HSC Maths Extension 1 vectors: deep dive (2026 guide)

HSCMaths Extension 1Study guide22 min readUpdated 25 May 2026

A complete deep dive into vectors for HSC Mathematics Extension 1. Vector arithmetic, magnitude and unit vectors, the scalar product, projections, parametric vector equations of lines, geometric proofs, and an embedded practice question set with full solutions.

Jump to a section

What this guide covers
Vector arithmetic
The scalar (dot) product
Projections
Parametric vector equations of lines
Geometric proofs using vectors
Applications to physics-style problems
Common exam questions
Common traps
Exam strategy
Check your knowledge
Linked dot points

What this guide covers

Vectors were introduced into HSC Mathematics Extension 1 in the 2017 syllabus and are now examined every year. This guide covers:

Vector arithmetic, magnitude and unit vectors.
The scalar (dot) product, the angle between vectors, orthogonality.
Scalar and vector projection.
Parametric vector equations of lines and intersection problems.
Geometric proofs using vectors.
Applications to physics-style problems (forces, velocity).

Vector arithmetic

A two-dimensional vector $\mathbf{a} = (a_1, a_2)$ can be written in column form, component form, or unit-vector form ( $a_1 \mathbf{i} + a_2 \mathbf{j}$ with $\mathbf{i} = (1, 0)$ , $\mathbf{j} = (0, 1)$ ).

Addition, subtraction and scalar multiplication

\mathbf{a} + \mathbf{b} = (a_1 + b_1, a_2 + b_2), \qquad \lambda \mathbf{a} = (\lambda a_1, \lambda a_2).

Geometrically: $\mathbf{a} + \mathbf{b}$ is the parallelogram-rule diagonal; $\lambda \mathbf{a}$ scales magnitude by $|\lambda|$ and may reverse direction if $\lambda < 0$ .

Magnitude and unit vector

$|\mathbf{a}| = \sqrt{a_1^2 + a_2^2}$ , and the unit vector $\hat{\mathbf{a}} = \frac{1}{|\mathbf{a}|} \mathbf{a}$ .

Position vector versus displacement vector

If $A = (a_1, a_2)$ is a point, $\mathbf{OA} = \mathbf{a}$ is its position vector (origin to $A$ ). The displacement from $A$ to $B$ is $\mathbf{AB} = \mathbf{b} - \mathbf{a}$ .

The scalar (dot) product

\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 = |\mathbf{a}| |\mathbf{b}| \cos \theta.

Angle between two vectors

\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}.

For non-zero vectors, $\theta \in [0, \pi]$ .

Orthogonality test

$\mathbf{a} \perp \mathbf{b}$ iff $\mathbf{a} \cdot \mathbf{b} = 0$ .

This is the most common HSC use of the scalar product.

Properties

The dot product is commutative, bilinear, and $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$ .

(\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) = |\mathbf{a}|^2 + 2 \mathbf{a} \cdot \mathbf{b} + |\mathbf{b}|^2.

This is the vector form of the cosine rule.

Find the angle between

\mathbf{a} = (3, 4)

and

\mathbf{b} = (1, 2)

in radians, correct to three decimal places

Step 1. Compute the dot product. Multiply component-wise and sum.

\mathbf{a} \cdot \mathbf{b} = 3 \cdot 1 + 4 \cdot 2 = 3 + 8 = 11.

Step 2. Compute the magnitudes. Use $|\mathbf{v}| = \sqrt{v_1^2 + v_2^2}$ .

|\mathbf{a}| = \sqrt{9 + 16} = \sqrt{25} = 5, \qquad |\mathbf{b}| = \sqrt{1 + 4} = \sqrt{5}.

Step 3. Apply the cosine formula.

\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} = \frac{11}{5 \sqrt{5}} = \frac{11 \sqrt{5}}{25}.

Step 4. Take the inverse cosine. $\frac{11 \sqrt{5}}{25} \approx 0.98387$ , so $\theta = \arccos(0.98387) \approx 0.179$ radians (3 d.p.).

