The scalar projection of a onto b (the length of the shadow of a along b) is a·b|b|. The vector projection multiplies this by the unit vector b|b| to give a vector along b.

What is sign of the angle?

A negative dot product gives an obtuse angle (cosine negative); keep the sign through to the inverse cosine.

Compute pmatrix 1 \\ 3 \\ -2 pmatrix · pmatrix 4 \\ -1 \\ 1 pmatrix. [2 marks]

Show that pmatrix 2 \\ 1 \\ 0 pmatrix and pmatrix -1 \\ 2 \\ 5 pmatrix are perpendicular. [2 marks]

State the geometric meaning of a·b = 0 for non-zero vectors. [1 mark]

SEAB Original-style practice: Find the angle between the vectors a = pmatrix 1 \\ 2 \\ 2 pmatrix and b = pmatrix 2 \\ 0 \\ -1 pmatrix.

a · b = (1)(2) + (2)(0) + (2)(-1) = 2 + 0 - 2 = 0. Since the dot product is 0, the vectors are perpendicular: the angle is 90^. (Had the dot product been nonzero, use cosθ = a·b|a||b|.) Markers reward computing the dot product and recognising that a zero dot product means a right angle.

SEAB Original-style practice: Given a = pmatrix 3 \\ 4 \\ 0 pmatrix and b = pmatrix 1 \\ 0 \\ 0 pmatrix, find the length of the projection of a onto b.

The (scalar) projection of a onto b is a·b|b|. a·b = 3(1) + 4(0) + 0 = 3, and |b| = 1. So the projection length is 31 = 3. Markers reward the projection formula, the dot product, and dividing by the magnitude of b.

SingaporeMathsSyllabus dot point

How does the scalar product measure the angle between vectors and project one onto another?

Define and compute the scalar (dot) product, use it to find angles between vectors, test for perpendicularity, and find the projection of one vector onto another

A focused answer to the H2 Mathematics outcome on the scalar product. The algebraic and geometric definitions, finding the angle between vectors, the perpendicularity test, and projecting one vector onto another.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to define the scalar (dot) product both algebraically and geometrically, use it to find the angle between two vectors, test whether vectors are perpendicular, and compute the projection of one vector onto another.

The answer

Two definitions

The scalar product of $\mathbf{a}$ and $\mathbf{b}$ is a number, defined equivalently by:

\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \qquad \text{(algebraic)}

\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta \qquad \text{(geometric)}

where $\theta$ is the angle between them. Equating the two forms is the source of every angle calculation.

Finding the angle

Rearranging the geometric form:

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

Compute the dot product and the two magnitudes, then take the inverse cosine.

The perpendicularity test

Since $\cos 90^\circ = 0$ , two non-zero vectors are perpendicular if and only if their dot product is zero. This is the quickest test for a right angle and underpins normal vectors to planes.

Projection

The scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ (the length of the shadow of $\mathbf{a}$ along $\mathbf{b}$ ) is $\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}$ . The vector projection multiplies this by the unit vector $\dfrac{\mathbf{b}}{|\mathbf{b}|}$ to give a vector along $\mathbf{b}$ .

Worked example

Find the angle between $\mathbf{a} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$ , to the nearest degree.

Step 1: Compute the dot product

\mathbf{a}\cdot\mathbf{b} = (2)(1) + (-1)(2) + (2)(2) = 2 - 2 + 4 = 4

Step 2: Compute the magnitudes

|\mathbf{a}| = \sqrt{4 + 1 + 4} = 3, \qquad |\mathbf{b}| = \sqrt{1 + 4 + 4} = 3

Step 3: Apply the cosine formula

\cos\theta = \frac{4}{3 \times 3} = \frac{4}{9}

Step 4: Take the inverse cosine

\theta = \cos^{-1}\left(\frac{4}{9}\right) \approx 63.6^\circ \approx 64^\circ

Examples in context

Example 1. Work done by a force. The work done by a constant force $\mathbf{F}$ over a displacement $\mathbf{d}$ is $W = \mathbf{F}\cdot\mathbf{d} = |\mathbf{F}||\mathbf{d}|\cos\theta$ , so only the component of force along the motion does work, which is why a force perpendicular to motion does none.

Example 2. Checking a right angle in geometry. To verify that a triangle with vertices given by position vectors has a right angle at $B$ , compute $\overrightarrow{BA}\cdot\overrightarrow{BC}$ ; a result of zero confirms the right angle without measuring.

Try this

Q1. Compute $\begin{pmatrix} 1 \\ 3 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix}$ . [2 marks]

Cue. $4 - 3 - 2 = -1$ .

Q2. Show that $\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} -1 \\ 2 \\ 5 \end{pmatrix}$ are perpendicular. [2 marks]

Cue. Dot product $= -2 + 2 + 0 = 0$ , so perpendicular.

Q3. State the geometric meaning of $\mathbf{a}\cdot\mathbf{b} = 0$ for non-zero vectors. [1 mark]

Cue. The vectors are perpendicular (the angle between them is $90^\circ$ ).

Exam-style practice questions

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

Original4 marksFind the angle between the vectors

\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}

and

\mathbf{b} = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}

Show worked answer →

$\mathbf{a} \cdot \mathbf{b} = (1)(2) + (2)(0) + (2)(-1) = 2 + 0 - 2 = 0$ .

Since the dot product is $0$ , the vectors are perpendicular: the angle is $90^\circ$ .

(Had the dot product been nonzero, use $\cos\theta = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ .)

Markers reward computing the dot product and recognising that a zero dot product means a right angle.

Original4 marksGiven

\mathbf{a} = \begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix}

and

\mathbf{b} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}

, find the length of the projection of

\mathbf{a}

onto

\mathbf{b}

Show worked answer →

The (scalar) projection of $\mathbf{a}$ onto $\mathbf{b}$ is $\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}$ .

$\mathbf{a}\cdot\mathbf{b} = 3(1) + 4(0) + 0 = 3$ , and $|\mathbf{b}| = 1$ .

So the projection length is $\dfrac{3}{1} = 3$ .

Markers reward the projection formula, the dot product, and dividing by the magnitude of $\mathbf{b}$ .