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
VCAA
Paper 1
3 marks
Q4
3 marks

Find the coordinates of the point where the line given by the equation r∼=4i∼−2j∼+2k∼+t(2i∼−j∼+3k∼)\underset{\sim}{r} = 4\underset{\sim}{i} - 2\underset{\sim}{j} + 2\underset{\sim}{k} + t(2\underset{\sim}{i} - \underset{\sim}{j} + 3\underset{\sim}{k}), where t∈Rt \in R, intersects the plane with the equation 2x+3y−z=62x + 3y - z = 6.

Reveal Answer

The parametric equations of the line are x=4+2tx=4+2t, y=−2−ty=-2-t and z=2+3tz=2+3t. Substituting these into the equation of the plane:

2(4+2t)+3(−2−t)−(2+3t)=62(4+2t)+3(-2-t)-(2+3t)=6

and so t=−3t=-3.

The point of intersection is (−2,1,−7)(-2,1,-7).

Marking Criteria
DescriptorMarks

Substitutes the parametric equations of the line (x=4+2tx=4+2t, y=−2−ty=-2-t, z=2+3tz=2+3t) into the equation of the plane.

1

Solves the resulting equation to find the correct parameter value, t=−3t = -3.

1

Calculates the correct coordinates of the point of intersection, (−2,1,−7)(-2, 1, -7).

1
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}}.

Q15
2024
VCAA
Paper 2
1 mark
Q15
1 mark

A normal vector to the plane that contains the points (2,−1,−1)(2, -1, -1), (3,1,1)(3, 1, 1) and (−1,−1,2)(-1, -1, 2) is

A

2i∼+3j∼−3k∼2\underset{\sim}{i} + 3\underset{\sim}{j} - 3\underset{\sim}{k}

B

−2i∼−7j∼−10k∼-2\underset{\sim}{i} - 7\underset{\sim}{j} - 10\underset{\sim}{k}

C

−2i∼−3j∼−2k∼-2\underset{\sim}{i} - 3\underset{\sim}{j} - 2\underset{\sim}{k}

D

2i∼−3j∼+2k∼2\underset{\sim}{i} - 3\underset{\sim}{j} + 2\underset{\sim}{k}

E

2i∼+7j∼+10k∼2\underset{\sim}{i} + 7\underset{\sim}{j} + 10\underset{\sim}{k}

Reveal Answer
A

2i∼+3j∼−3k∼2\underset{\sim}{i} + 3\underset{\sim}{j} - 3\underset{\sim}{k}

This vector is incorrect because it is not orthogonal to the plane. Its dot product with the vector (1,2,2)(1, 2, 2) formed by the first two points is not zero.

B

−2i∼−7j∼−10k∼-2\underset{\sim}{i} - 7\underset{\sim}{j} - 10\underset{\sim}{k}

This vector is not perpendicular to the plane. You can verify this by taking the dot product with a vector between two of the given points, which does not equal zero.

C

−2i∼−3j∼−2k∼-2\underset{\sim}{i} - 3\underset{\sim}{j} - 2\underset{\sim}{k}

This is incorrect because a true normal vector must be orthogonal to all vectors in the plane, but the dot product of this vector with the vector (1,2,2)(1, 2, 2) between the first two points is not zero.

D

2i∼−3j∼+2k∼2\underset{\sim}{i} - 3\underset{\sim}{j} + 2\underset{\sim}{k}

Correct Answer

To find the normal vector, calculate the cross product of two vectors in the plane, such as (1,2,2)(1, 2, 2) and (−3,0,3)(-3, 0, 3) formed by the given points. Their cross product is (6,−9,6)(6, -9, 6), and dividing by the scalar 3 gives the parallel normal vector 2i∼−3j∼+2k∼2\underset{\sim}{i} - 3\underset{\sim}{j} + 2\underset{\sim}{k}.