The angle between $\mathbf{a}$ and $\mathbf{b}$ is approximately $0.179$ radians.

Find the value of

k

such that

\mathbf{a} = (3, k)

and

\mathbf{b} = (k - 2, 5)

are perpendicular

Step 1. Apply the orthogonality test. Two vectors are perpendicular iff their dot product is zero.

\mathbf{a} \cdot \mathbf{b} = 3(k - 2) + k \cdot 5 = 0.

Step 2. Expand and simplify.

3k - 6 + 5k = 0 \implies 8k = 6.

Step 3. Solve for $k$ . $k = \dfrac{6}{8} = \dfrac{3}{4}$ .

Step 4. Verify. With $k = 3/4$ : $\mathbf{a} = (3, 3/4)$ and $\mathbf{b} = (-5/4, 5)$ . Dot product $= 3 \cdot (-5/4) + (3/4) \cdot 5 = -15/4 + 15/4 = 0$ . Confirmed.

So $k = \dfrac{3}{4}$ .

Projections

Scalar projection

\text{proj}_{\mathbf{b}}^{\text{scalar}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}.

Signed length: positive if $\mathbf{a}$ has a component in the direction of $\mathbf{b}$ , negative if opposite.

Vector projection

\text{proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2} \mathbf{b}.

This is the scalar projection times the unit vector $\hat{\mathbf{b}}$ .

Decomposition

Any vector $\mathbf{a}$ can be split into a component parallel to $\mathbf{b}$ (the vector projection) and a component perpendicular to $\mathbf{b}$ .

\mathbf{a} = \text{proj}_{\mathbf{b}} \mathbf{a} + \mathbf{a}_{\perp}, \quad \mathbf{a}_{\perp} = \mathbf{a} - \text{proj}_{\mathbf{b}} \mathbf{a}.

By construction $\mathbf{a}_{\perp} \cdot \mathbf{b} = 0$ .

Find the vector projection of

\mathbf{a} = (4, 3)

onto

\mathbf{b} = (2, -1)

, and decompose

\mathbf{a}

into components parallel and perpendicular to

\mathbf{b}

Step 1. Compute the dot product: $\mathbf{a} \cdot \mathbf{b} = 4 \cdot 2 + 3 \cdot (-1) = 8 - 3 = 5$ .
Step 2. Compute $|\mathbf{b}|^2$: Avoid the square root by squaring directly: $|\mathbf{b}|^2 = 2^2 + (-1)^2 = 5$ .
Step 3. Apply the vector projection formula

\text{proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2} \mathbf{b} = \frac{5}{5} (2, -1) = (2, -1).

Step 4. Find the perpendicular component. Subtract the parallel part from $\mathbf{a}$ .

\mathbf{a}_{\perp} = \mathbf{a} - \text{proj}_{\mathbf{b}} \mathbf{a} = (4, 3) - (2, -1) = (2, 4).

Step 5. Verify $\mathbf{a}_{\perp} \perp \mathbf{b}$ . Dot product: $(2, 4) \cdot (2, -1) = 4 - 4 = 0$ . Confirmed.

So $\mathbf{a} = (2, -1) + (2, 4)$ , where $(2, -1)$ is parallel to $\mathbf{b}$ and $(2, 4)$ is perpendicular to $\mathbf{b}$ .

Parametric vector equations of lines

Point-direction form

A line through $A$ (position vector $\mathbf{a}$ ) with direction $\mathbf{d}$ :

\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}, \quad \lambda \in \mathbb{R}.

Line through two points

Direction $\mathbf{AB} = \mathbf{b} - \mathbf{a}$ , so $\mathbf{r} = \mathbf{a} + \lambda (\mathbf{b} - \mathbf{a}) = (1 - \lambda) \mathbf{a} + \lambda \mathbf{b}$ .

At $\lambda = 0$ : at $A$ . At $\lambda = 1$ : at $B$ .

Convert to Cartesian

If $\mathbf{d} = (d_1, d_2)$ with $d_1 \neq 0$ , $\lambda = \frac{x - a_1}{d_1}$ , so $y = a_2 + \frac{x - a_1}{d_1} d_2$ . Slope: $\frac{d_2}{d_1}$ .

