁡(2t)i∼−5(cos⁡(2t)+1)j∼+3(e−2t+1)k∼\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

v∼(t)=2sin⁡(2t)i∼−5(cos⁡(2t)−1)j∼+3(e−2t−1)k∼\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

v∼(t)=−8sin⁡(2t)i∼+20cos⁡(2t)j∼+12e−2tk∼\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

v∼(t)=2sin⁡(2t)i∼−5cos⁡(2t)j∼+3e−2tk∼\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 v∼(0)=0\underset{\sim}{\text{v}}(0) = 0. It fails to include the necessary constants of integration.

B

v∼(t)=2sin⁡(2t)i∼−5(cos⁡(2t)+1)j∼+3(e−2t+1)k∼\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=0t=0 gives −10j∼+6k∼-10\underset{\sim}{\text{j}} + 6\underset{\sim}{\text{k}} instead of the required zero vector.

C

v∼(t)=2sin⁡(2t)i∼−5(cos⁡(2t)−1)j∼+3(e−2t−1)k∼\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 v∼(t)=(2sin⁡(2t)+C1)i∼+(−5cos⁡(2t)+C2)j∼+(3e−2t+C3)k∼\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 v∼(0)=0\underset{\sim}{\text{v}}(0) = 0 yields the correct constants: C1=0C_1=0, C2=5C_2=5, and C3=−3C_3=-3.

D

v∼(t)=−8sin⁡(2t)i∼+20cos⁡(2t)j∼+12e−2tk∼\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.

Q14
2024
VCAA
Paper 2
1 mark
Q14
1 mark

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

A

−18-18

B

−2-2

C

22

D

33

Reveal Answer
A

−18-18

This is the dot product r∼⋅s∼\underset{\sim}{\mathrm{r}} \cdot \underset{\sim}{\mathrm{s}}, not the scalar resolute. The dot product is found by multiplying the scalar resolute of r∼\underset{\sim}{\mathrm{r}} on s∼\underset{\sim}{\mathrm{s}} (−6-6) by ∣s∼∣|\underset{\sim}{\mathrm{s}}| (33).

B

−2-2

Correct Answer

The vector resolute of r∼\underset{\sim}{\mathrm{r}} on s∼\underset{\sim}{\mathrm{s}} gives a scalar resolute of −6-6, meaning r∼⋅s∼=−6×∣s∼∣=−18\underset{\sim}{\mathrm{r}} \cdot \underset{\sim}{\mathrm{s}} = -6 \times |\underset{\sim}{\mathrm{s}}| = -18. The scalar resolute of s∼\underset{\sim}{\mathrm{s}} on r∼\underset{\sim}{\mathrm{r}} is then r∼⋅s∼∣r∼∣=−189=−2\frac{\underset{\sim}{\mathrm{r}} \cdot \underset{\sim}{\mathrm{s}}}{|\underset{\sim}{\mathrm{r}}|} = \frac{-18}{9} = -2.

C

22

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

D

33

This is the magnitude of vector s∼\underset{\sim}{\mathrm{s}} (∣s∼∣=22+(−2)2+12=3|\underset{\sim}{\mathrm{s}}| = \sqrt{2^2 + (-2)^2 + 1^2} = 3), not its scalar resolute in the direction of r∼\underset{\sim}{\mathrm{r}}.

Q5
2023
SCSA
Paper 1
5 marks
Q5

Consider two planes given by their Cartesian equations:
x−3y+3z=9x - 3y + 3z = 9
2x+y−z=42x + y - z = 4

Q5a
1 mark

Explain why these planes are not parallel.

Reveal Answer

The normal vectors for each plane (1−33)\begin{pmatrix} 1 \\ -3 \\ 3 \end{pmatrix} and (21−1)\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
DescriptorMarks

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
DescriptorMarks

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

Consider (1)+3×(2)(1) + 3 \times (2):

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

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

3−3y+3z=9i.e.z=y+2where y∈R\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,zx, y, z.
Hence the intersection of the two planes is a line in space.