E

2i∼+7j∼+10k∼2\underset{\sim}{i} + 7\underset{\sim}{j} + 10\underset{\sim}{k}

This vector is incorrect because its dot product with the vector (1,2,2)(1, 2, 2) formed by the first two points is 2(1)+7(2)+10(2)=362(1) + 7(2) + 10(2) = 36, not zero.

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
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
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.

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
2020
VCAA
Paper 2
1 mark
Q14
1 mark

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

A

9219\frac{92}{19}

B

927\frac{92}{7}

C

1247\frac{124}{7}

D

9211\frac{92}{11}

E

187\frac{18}{7}

Reveal Answer
A

9219\frac{92}{19}

Incorrect. This option uses an incorrect magnitude for the direction vector d\mathbf{d}. The correct magnitude is 22+(−3)2+(−6)2=7\sqrt{2^2 + (-3)^2 + (-6)^2} = 7.

B

927\frac{92}{7}

Correct Answer

Correct. The component of F\mathbf{F} in the direction of d\mathbf{d} is found using F⋅d∣d∣\frac{\mathbf{F} \cdot \mathbf{d}}{|\mathbf{d}|}. Calculating this gives (1)(2)+(6)(−3)+(−18)(−6)22+(−3)2+(−6)2=927\frac{(1)(2) + (6)(-3) + (-18)(-6)}{\sqrt{2^2 + (-3)^2 + (-6)^2}} = \frac{92}{7}.

C

1247\frac{124}{7}

Incorrect. This answer likely stems from an arithmetic error when calculating the dot product F⋅d\mathbf{F} \cdot \mathbf{d}, which should be 9292.

D

9211\frac{92}{11}

Incorrect. While the dot product is correctly calculated as 9292, the magnitude of the direction vector ∣d∣|\mathbf{d}| is 77, not 1111.

E

187\frac{18}{7}

Incorrect. This answer comes from incorrectly calculating the dot product F⋅d\mathbf{F} \cdot \mathbf{d}. The correct dot product is 9292.

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
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
Q12
2022
VCAA
Paper 2
1 mark
Q12
1 mark

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

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

A

π6\frac{\pi}{6} and 7π6\frac{7\pi}{6}

B

π3\frac{\pi}{3} and 4π3\frac{4\pi}{3}

C

5π6\frac{5\pi}{6} and 11π6\frac{11\pi}{6}

D

2π3\frac{2\pi}{3} and 5π3\frac{5\pi}{3}

E

π6\frac{\pi}{6} and 5π6\frac{5\pi}{6}

Reveal Answer
A

π6\frac{\pi}{6} and 7π6\frac{7\pi}{6}

Correct Answer

Setting the dot product to zero gives −cot⁡(x)+3=0-\cot(x) + \sqrt{3} = 0, which simplifies to tan⁡(x)=13\tan(x) = \frac{1}{\sqrt{3}}. The solutions for this equation are x=π6x = \frac{\pi}{6} and 7π6\frac{7\pi}{6}.

B

π3\frac{\pi}{3} and 4π3\frac{4\pi}{3}

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

C

5π6\frac{5\pi}{6} and 11π6\frac{11\pi}{6}

These values correspond to tan⁡(x)=−13\tan(x) = -\frac{1}{\sqrt{3}}. This would be the solution if the dot product equation was cot⁡(x)+3=0\cot(x) + \sqrt{3} = 0, missing the negative sign.

D

2π3\frac{2\pi}{3} and 5π3\frac{5\pi}{3}

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

E

π6\frac{\pi}{6} and 5π6\frac{5\pi}{6}

While π6\frac{\pi}{6} is a valid solution, 5π6\frac{5\pi}{6} yields tan⁡(x)=−13\tan(x) = -\frac{1}{\sqrt{3}}, which does not satisfy the condition for perpendicular vectors.