Intersection of two lines

Set parametric equations equal:

\mathbf{a}_1 + \lambda \mathbf{d}_1 = \mathbf{a}_2 + \mu \mathbf{d}_2.

Solve for $\lambda$ and $\mu$ . If a unique solution, lines meet at a point. If the system is inconsistent, lines are parallel and disjoint.

Find the point of intersection of the lines

\mathbf{r}_1 = (1, 2) + \lambda (2, 1)

and

\mathbf{r}_2 = (4, -1) + \mu (-1, 3)

Step 1. Set the position vectors equal. Equate components.

1 + 2 \lambda = 4 - \mu \quad \text{and} \quad 2 + \lambda = -1 + 3 \mu.

Step 2. Tidy the system. Rearrange to standard form.

2 \lambda + \mu = 3 \tag{i}

\lambda - 3 \mu = -3 \tag{ii}

Step 3. Solve the system. From (ii), $\lambda = 3 \mu - 3$ . Substitute into (i):

2(3 \mu - 3) + \mu = 3 \implies 6 \mu - 6 + \mu = 3 \implies 7 \mu = 9 \implies \mu = \frac{9}{7}.

Then $\lambda = 3 \cdot \frac{9}{7} - 3 = \frac{27}{7} - \frac{21}{7} = \frac{6}{7}$ .

Step 4. Substitute back to find the point. Use $\mathbf{r}_1$ at $\lambda = 6/7$ :

\mathbf{r}_1 = \left(1 + 2 \cdot \frac{6}{7},\, 2 + \frac{6}{7}\right) = \left(\frac{19}{7},\, \frac{20}{7}\right).

Step 5. Check using $\mathbf{r}_2$ . With $\mu = 9/7$ : $\mathbf{r}_2 = (4 - 9/7,\, -1 + 27/7) = (19/7,\, 20/7)$ . Confirmed.

The lines meet at the point $\left(\dfrac{19}{7}, \dfrac{20}{7}\right)$ .

Geometric proofs using vectors

The recipe

Assign position vectors to labelled points.
Express the relevant displacement vectors algebraically.
Compute the geometric relation (equality, scalar multiple, dot product zero, etc.).
Conclude the geometric statement.

In triangle

ABC

, let

P

and

Q

be midpoints of

AB

and

AC

. Show that

\mathbf{PQ}

is parallel to

\mathbf{BC}

and equal to half its length

Step 1. Assign position vectors. Let $A$ , $B$ , $C$ have position vectors $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ relative to a fixed origin.

Step 2. Express the midpoints. The midpoint of a segment is the average of its endpoints.

P = \frac{\mathbf{a} + \mathbf{b}}{2}, \qquad Q = \frac{\mathbf{a} + \mathbf{c}}{2}.

Step 3. Compute $\mathbf{PQ}$ . Use $\mathbf{PQ} = Q - P$ .

\mathbf{PQ} = \frac{\mathbf{a} + \mathbf{c}}{2} - \frac{\mathbf{a} + \mathbf{b}}{2} = \frac{\mathbf{c} - \mathbf{b}}{2}.

Step 4. Relate to $\mathbf{BC}$ . Since $\mathbf{BC} = \mathbf{c} - \mathbf{b}$ , we have

\mathbf{PQ} = \frac{1}{2} \mathbf{BC}.

Step 5. Interpret. $\mathbf{PQ}$ is a positive scalar multiple of $\mathbf{BC}$ , so it is parallel to $\mathbf{BC}$ and points in the same direction. Its magnitude is half: $|\mathbf{PQ}| = \frac{1}{2} |\mathbf{BC}|$ .

Show that the diagonals of a parallelogram bisect each other

Step 1. Set up the parallelogram. Let $ABCD$ be a parallelogram with vertices having position vectors $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ , $\mathbf{d}$ . The parallelogram condition is $\mathbf{AB} = \mathbf{DC}$ , equivalently $\mathbf{b} - \mathbf{a} = \mathbf{c} - \mathbf{d}$ , giving $\mathbf{a} + \mathbf{c} = \mathbf{b} + \mathbf{d}$ .