Vector equation for this line:
r∼=(3λλ+2)=(302)+λ(011)\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
DescriptorMarks

eliminates a variable correctly from the pair of equations

1

obtains the relationship z=y+2z = y+2

1

forms the vector equation of the line using a parameter correctly

1
Q17
2023
QCAA
Paper 2
6 marks
Q17
6 marks

An object is projected upwards from ground level with an initial velocity of 15 m s−115 \text{ m s}^{-1} at an angle of 54∘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 m20 \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 i^\hat{i} and j^\hat{j} be the horizontal and vertical unit vectors respectively. Let tt represent the time in seconds after the projection of the object.
a(t)=−9.8j^a(t) = -9.8\hat{j}
v(t)=∫a(t)dt=−9.8tj^+cv(t) = \int a(t) dt = -9.8t\hat{j} + c
Given v(0)=15cos⁡(54∘)i^+15sin⁡(54∘)j^v(0) = 15\cos(54^\circ)\hat{i} + 15\sin(54^\circ)\hat{j}
v(t)=15cos⁡(54∘)i^+(15sin⁡(54∘)−9.8t)j^v(t) = 15\cos(54^\circ)\hat{i} + (15\sin(54^\circ) - 9.8t)\hat{j}
r(t)=∫v(t)dtr(t) = \int v(t) dt
=15cos⁡(54∘)ti^+(15sin⁡(54∘)t−4.9t2)j^+c= 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)=0i^+0j^r(0) = 0\hat{i} + 0\hat{j}
r(t)=15cos⁡(54∘)ti^+(15sin⁡(54∘)t−4.9t2)j^r(t) = 15\cos(54^\circ)t\hat{i} + (15\sin(54^\circ)t - 4.9t^2)\hat{j}
When rx=20⇒15cos⁡(54∘)t=20r_x = 20 \Rightarrow 15\cos(54^\circ)t = 20
Time object just passes drone: t=2.27st = 2.27\text{s}
Finding maximum value of ry:15sin⁡(54∘)t−4.9t2r_y: 15\sin(54^\circ)t - 4.9t^2
Using GDC
Time object reaches maximum height: t=1.24st = 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
DescriptorMarks

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
Q3
2024
QCAA
Paper 2
1 mark
Q3
1 mark

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

A

i^−2j^−2k^\hat{i} - 2\hat{j} - 2\hat{k}

B

i^−2j^+2k^\hat{i} - 2\hat{j} + 2\hat{k}

C

i^+2j^−2k^\hat{i} + 2\hat{j} - 2\hat{k}

D

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

Reveal Answer
A

i^−2j^−2k^\hat{i} - 2\hat{j} - 2\hat{k}

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

B

i^−2j^+2k^\hat{i} - 2\hat{j} + 2\hat{k}

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

C

i^+2j^−2k^\hat{i} + 2\hat{j} - 2\hat{k}

Correct Answer

Using the determinant method with rows i^,j^,k^\hat{i}, \hat{j}, \hat{k}; 0,1,10, 1, 1; and 2,0,12, 0, 1, the result is i^(1−0)−j^(0−2)+k^(0−2)=i^+2j^−2k^\hat{i}(1-0) - \hat{j}(0-2) + \hat{k}(0-2) = \hat{i} + 2\hat{j} - 2\hat{k}.

D

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

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

Q11
2025
SCSA
Paper 2
6 marks
Q11

Two lines in space are defined by: r1∼=(21−3)+λ(1−41),r2∼=(48−16)+μ(−112)\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 (2+λ1−4λ−3+λ)=(4−μ8+μ−16+2μ)\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+λ=4−μ...(1)2+\lambda = 4-\mu \quad ...(1)
1−4λ=8+μ...(2)1-4\lambda = 8+\mu \quad ...(2)
yields λ=−3\lambda = -3, μ=5\mu = 5

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

Intersect at r2∼=(48−16)+5(−112)=(−113−6)\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
DescriptorMarks

forms the equation(s) that produce the intersection of lines

1

solves for the parameters λ\lambda and μ\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 n∼⊥d1∼\underset{\sim}{n} \perp \underset{\sim}{d_1} and n∼⊥d2∼\underset{\sim}{n} \perp \underset{\sim}{d_2}.

Hence n∼∥(d1∼×d2∼)\underset{\sim}{n} \parallel (\underset{\sim}{d_1}\times \underset{\sim}{d_2})\quad i.e. use n∼=(1−41)×(−112)=(−9−3−3)\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 n∼=(311)\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)(2,1,-3)

Equation of the plane is given by: (xyz)⋅(311)=(311)⋅(21−3)=6+1−3\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=43x+y+z=4

Marking Criteria
DescriptorMarks

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
Q6
2025
QCAA
Paper 1
1 mark
Q6
1 mark

At time tt, a particle travels with a velocity of v=(21+t2)i^−2tj^v = \left(\frac{2}{1+t^2}\right)\hat{i} - 2t\hat{j}.

Determine a general expression for the position vector, rr, of the particle during this motion.