Q6
2025
VCAA
Paper 1
4 marks
Q6

Consider the lines L1L_1 and L2L_2 with respective equations

r∼1(t)=i∼−2j∼+4k∼+t(i∼+3j∼−k∼)\underset{\sim}{r}_1(t) = \underset{\sim}{i} - 2\underset{\sim}{j} + 4\underset{\sim}{k} + t(\underset{\sim}{i} + 3\underset{\sim}{j} - \underset{\sim}{k}) and r∼2(s)=3j∼−3k∼+s(2i∼+j∼+4k∼)\underset{\sim}{r}_2(s) = 3\underset{\sim}{j} - 3\underset{\sim}{k} + s(2\underset{\sim}{i} + \underset{\sim}{j} + 4\underset{\sim}{k}), where s,t∈Rs, t \in R.

Q6a
2 marks

Show that L1L_1 and L2L_2 are skew, that is they do not intersect and they are not parallel.

Reveal Answer

Suppose the lines meet. Then for some t,st, s, we have the system of equations:

1+t=2s1 + t = 2s
−2+3t=3+s-2 + 3t = 3 + s
4−t=−3+4s4 - t = -3 + 4s

Solving the first two equations gives t=115t = \frac{11}{5} and s=85s = \frac{8}{5}. These values do not satisfy the third equation and so the lines do not meet.

The direction vectors i∼+3j∼−k∼\underset{\sim}{i} + 3\underset{\sim}{j} - \underset{\sim}{k} and 2i∼+j∼+4k∼2\underset{\sim}{i} + \underset{\sim}{j} + 4\underset{\sim}{k} are not parallel.

Therefore, the lines are shown to be skew as required.

Marking Criteria
DescriptorMarks

Demonstrates that the lines do not intersect by setting up a system of equations and showing they are inconsistent.

1

Demonstrates that the lines are not parallel by showing their direction vectors are not scalar multiples of each other.

1
Q6b
2 marks

Find the shortest distance between L1L_1 and L2L_2.

Reveal Answer

n∼=∣i∼j∼k∼13−1214∣=13i∼−6j∼−5k∼ and ∣n∼∣=169+36+25=230\underset{\sim}{n} = \begin{vmatrix} \underset{\sim}{i} & \underset{\sim}{j} & \underset{\sim}{k} \\ 1 & 3 & -1 \\ 2 & 1 & 4 \end{vmatrix} = 13\underset{\sim}{i} - 6\underset{\sim}{j} - 5\underset{\sim}{k} \text{ and } |\underset{\sim}{n}| = \sqrt{169 + 36 + 25} = \sqrt{230}

A(1,−2,4)A(1, -2, 4) and B(0,3,−3)B(0, 3, -3) are points on L1L_1 and L2L_2, respectively. AB⃗=−i∼+5j∼−7k∼\vec{AB} = -\underset{\sim}{i} + 5\underset{\sim}{j} - 7\underset{\sim}{k}

distance=∣AB⃗⋅n∼^∣=∣1230(−13−30+35)∣=∣−8230∣\text{distance} = \left|\vec{AB} \cdot \hat{\underset{\sim}{n}}\right| = \left|\frac{1}{\sqrt{230}}(-13 - 30 + 35)\right| = \left|\frac{-8}{\sqrt{230}}\right|

Therefore, the distance between the lines is 8230\frac{8}{\sqrt{230}}.

Marking Criteria
DescriptorMarks

Calculates the normal vector to both lines using the cross product of their direction vectors, e.g., n∼=13i∼−6j∼−5k∼\underset{\sim}{n} = 13\underset{\sim}{i} - 6\underset{\sim}{j} - 5\underset{\sim}{k}.

1

Calculates the correct shortest distance between the lines, e.g., 8230\frac{8}{\sqrt{230}}.