Step 2. Compute the midpoint of diagonal $AC$ . Average the endpoints.

M_{AC} = \frac{\mathbf{a} + \mathbf{c}}{2}.

Step 3. Compute the midpoint of diagonal $BD$ .

M_{BD} = \frac{\mathbf{b} + \mathbf{d}}{2}.

Step 4. Use the parallelogram identity. From Step 1, $\mathbf{a} + \mathbf{c} = \mathbf{b} + \mathbf{d}$ , so $M_{AC} = M_{BD}$ .

Step 5. Interpret. The two diagonals share the same midpoint, hence they bisect each other.

A parallelogram has perpendicular diagonals. Show that all four sides have equal length (the parallelogram is a rhombus)

Step 1. Choose convenient vectors. Let $\mathbf{AB} = \mathbf{u}$ and $\mathbf{AD} = \mathbf{v}$ . The diagonals are $\mathbf{AC} = \mathbf{u} + \mathbf{v}$ and $\mathbf{BD} = \mathbf{v} - \mathbf{u}$ .

Step 2. Apply the perpendicularity condition. Perpendicular diagonals means $\mathbf{AC} \cdot \mathbf{BD} = 0$ :

(\mathbf{u} + \mathbf{v}) \cdot (\mathbf{v} - \mathbf{u}) = 0.

Step 3. Expand the dot product. Use distributivity and $\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}$ .

\mathbf{u} \cdot \mathbf{v} - \mathbf{u} \cdot \mathbf{u} + \mathbf{v} \cdot \mathbf{v} - \mathbf{v} \cdot \mathbf{u} = |\mathbf{v}|^2 - |\mathbf{u}|^2 = 0.

Step 4. Conclude $|\mathbf{u}| = |\mathbf{v}|$ . Taking square roots (lengths are non-negative) gives $|\mathbf{AB}| = |\mathbf{AD}|$ .

Step 5. Interpret. In a parallelogram opposite sides are already equal, so $|\mathbf{AB}| = |\mathbf{CD}|$ and $|\mathbf{AD}| = |\mathbf{BC}|$ . Combined with $|\mathbf{AB}| = |\mathbf{AD}|$ , all four sides are equal, so $ABCD$ is a rhombus.

Applications to physics-style problems

Vectors model displacement, velocity and force. HSC problems often ask you to resolve a force along a given direction (scalar projection) or to find the resultant velocity / displacement.

A box of weight

40

N sits on a slope inclined at

30^\circ

above horizontal. Find the magnitude of the component of weight acting parallel to the slope, directed down the slope

Step 1. Set up vectors. Take the unit vector pointing down the slope as $\hat{\mathbf{s}} = (\cos 30^\circ, -\sin 30^\circ)$ (the slope drops by $\sin 30^\circ$ for each unit of horizontal travel). The weight acts straight down: $\mathbf{W} = (0, -40)$ N.

Step 2. Compute the scalar projection of $\mathbf{W}$ onto $\hat{\mathbf{s}}$ . Since $\hat{\mathbf{s}}$ is a unit vector, the scalar projection equals $\mathbf{W} \cdot \hat{\mathbf{s}}$ .

\mathbf{W} \cdot \hat{\mathbf{s}} = 0 \cdot \cos 30^\circ + (-40) \cdot (-\sin 30^\circ) = 40 \sin 30^\circ.

Step 3. Evaluate. $\sin 30^\circ = \tfrac{1}{2}$ , so $\mathbf{W} \cdot \hat{\mathbf{s}} = 40 \cdot \tfrac{1}{2} = 20$ .

Step 4. Interpret. The component of weight down the slope has magnitude $20$ N (the sign is positive because the projection is in the chosen down-slope direction).

The magnitude of the parallel weight component is $20$ N.