A

r=2tan⁡−1(t)i^−2j^+cr = 2\tan^{-1}(t)\hat{i} - 2\hat{j} + c

B

r=2tan⁡−1(t)i^−t2j^+cr = 2\tan^{-1}(t)\hat{i} - t^2\hat{j} + c

C

r=12tan⁡−1(t)i^−2j^+cr = \frac{1}{2}\tan^{-1}(t)\hat{i} - 2\hat{j} + c

D

r=12tan⁡−1(t)i^−t2j^+cr = \frac{1}{2}\tan^{-1}(t)\hat{i} - t^2\hat{j} + c

Reveal Answer
A

r=2tan⁡−1(t)i^−2j^+cr = 2\tan^{-1}(t)\hat{i} - 2\hat{j} + c

This is incorrect because the integral of the j^\hat{j} component, −2t-2t, is −t2-t^2, not −2-2.

B

r=2tan⁡−1(t)i^−t2j^+cr = 2\tan^{-1}(t)\hat{i} - t^2\hat{j} + c

Correct Answer

This is correct. The position vector is the integral of the velocity vector with respect to time. Integrating 21+t2\frac{2}{1+t^2} yields 2tan⁡−1(t)2\tan^{-1}(t), and integrating −2t-2t yields −t2-t^2.

C

r=12tan⁡−1(t)i^−2j^+cr = \frac{1}{2}\tan^{-1}(t)\hat{i} - 2\hat{j} + c

This is incorrect because both components are integrated improperly. The integral of 21+t2\frac{2}{1+t^2} is 2tan⁡−1(t)2\tan^{-1}(t), and the integral of −2t-2t is −t2-t^2.

D

r=12tan⁡−1(t)i^−t2j^+cr = \frac{1}{2}\tan^{-1}(t)\hat{i} - t^2\hat{j} + c

This is incorrect because the integral of the i^\hat{i} component, 21+t2\frac{2}{1+t^2}, is 2tan⁡−1(t)2\tan^{-1}(t), not 12tan⁡−1(t)\frac{1}{2}\tan^{-1}(t).

Q14
2025
VCAA
Paper 2
1 mark
Q14
1 mark

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

A

00

B

π4\frac{\pi}{4}

C

π2\frac{\pi}{2}

D

3π4\frac{3\pi}{4}

Reveal Answer
A

00

If the angle is 00, the dot product is maximized (cos⁡(0)=1\cos(0)=1) and the cross product is zero (sin⁡(0)=0\sin(0)=0), so they are not equal.

B

π4\frac{\pi}{4}

Correct Answer

Using the formulas a∼⋅b∼=∣a∼∣∣b∼∣cos⁡θ\underset{\sim}{a} \cdot \underset{\sim}{b} = |\underset{\sim}{a}||\underset{\sim}{b}|\cos\theta and ∣a∼×b∼∣=∣a∼∣∣b∼∣sin⁡θ|\underset{\sim}{a} \times \underset{\sim}{b}| = |\underset{\sim}{a}||\underset{\sim}{b}|\sin\theta, equating them gives cos⁡θ=sin⁡θ\cos\theta = \sin\theta. This simplifies to tan⁡θ=1\tan\theta = 1, which means the angle θ\theta is π4\frac{\pi}{4}.

C

π2\frac{\pi}{2}

If the angle is π2\frac{\pi}{2}, the dot product is zero (cos⁡(π2)=0\cos(\frac{\pi}{2})=0) while the magnitude of the cross product is maximized (sin⁡(π2)=1\sin(\frac{\pi}{2})=1).

D

3π4\frac{3\pi}{4}

If the angle is 3π4\frac{3\pi}{4}, the dot product is negative (cos⁡(3π4)<0\cos(\frac{3\pi}{4}) < 0) while the magnitude of the cross product is positive, so they cannot be equal.

Q2
2020
SCSA
Paper 1
5 marks
Q2

Plane Π\Pi has vector equation r∼=(004)+λ(301)+μ(1−12).\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 n∼\underset{\sim}{n} for plane Π\Pi.

Reveal Answer

Normal n∼=(301)×(1−12)=(0(2)−1(−1)1(1)−3(2)3(−1)−0(1))=(1−5−3)\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
DescriptorMarks

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 Π\Pi.

