NSWMaths Extension 2Syllabus dot point

How do we resolve one vector along the direction of another, and what do the scalar and vector projections measure geometrically?

Compute the projection of one vector onto another, distinguishing the scalar projection from the vector projection, and resolve a vector into parallel and perpendicular components

A focused answer to the HSC Maths Extension 2 dot point on vector projection. The scalar projection as a signed length, the vector projection formula, resolving a vector into components parallel and perpendicular to another, with verified worked examples.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The scalar projection
The vector projection
Resolving a vector into components
Geometric meaning

What this dot point is asking

NESA wants you to project one vector onto the direction of another. You must compute the scalar projection (a signed number measuring how much of $\mathbf{a}$ points along $\mathbf{b}$ ), the vector projection (the actual vector component of $\mathbf{a}$ along $\mathbf{b}$ ), and use these to split $\mathbf{a}$ into a part parallel to $\mathbf{b}$ and a part perpendicular to $\mathbf{b}$ .

The scalar projection

Given vectors $\mathbf{a}$ and $\mathbf{b}$ with $\mathbf{b} \neq \mathbf{0}$ , the scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ measures the signed length of the shadow $\mathbf{a}$ casts on the line through $\mathbf{b}$ . Since $\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$ , the scalar projection is

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

It is positive when the angle $\theta$ between the vectors is acute, zero when they are perpendicular, and negative when $\theta$ is obtuse. The sign carries genuine information about direction.

The vector projection

The vector projection is the scalar projection multiplied by the unit vector in the direction of $\mathbf{b}$ , namely $\hat{\mathbf{b}} = \dfrac{\mathbf{b}}{|\mathbf{b}|}$ . Thus

\operatorname{proj}_{\mathbf{b}}\mathbf{a} = \left(\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\right)\frac{\mathbf{b}}{|\mathbf{b}|} = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\,\mathbf{b} = \frac{\mathbf{a}\cdot\mathbf{b}}{\mathbf{b}\cdot\mathbf{b}}\,\mathbf{b}.

This is a vector, parallel to $\mathbf{b}$ , representing the component of $\mathbf{a}$ lying along $\mathbf{b}$ . Its magnitude equals the absolute value of the scalar projection, and its direction is along $\mathbf{b}$ (or against it, when the scalar projection is negative).

Resolving a vector into components

Every vector $\mathbf{a}$ can be split into a part parallel to $\mathbf{b}$ and a part perpendicular to $\mathbf{b}$ :

\mathbf{a} = \mathbf{a}_{\parallel} + \mathbf{a}_{\perp}, \qquad \mathbf{a}_{\parallel} = \operatorname{proj}_{\mathbf{b}}\mathbf{a}, \qquad \mathbf{a}_{\perp} = \mathbf{a} - \operatorname{proj}_{\mathbf{b}}\mathbf{a}.

The perpendicular component is genuinely perpendicular to $\mathbf{b}$ : checking the scalar product,

\mathbf{a}_{\perp}\cdot\mathbf{b} = \mathbf{a}\cdot\mathbf{b} - \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}(\mathbf{b}\cdot\mathbf{b}) = \mathbf{a}\cdot\mathbf{b} - \mathbf{a}\cdot\mathbf{b} = 0.

This decomposition is the foundation for finding the distance from a point to a line and for resolving forces in mechanics.

Geometric meaning

The vector projection answers the question: if I slide $\mathbf{a}$ straight down onto the line carrying $\mathbf{b}$ , where does it land? The scalar projection answers: how far along that line, and in which direction? Because both depend only on the direction of $\mathbf{b}$ and not its length, scaling $\mathbf{b}$ leaves the projection of $\mathbf{a}$ onto it unchanged.

Worked example

Project $\mathbf{a} = 3\mathbf{i} + 4\mathbf{j}$ onto $\mathbf{b} = \mathbf{i} + 2\mathbf{j}$

Scalar product: $\mathbf{a}\cdot\mathbf{b} = (3)(1) + (4)(2) = 3 + 8 = 11$ .

Squared magnitude of $\mathbf{b}$ : $|\mathbf{b}|^2 = 1^2 + 2^2 = 5$ .