A boat heads at

5

m/s in still water on a bearing measured by the direction vector

(3, 4)/5

(so

3/5

east and

4/5

north per unit velocity). A current adds a velocity of

(1, -2)

m/s. Find the resultant speed and the angle of the resultant from due east (positive

x

-axis), correct to one decimal place

Step 1. Write the boat's velocity in components: Direction $(3/5, 4/5)$ with speed $5$ gives $\mathbf{v}_{\text{boat}} = 5 \cdot (3/5, 4/5) = (3, 4)$ m/s.
Step 2. Add the current velocity: $\mathbf{v}_{\text{result}} = \mathbf{v}_{\text{boat}} + \mathbf{v}_{\text{current}} = (3, 4) + (1, -2) = (4, 2)$ m/s.
Step 3. Find the magnitude: $|\mathbf{v}_{\text{result}}| = \sqrt{4^2 + 2^2} = \sqrt{20} = 2 \sqrt{5}$ m/s, approximately $4.5$ m/s.
Step 4. Find the angle from the positive $x$ -axis: $\tan \theta = \dfrac{2}{4} = \dfrac{1}{2}$ , so $\theta = \arctan(1/2) \approx 26.6^\circ$ .

The resultant speed is $2 \sqrt{5} \approx 4.5$ m/s, and the direction is approximately $26.6^\circ$ north of east.

Common exam questions

Question type A: angle and orthogonality

Find the angle between two given vectors. Find $k$ so that two vectors are perpendicular. These are 2-3 mark items.

Question type B: projections

Find the scalar or vector projection. Decompose a vector into parallel and perpendicular components. These are 3-4 mark items.

Question type C: parametric vector equation of a line

Write the vector equation, convert to Cartesian, or find an intersection. 3-4 marks.

Question type D: geometric proof using vectors

Show two lines are parallel, two are perpendicular, or a figure is a parallelogram. 4-5 marks.

Question type E: applications

Resolve a force into components along a given direction. Find collision conditions for two moving particles. 4-5 marks.

Common traps

Trap: $\mathbf{PQ} = Q - P$ , not $P - Q$

The displacement from $P$ to $Q$ is the position vector of $Q$ minus the position vector of $P$ . Reversing gives the opposite vector.

Trap: Confusing the dot product with the length

$|\mathbf{a}|^2 = \mathbf{a} \cdot \mathbf{a}$ , so $|\mathbf{a} + \mathbf{b}|^2 = |\mathbf{a}|^2 + 2 \mathbf{a} \cdot \mathbf{b} + |\mathbf{b}|^2$ . This expansion is essential for the vector cosine rule.

Trap: $|\mathbf{b}|$ vs $|\mathbf{b}|^2$

Scalar projection has $|\mathbf{b}|$ in the denominator. Vector projection has $|\mathbf{b}|^2$ .

Trap: Parallel parameters

Two lines with the same direction vector are parallel. They may or may not coincide. Check if a point on one is also on the other.

Trap: Different parameters for different lines

When intersecting two lines, use $\lambda$ for one and $\mu$ for another. Reusing the same parameter creates confusion.

Exam strategy

Vector questions typically run 2-5 marks each. Allocate $3-6$ minutes per question. Steps:

Read carefully: scalar or vector projection? parallel or equal?
Write down position or displacement vectors.
Compute the relevant quantity algebraically.
State the answer with units (if appropriate) and direction (positive, negative, parallel, perpendicular).

For geometric proofs, work the algebra one line at a time and cite each step. Markers reward clear structure.

Check your knowledge

Try these questions before reading the solutions. They are roughly arranged from easier to harder.