Reveal Answer

Equation given by (xyz)∙(1−5−3)=(004)∙(1−5−3)=−12 \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

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

Marking Criteria
DescriptorMarks

forms the equation using the normal vector n∼\underset{\sim}{n} correctly

1

determines the Cartesian equation 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^{-1}) at time tt (in s) are given respectively by

vA=6sin⁡(3t)i^+6cos⁡(3t)j^\boldsymbol{v}_{\mathrm{A}} = 6\sin(3t)\hat{\boldsymbol{i}} + 6\cos(3t)\hat{\boldsymbol{j}}
vB=cos⁡(t)i^−sin⁡(t)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)(-2, 0, 2) and (0,1,−1)(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

vA=6sin⁡(3t)i+6cos⁡(3t)jv_A = 6\sin(3t)i + 6\cos(3t)j
vB=cos⁡(t)i−sin⁡(t)jv_B = \cos(t)i - \sin(t)j

rA=∫vAdt=−2cos⁡(3t)i+2sin⁡(3t)j+cAr_A = \int v_A dt = -2\cos(3t)i + 2\sin(3t)j + c_A
When t=0t = 0
−2i+2k=−2cos⁡(0)i+2sin⁡(0)j+cA⇒cA=2k-2i + 2k = -2\cos(0)i + 2\sin(0)j + c_A \Rightarrow c_A = 2k
∴rA=−2cos⁡(3t)i+2sin⁡(3t)j+2k\therefore r_A = -2\cos(3t)i + 2\sin(3t)j + 2k

rB=∫vBdt=sin⁡(t)i+cos⁡(t)j+cBr_B = \int v_B dt = \sin(t)i + \cos(t)j + c_B
When t=0t = 0
j−k=sin⁡(0)i+cos⁡(0)j+cB⇒cB=−kj - k = \sin(0)i + \cos(0)j + c_B \Rightarrow c_B = -k
∴rB=sin⁡(t)i+cos⁡(t)j−k\therefore r_B = \sin(t)i + \cos(t)j - k

rB−rAr_B - r_A
=(sin⁡(t)i+cos⁡(t)j−k)…= (\sin(t)i + \cos(t)j - k) \dots
⋯−(−2cos⁡(3t)i+2sin⁡(3t)j+2k)\dots - (-2\cos(3t)i + 2\sin(3t)j + 2k)
=(sin⁡(t)+2cos⁡(3t))i+(cos⁡(t)−2sin⁡(3t))j−3k= (\sin(t) + 2\cos(3t))i + (\cos(t) - 2\sin(3t))j - 3k

∣rB−rA∣|r_B - r_A|
=sin⁡2(t)+4sin⁡(t)cos⁡(3t)+4cos⁡2(3t)+…= \sqrt{\sin^2(t) + 4\sin(t)\cos(3t) + 4\cos^2(3t) + \dots}
…cos⁡2(t)−4cos⁡(t)sin⁡(3t)+4sin⁡2(3t)+9‾\overline{\dots \cos^2(t) - 4\cos(t)\sin(3t) + 4\sin^2(3t) + 9}
=14−4(sin⁡(3t)cos⁡(t)−cos⁡(3t)sin⁡(t))= \sqrt{14 - 4(\sin(3t)\cos(t) - \cos(3t)\sin(t))}
=14−4sin⁡(2t)= \sqrt{14 - 4\sin(2t)}

Given ∣rB−rA∣=4|r_B - r_A| = 4
14−4sin⁡(2t)=4\sqrt{14 - 4\sin(2t)} = 4
sin⁡(2t)=−12\sin(2t) = -\frac{1}{2}
2t=7π62t = \frac{7\pi}{6}
t=7π12t = \frac{7\pi}{12} s (first positive solution)

Position of A
rA=−2cos⁡(3t)i+2sin⁡(3t)j+2kr_A = -2\cos(3t)i + 2\sin(3t)j + 2k
=−2cos⁡(7π4)i+2sin⁡(7π4)j+2k= -2\cos(\frac{7\pi}{4})i + 2\sin(\frac{7\pi}{4})j + 2k
=−2i−2j+2k= -\sqrt{2}i - \sqrt{2}j + 2k (m)

Marking Criteria
DescriptorMarks

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
Q13
2024
VCAA
Paper 2
1 mark
Q13
1 mark

If the angle between the vectors 2i∼−j∼+2k∼2\underset{\sim}{\mathrm{i}} - \underset{\sim}{\mathrm{j}} + 2\underset{\sim}{\mathrm{k}} and 2i∼+mj∼+6k∼2\underset{\sim}{\mathrm{i}} + m\underset{\sim}{\mathrm{j}} + 6\underset{\sim}{\mathrm{k}} is cos⁡−1(1321)\cos^{-1}\left(\frac{13}{21}\right), then the value of mm, where m∈R+m \in R^+, is

A

22

B

33

C

44

D

55

Reveal Answer
A

22

Substituting m=2m=2 into the dot product formula yields cos⁡θ=14344\cos \theta = \frac{14}{3\sqrt{44}}, which does not simplify to 1321\frac{13}{21}.