Vector projection:

\operatorname{proj}_{\mathbf{b}}\mathbf{a} = \frac{11}{5}(\mathbf{i} + 2\mathbf{j}) = \frac{11}{5}\mathbf{i} + \frac{22}{5}\mathbf{j}.

Scalar projection: $\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|} = \dfrac{11}{\sqrt{5}} = \dfrac{11\sqrt{5}}{5}$ .

Perpendicular component:

\mathbf{a}_{\perp} = (3\mathbf{i} + 4\mathbf{j}) - \left(\tfrac{11}{5}\mathbf{i} + \tfrac{22}{5}\mathbf{j}\right) = \tfrac{4}{5}\mathbf{i} - \tfrac{2}{5}\mathbf{j}.

Check perpendicularity: $\mathbf{a}_{\perp}\cdot\mathbf{b} = \tfrac{4}{5}(1) + (-\tfrac{2}{5})(2) = \tfrac{4}{5} - \tfrac{4}{5} = 0$ . Correct.

Common traps

Dividing by $|\mathbf{b}|$ instead of $|\mathbf{b}|^2$ in the vector projection. The vector projection has $\mathbf{b}\cdot\mathbf{b} = |\mathbf{b}|^2$ in the denominator. Only the scalar projection divides by $|\mathbf{b}|$ .

Projecting onto the wrong vector: $\operatorname{proj}_{\mathbf{b}}\mathbf{a}$ and $\operatorname{proj}_{\mathbf{a}}\mathbf{b}$ are different vectors. Read carefully which is projected onto which.
Dropping the sign of the scalar projection: A negative scalar projection means $\mathbf{a}$ points partly against $\mathbf{b}$ . Reporting only the magnitude loses information.
Forgetting the perpendicular check: Confirming $\mathbf{a}_{\perp}\cdot\mathbf{b} = 0$ catches arithmetic slips immediately.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 HSC3 marksThe vector a is (1, 2, 3) and the vector b is (2, 0, -4). (i) Find (a . b / (b . b)) b. (ii) Show that a - (a . b / (b . b)) b is perpendicular to b.

Show worked answer →

The expression (a . b / (b . b)) b is the vector projection of a onto b; subtracting it from a leaves the component of a perpendicular to b.

Part (i). Compute the dot products:
a . b = (1)(2) + (2)(0) + (3)(-4) = 2 + 0 - 12 = -10.
b . b = (2)^2 + 0^2 + (-4)^2 = 4 + 0 + 16 = 20.

So a . b / (b . b) = -10/20 = -1/2, and the projection is

(-1/2)(2, 0, -4) = (-1, 0, 2).

Part (ii). The perpendicular component is
a - (-1, 0, 2) = (1 - (-1), 2 - 0, 3 - 2) = (2, 2, 1).

Check it is perpendicular to b by taking the dot product:
(2, 2, 1) . (2, 0, -4) = (2)(2) + (2)(0) + (1)(-4) = 4 + 0 - 4 = 0.

Since the dot product is 0, a - (a . b / (b . b)) b is perpendicular to b.

Mark notes: part (i) 1 mark for the projection (-1, 0, 2). Part (ii): 1 mark for the perpendicular component (2, 2, 1), 1 mark for showing its dot product with b is 0.

2021 HSC1 marksConsider the two non-zero complex numbers z and w as vectors. Which of the following expressions is the projection of z onto w? A. (Re(z times conjugate of w)/|w|^2) w. B. (z/|w|) w. C. Re(z/w) w. D. (Re(z)/|w|^2) w.

Show worked answer →

The vector projection of one vector onto another is (dot product divided by the squared length of the target) times the target: proj = ((z . w)/(w . w)) w.

When complex numbers are treated as vectors, the dot product z . w equals Re(z times conjugate of w), and w . w = |w|^2.

Substituting gives proj = (Re(z times conjugate of w)/|w|^2) w, which is option A.

Options B, C and D do not give a real scalar multiple equal to the projection coefficient (Re(z times conjugate of w))/|w|^2, so they are incorrect.

The answer is A.