Find the magnitude of $\mathbf{a} = (5, -12)$ and a unit vector in the direction of $\mathbf{a}$ .
Compute the dot product $\mathbf{a} \cdot \mathbf{b}$ and the angle between $\mathbf{a} = (1, 2)$ and $\mathbf{b} = (3, -1)$ , giving the angle in radians correct to three decimal places.
Find the value of $t$ such that $\mathbf{u} = (t, 2)$ is perpendicular to $\mathbf{v} = (4, -t + 1)$ .
Find the scalar projection of $\mathbf{a} = (6, 2)$ onto $\mathbf{b} = (3, 4)$ , and the vector projection.
Find the point where the lines $\mathbf{r}_1 = (2, 1) + \lambda (1, -1)$ and $\mathbf{r}_2 = (-1, 0) + \mu (2, 1)$ intersect.
Decompose $\mathbf{a} = (5, 1)$ into components parallel and perpendicular to $\mathbf{b} = (1, 2)$ .
The position vectors of $A$ , $B$ , $C$ , $D$ are $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ , $\mathbf{d}$ . Let $P$ , $Q$ , $R$ , $S$ be the midpoints of sides $AB$ , $BC$ , $CD$ , $DA$ respectively. Prove that $PQRS$ is a parallelogram (Varignon's theorem).
Two particles travel in straight lines. Particle $P$ has position $\mathbf{r}_P(t) = (1, 2) + t (3, 1)$ and particle $Q$ has position $\mathbf{r}_Q(t) = (10, 6) + t (-1, -1)$ , where $t$ is time in seconds. Determine whether the two particles collide. If they do, give the time and position of collision; if they do not, justify your answer.

Solutions

Q1
Step 1. Magnitude: $|\mathbf{a}| = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ .
Step 2. Unit vector: $\hat{\mathbf{a}} = \dfrac{1}{13} (5, -12) = \left( \dfrac{5}{13}, -\dfrac{12}{13} \right)$ .
Q2
Step 1. Dot product: $\mathbf{a} \cdot \mathbf{b} = 1 \cdot 3 + 2 \cdot (-1) = 3 - 2 = 1$ .
Step 2. Magnitudes: $|\mathbf{a}| = \sqrt{1 + 4} = \sqrt{5}$ and $|\mathbf{b}| = \sqrt{9 + 1} = \sqrt{10}$ .
Step 3. Apply the cosine formula: $\cos \theta = \dfrac{1}{\sqrt{5} \sqrt{10}} = \dfrac{1}{\sqrt{50}} = \dfrac{1}{5 \sqrt{2}}$ .
Step 4. Evaluate: $\dfrac{1}{5\sqrt{2}} \approx 0.14142$ , so $\theta = \arccos(0.14142) \approx 1.429$ radians (3 d.p.).
Q3
Step 1. Apply orthogonality: $\mathbf{u} \cdot \mathbf{v} = 4 t + 2(-t + 1) = 0$ .
Step 2. Simplify: $4 t - 2 t + 2 = 0 \implies 2 t + 2 = 0 \implies t = -1$ .
Step 3. Verify: $\mathbf{u} = (-1, 2)$ and $\mathbf{v} = (4, 2)$ . Dot product $= -4 + 4 = 0$ . Confirmed.

So $t = -1$ .

Q4
Step 1. Dot product: $\mathbf{a} \cdot \mathbf{b} = 6 \cdot 3 + 2 \cdot 4 = 18 + 8 = 26$ .
Step 2. Magnitude $|\mathbf{b}|$: $|\mathbf{b}| = \sqrt{9 + 16} = 5$ .
Step 3. Scalar projection: $\dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|} = \dfrac{26}{5}$ .
Step 4. Vector projection: $|\mathbf{b}|^2 = 25$ , so $\dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2} \mathbf{b} = \dfrac{26}{25} (3, 4) = \left( \dfrac{78}{25}, \dfrac{104}{25} \right)$ .
Q5
Step 1. Equate components: $2 + \lambda = -1 + 2 \mu$ and $1 - \lambda = \mu$ .
Step 2. Solve the system: From the second equation, $\mu = 1 - \lambda$ . Substitute into the first: $2 + \lambda = -1 + 2(1 - \lambda) = 1 - 2 \lambda$ , so $3 \lambda = -1$ and $\lambda = -\dfrac{1}{3}$ . Then $\mu = 1 - (-1/3) = \dfrac{4}{3}$ .
Step 3. Find the point: Use $\mathbf{r}_1$ : $(2 + (-1/3),\, 1 - (-1/3)) = \left( \dfrac{5}{3}, \dfrac{4}{3} \right)$ .
Step 4. Check using $\mathbf{r}_2$: $(-1 + 2 \cdot 4/3,\, 0 + 4/3) = (-1 + 8/3,\, 4/3) = (5/3,\, 4/3)$ . Confirmed.