B

33

Correct Answer

Using the dot product formula cos⁡θ=a⋅b∣a∣∣b∣\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}, we get 16−m3m2+40=1321\frac{16 - m}{3\sqrt{m^2 + 40}} = \frac{13}{21}. Solving this equation for m>0m > 0 yields m=3m = 3.

C

44

Substituting m=4m=4 into the dot product formula yields cos⁡θ=12356\cos \theta = \frac{12}{3\sqrt{56}}, which does not simplify to 1321\frac{13}{21}.

D

55

Substituting m=5m=5 into the dot product formula yields cos⁡θ=11365\cos \theta = \frac{11}{3\sqrt{65}}, which does not simplify to 1321\frac{13}{21}.

Q15
2025
VCAA
Paper 2
1 mark
Q15
1 mark

Consider the two planes described by the equations 2x+2y+z=22x + 2y + z = 2 and ax+4z=1ax + 4z = 1, where aa is a positive constant.

The angle between the two planes is cos⁡−1(23)\cos^{-1}\left(\frac{2}{3}\right).

The value of aa satisfies the equation

A

a+2=a2+16a + 2 = \sqrt{a^2 + 16}

B

2a+43a2+16=32\frac{2a + 4}{3\sqrt{a^2 + 16}} = \frac{3}{2}

C

2a+4=3a2+162a + 4 = 3\sqrt{a^2 + 16}

D

2a+4a2+16=23\frac{2a + 4}{\sqrt{a^2 + 16}} = \frac{2}{3}

Reveal Answer
A

a+2=a2+16a + 2 = \sqrt{a^2 + 16}

Correct Answer

The normal vectors are n1⃗=⟨2,2,1⟩\vec{n_1} = \langle 2, 2, 1 \rangle and n2⃗=⟨a,0,4⟩\vec{n_2} = \langle a, 0, 4 \rangle. Using the dot product formula cos⁡θ=n1⃗⋅n2⃗∣n1⃗∣∣n2⃗∣\cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}||\vec{n_2}|}, we get 2a+43a2+16=23\frac{2a + 4}{3\sqrt{a^2 + 16}} = \frac{2}{3}, which simplifies to a+2=a2+16a + 2 = \sqrt{a^2 + 16}.

B

2a+43a2+16=32\frac{2a + 4}{3\sqrt{a^2 + 16}} = \frac{3}{2}

This incorrectly sets the cosine of the angle to 32\frac{3}{2} instead of the given 23\frac{2}{3}.

C

2a+4=3a2+162a + 4 = 3\sqrt{a^2 + 16}

This equation results from an algebraic error when simplifying 2a+43a2+16=23\frac{2a + 4}{3\sqrt{a^2 + 16}} = \frac{2}{3}, likely by multiplying the right side by 32\frac{3}{2} instead of correctly cross-multiplying.

D

2a+4a2+16=23\frac{2a + 4}{\sqrt{a^2 + 16}} = \frac{2}{3}

This equation incorrectly omits the magnitude of the first normal vector (∣n1⃗∣=3|\vec{n_1}| = 3) in the denominator of the dot product formula.

Q16
2025
VCAA
Paper 2
1 mark
Q16
1 mark

The position vector of a particle at time tt is given by r∼(t)=ne−2ti∼−t2j∼\underset{\sim}{\text{r}}(t) = ne^{-2t}\underset{\sim}{\text{i}} - t^2\underset{\sim}{\text{j}}, where nn is a positive constant.

For what value of nn is the particle's acceleration perpendicular to its velocity when t=12t = \frac{1}{2}?

A

2e2e

B

e0.52\frac{e^{0.5}}{2}

C

e2\frac{e}{2}

D

e22\frac{e}{2\sqrt{2}}

Reveal Answer
A

2e2e

Incorrect. This value results from an algebraic error when solving the dot product equation −8n2e−2+2=0-8n^2e^{-2} + 2 = 0, perhaps by incorrectly isolating nn.

B

e0.52\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

e2\frac{e}{2}

Correct Answer

Correct. The velocity is v(t)=−2ne−2ti−2tj\mathbf{v}(t) = -2ne^{-2t}\mathbf{i} - 2t\mathbf{j} and acceleration is a(t)=4ne−2ti−2j\mathbf{a}(t) = 4ne^{-2t}\mathbf{i} - 2\mathbf{j}. Setting their dot product at t=1/2t=1/2 to zero gives −8n2e−2+2=0-8n^2e^{-2} + 2 = 0, which yields n=e2n = \frac{e}{2}.

D

e22\frac{e}{2\sqrt{2}}

Incorrect. This answer stems from a miscalculation in the dot product, such as setting 8n2e−2=18n^2e^{-2} = 1 instead of 22, leading to n=e22n = \frac{e}{2\sqrt{2}}.

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