1
Q9
2023
VCAA
Paper 1
5 marks
Q9

The position vectors of two cyclists, AA and BB, after tt seconds are given by r∼A=(6t+a)i∼+(7t+2−3a)j∼\underset{\sim}{\text{r}}_A = (6t + a)\underset{\sim}{\text{i}} + (7t + 2 - 3a)\underset{\sim}{\text{j}} and r∼B=(5t+4a)i∼+(6t+2a−2)j∼\underset{\sim}{\text{r}}_B = (5t + 4a)\underset{\sim}{\text{i}} + (6t + 2a - 2)\underset{\sim}{\text{j}} respectively for t≥0t \ge 0, where aa is a positive constant and the components are measured in metres. The distance between the cyclists is initially 626\sqrt{2} metres.

Q9a
2 marks

Find a quadratic equation in the form f(a)=0f(a) = 0 that, when solved, gives the value of aa.
A solution to this equation is not required.

Reveal Answer

rA(0)=ai+(2−3a)j\mathbf{r}_A(0)=a\mathbf{i}+(2-3a)\mathbf{j}
rB(0)=4ai+(2a−2)j\mathbf{r}_B(0)=4a\mathbf{i}+(2a-2)\mathbf{j}

and so

∣rA(0)−rB(0)∣=(3a)2+(5a−4)2\left|\mathbf{r}_A(0)-\mathbf{r}_B(0)\right|=\sqrt{(3a)^2+(5a-4)^2}
=9a2+25a2−40a+16=\sqrt{9a^2+25a^2-40a+16}
=34a2−40a+16=\sqrt{34a^2-40a+16}

Distance between starting positions is 626\sqrt{2}:

34a2−40a+16=(62)2=7234a^2-40a+16=(6\sqrt{2})^2=72
⇒34a2−40a−56=0\Rightarrow 34a^2-40a-56=0
⇒17a2−20a−28=0\Rightarrow 17a^2-20a-28=0

Marking Criteria
DescriptorMarks

Sets up a correct equation for the initial distance or distance squared between cyclists AA and BB

1

Derives the correct quadratic equation in the form f(a)=0f(a) = 0, such as 17a2−20a−28=017a^2-20a-28=0

1
Q9b
3 marks

Another cyclist, CC, is moving with velocity given by r˙∼C=2ti∼+(n−3)j∼\underset{\sim}{\dot{\text{r}}}_C = 2t\underset{\sim}{\text{i}} + (n - 3)\underset{\sim}{\text{j}}, where n∈Rn \in R, with initial position r∼C=2i∼\underset{\sim}{\text{r}}_C = 2\underset{\sim}{\text{i}}.

Given that a=2a = 2, find the value of nn such that the three cyclists, AA, BB and CC, meet at the same point at the same time.

Reveal Answer

Integrate to find rC\mathbf{r}_C:

rC=t2i+(n−3)tj+c\mathbf{r}_C=t^2\mathbf{i}+(n-3)t\mathbf{j}+\mathbf{c},

rC(0)=c=2i\mathbf{r}_C(0)=\mathbf{c}=2\mathbf{i},

rC=(t2+2)i+(n−3)tj\mathbf{r}_C=(t^2+2)\mathbf{i}+(n-3)t\mathbf{j}

As cyclists AA and BB meet, they must meet when

6t+2=5t+86t+2=5t+8
⇒t=6\Rightarrow t=6

Then

rA(6)=rB(6)=38i+38j\mathbf{r}_A(6)=\mathbf{r}_B(6)=38\mathbf{i}+38\mathbf{j}
rC(6)=38i+6(n−3)j\mathbf{r}_C(6)=38\mathbf{i}+6(n-3)\mathbf{j}

and so

38=6(n−3)38=6(n-3)

⇒n=386+3\Rightarrow n=\frac{38}{6}+3
=193+93=\frac{19}{3}+\frac{9}{3}
=283=\frac{28}{3}

Marking Criteria
DescriptorMarks

Integrates the velocity vector to find the correct position vector of cyclist CC, r∼C\underset{\sim}{\text{r}}_C

1

Determines the correct time t=6t=6 when cyclists AA and BB meet

1

Equates the relevant components to find the correct value of n=283n = \frac{28}{3}

1

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