The lines meet at $\left( \dfrac{5}{3}, \dfrac{4}{3} \right)$ .

Q6
Step 1. Dot product: $\mathbf{a} \cdot \mathbf{b} = 5 \cdot 1 + 1 \cdot 2 = 7$ .
Step 2. $|\mathbf{b}|^2$: $|\mathbf{b}|^2 = 1 + 4 = 5$ .
Step 3. Vector projection (parallel component): $\text{proj}_{\mathbf{b}} \mathbf{a} = \dfrac{7}{5} (1, 2) = \left( \dfrac{7}{5}, \dfrac{14}{5} \right)$ .
Step 4. Perpendicular component: $\mathbf{a}_{\perp} = \mathbf{a} - \text{proj}_{\mathbf{b}} \mathbf{a} = (5 - 7/5,\, 1 - 14/5) = \left( \dfrac{18}{5}, -\dfrac{9}{5} \right)$ .
Step 5. Verify perpendicularity: $\left( \dfrac{18}{5}, -\dfrac{9}{5} \right) \cdot (1, 2) = \dfrac{18}{5} - \dfrac{18}{5} = 0$ . Confirmed.

So $\mathbf{a} = \left( \dfrac{7}{5}, \dfrac{14}{5} \right) + \left( \dfrac{18}{5}, -\dfrac{9}{5} \right)$ .

Q7.

Step 1. Express the midpoints as averages of position vectors.

P = \frac{\mathbf{a} + \mathbf{b}}{2}, \quad Q = \frac{\mathbf{b} + \mathbf{c}}{2}, \quad R = \frac{\mathbf{c} + \mathbf{d}}{2}, \quad S = \frac{\mathbf{d} + \mathbf{a}}{2}.

Step 2. Compute $\mathbf{PQ}$ .

\mathbf{PQ} = Q - P = \frac{\mathbf{b} + \mathbf{c}}{2} - \frac{\mathbf{a} + \mathbf{b}}{2} = \frac{\mathbf{c} - \mathbf{a}}{2}.

Step 3. Compute $\mathbf{SR}$ .

\mathbf{SR} = R - S = \frac{\mathbf{c} + \mathbf{d}}{2} - \frac{\mathbf{d} + \mathbf{a}}{2} = \frac{\mathbf{c} - \mathbf{a}}{2}.

Step 4. Conclude one pair of opposite sides is equal and parallel: Since $\mathbf{PQ} = \mathbf{SR}$ , the sides $PQ$ and $SR$ of quadrilateral $PQRS$ are equal in length and parallel (same vector).
Step 5. Apply the parallelogram criterion: A quadrilateral in which one pair of opposite sides is both equal and parallel is a parallelogram. So $PQRS$ is a parallelogram.
Q8
Step 1. Set the position vectors equal: $\mathbf{r}_P(t) = \mathbf{r}_Q(t)$ gives

1 + 3 t = 10 - t \quad \text{and} \quad 2 + t = 6 - t.

Step 2. Solve each equation: From the first: $4 t = 9$ , so $t = 9/4$ . From the second: $2 t = 4$ , so $t = 2$ .
Step 3. Compare the two values: The two coordinate equations give different values of $t$ ( $9/4$ and $2$ ), so the particles never occupy the same point at the same instant.
Conclusion: The particles do not collide. Their trajectories may cross geometrically, but not simultaneously.

Linked dot points

For more detail, see the dot point pages at vector arithmetic and magnitude, scalar product, vector projection, parametric vector equations of lines, and geometric proofs with vectors.

For NESA past papers and marking guidelines, refer to educationstandards.nsw.edu.au.

Sources & how we know this

Mathematics Extension 1 Stage 6 Syllabus (2017) — NESA
Mathematics Extension 1 HSC exam papers — NESA