Why divide by the magnitude of b squared?

One factor normalises b to a unit vector, the other scales it to the correct length.

Why divide by the magnitude of b squared?

One factor normalises b to a unit vector, the other scales it to the correct length.

Year 12 Maths Extension 1 (2027) Vectors

Projection of a vector & perpendicular component

Q: What is the projection of a onto b?

The vector component of a along b, equal to (a dot b over the magnitude of b squared) times b.

Q: What is the difference between vector and scalar projection?

The vector projection is a vector; the scalar projection a dot b over the magnitude of b is just its signed length.

Q: How do you find the perpendicular component?

Subtract the projection from a, giving a minus proj b a.

Q: What happens for an obtuse angle?

The dot product is negative, so the projection points opposite to b.

20 practice questions 1 video lesson Theory + worked examples

Learn the projection of a vector and its perpendicular component for NSW Year 12 Mathematics Extension 1. Projection resolves one vector along the direction of another, splitting it into parallel and perpendicular parts.

You will learn to apply the projection formula, find the perpendicular component, and follow the proof behind the result — a resolving technique that underpins force and vector geometry problems in Extension 1.

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Practice questions

Every question with a fully worked solution.

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Y11-12 Mathematics: Defining Vector Projections Watch

Theory

The projection of $a$ onto $b$ is ${proj}_{b} a = \frac{a \cdot b}{| b |^{2}} b$ , the component of $a$ along $b$ ; the perpendicular part is $a - {proj}_{b} a$ . This NSW Year 12 Mathematics Extension 1 topic is NESA outcome ME1-12-02.

The projection of $a$ onto $b$ is the vector component of $a$ in the direction of $b$ — the "shadow" $a$ casts along $b$ .

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

The scalar projection (signed length) is $\frac{a \cdot b}{| b |}$ , and the perpendicular component is $a - {proj}_{b} a$ ; the two components add back to $a$ .

NESA link. Part of the Year 12 Introduction to vectors focus area, outcome ME1-12-02 ("operates with 2D and 3D vectors and uses 2D vectors to solve problems involving motion in two dimensions") with MAO-WM-01.

{proj}_{b} a

along

b

, plus the perpendicular component.

When

θ

is obtuse, the projection points opposite to

b

{proj}_{b} a = \frac{a \cdot b}{| b |^{2}} b, scalar proj = \frac{a \cdot b}{| b |}, perp = a - {proj}_{b} a .

proj_b a = (a.b / |b|^2) b; scalar projection = a.b / |b|; perpendicular = a - proj_b a

Divide by $| b |^{2}$ . The vector projection uses $| b |^{2}$ ; the scalar projection (a length) uses $| b |$ . The two components add back to $a$ .

How to project a vector

Compute $a \cdot b$ and $| b |^{2}$ .
Vector projection: ${proj}_{b} a = \frac{a \cdot b}{| b |^{2}} b$ .
Scalar projection: divide by $| b |$ instead of $| b |^{2}$ .
Perpendicular part: subtract the projection from $a$ .

Example 1 — Vector projection

Find

{proj}_{b} a

for

a = 4 i + 2 j

b = 3 i + 4 j

Solution

$a \cdot b = 20$ , $| b |^{2} = 25$ .

{proj}_{b} a

=

\frac{20}{25} (3 i + 4 j) = \frac{4}{5} (3 i + 4 j)

proj = (4/5)(3i + 4j)

${proj}_{b} a = \frac{4}{5} (3 i + 4 j)$ .

Example 2 — Scalar projection

Find the scalar projection of

a = 6 i + 3 j

onto

b = 4 i + 3 j

Solution

$a \cdot b = 24 + 9 = 33$ , $| b | = 5$ .

\frac{a \cdot b}{| b |}

=

\frac{33}{5} = 6.6

scalar projection = 6.6

Scalar projection $= 6.6$ .

Example 3 — Perpendicular component

For

a = 3 i + 5 j

b = i + 2 j

, find the component of

a

perpendicular to

b

Solution

${proj}_{b} a = \frac{13}{5} (i + 2 j)$ .

$a - {proj}_{b} a$	$=$	$(3 - \frac{13}{5}) i + (5 - \frac{26}{5}) j$
	$=$	$\frac{2}{5} i - \frac{1}{5} j$

perpendicular component = (2/5)i - (1/5)j

Perpendicular component $= \frac{2}{5} i - \frac{1}{5} j$ .

Example 4 — Projection in 3D

Find

{proj}_{b} a

for

a = i + 3 j + 2 k

b = 2 i + 2 j + k

Solution

$a \cdot b = 2 + 6 + 2 = 10$ , $| b |^{2} = 9$ .

{proj}_{b} a

=

\frac{10}{9} (2 i + 2 j + k)

proj = (10/9)(2i + 2j + k)

${proj}_{b} a = \frac{10}{9} (2 i + 2 j + k)$ .

Common pitfalls

Vector vs scalar. The vector projection is a vector; the scalar projection

\frac{a \cdot b}{| b |}

is only its signed length.

Which power of $| b |$ . The vector projection divides by

| b |^{2}

, not

| b |

Projecting the right way.

{proj}_{b} a

lies along

b

;

{proj}_{a} b

is different.

Perpendicular component. It is

a - {proj}_{b} a

, and the two parts add back to

a

Frequently asked questions

What is the projection of a onto b?

The vector component of $a$ along $b$ : ${proj}_{b} a = \frac{a \cdot b}{| b |^{2}} b$ .

What is the difference between vector and scalar projection?

The vector projection is a vector; the scalar projection $\frac{a \cdot b}{| b |}$ is just its signed length.

Why divide by |b| squared?

To turn $a \cdot b$ into the right multiple of $b$ ; one $| b |$ normalises $b$ , the other scales the length.

How do you find the perpendicular component?

Subtract the projection from $a$ : $a - {proj}_{b} a$ .

What happens for an obtuse angle?

$a \cdot b < 0$ , so the projection points opposite to $b$ .

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Vectors in geometry (proofs)

Projection of a vector & perpendicular component

📖 Theory

How to project a vector

Common pitfalls

Frequently asked questions

Theory