What are sign errors in components?

A negative component, squared in the magnitude or multiplied in the dot product, must keep its sign through the arithmetic.

SACE Board SACE 2022-style practice: Calculator-assumed. The points are A(1, 2, -3), B(5, 3, -2) and C(6, 7, -3). Find AB · AC and hence cos( BAC).

First the two vectors from A: AB = (4, 1, 1) and AC = (5, 5, 0). Dot product: AB · AC = (4)(5) + (1)(5) + (1)(0) = 25. [1 mark] Magnitudes: |AB| = sqrt(4^2 + 1^2 + 1^2) = sqrt(18) = 3sqrt(2) and |AC| = sqrt(5^2 + 5^2) = sqrt(50) = 5sqrt(2). [1 mark] So cos( BAC) = 253sqrt(2)× 5sqrt(2) = 2530 = 56. [1 mark]

SACE Board SACE 2021-style practice: Calculator-free. The rhombus OPQR has OP = p and OR = r. Show that OQ · PR = |r|^2 - |p|^2, then prove the diagonals are perpendicular.

The diagonals are OQ = p + r and PR = r - p. Expand: OQ · PR = (p + r) · (r - p) = p·r - p·p + r·r - r·p. Since p·r = r·p, the middle terms cancel, leaving |r|^2 - |p|^2. [1 mark] In a rhombus all sides are equal, so |p| = |r|, giving OQ · PR = 0. A zero dot product of two nonzero vectors means they are perpendicular, so the diagonals are perpendicular. [1 mark]

§-Syllabus dot point

SASpecialist MathematicsSyllabus dot point

How does the dot product measure the angle between two vectors and detect perpendicularity?

Represent vectors in three dimensions, compute magnitudes and unit vectors, and use the dot product to find angles and test for perpendicularity.

Three-dimensional vectors in component form, magnitude and unit vectors, the dot product in both its component and geometric forms, and using it to find angles, projections and perpendicularity.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

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

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What this dot point is asking
Vectors in three dimensions
The dot product
Finding the angle between vectors
Perpendicularity and projection
Properties of the dot product

What this dot point is asking

You need to represent vectors in three dimensions, find magnitudes and unit vectors, and use the dot (scalar) product to compute angles and test for perpendicularity.

Vectors in three dimensions

Write a vector in component form using the standard basis $\mathbf{i}, \mathbf{j}, \mathbf{k}$ :

\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}.

Addition and scalar multiplication act componentwise. The magnitude (length) is

|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}.

A unit vector in the direction of

\mathbf{a}

\hat{\mathbf{a}} = \dfrac{1}{|\mathbf{a}|}\mathbf{a}

, which has length

1

The dot product

The two forms are the key tool: the component form computes a number, and equating it to $|\mathbf{a}||\mathbf{b}|\cos\theta$ extracts the angle.

Finding the angle between vectors

Rearranging gives $\cos\theta = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ .

Angle between two 3D vectors

Find the angle between $\mathbf{a} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ and $\mathbf{b} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$

Step 1: Compute the dot product

Multiply corresponding components and add. This gives a single number that encodes how much the two vectors point in the same direction.

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

Step 2: Compute both magnitudes

The magnitude of each vector is the square root of the sum of squared components. We need both magnitudes to normalise the dot product into a cosine.

|\mathbf{a}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3, \qquad |\mathbf{b}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3.

Step 3: Find the angle using $\cos\theta = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$

Substituting the dot product and both magnitudes into the angle formula gives a value for $\cos\theta$ , which we then invert.

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

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

(about

1.11

radians).

Final answer: $\theta \approx 63.6^{\circ}$

Perpendicularity and projection

The scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ is $\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}$ , the signed length of the shadow of $\mathbf{a}$ along $\mathbf{b}$ . The vector projection is $\left(\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\right)\mathbf{b}$ . Projections appear when resolving one vector into components parallel and perpendicular to another, for instance splitting a force into a component along an incline and a component perpendicular to it.

Properties of the dot product

The dot product obeys familiar algebraic rules that make it easy to manipulate. It is commutative, $\mathbf{a}\cdot\mathbf{b} = \mathbf{b}\cdot\mathbf{a}$ , and distributive over addition, $\mathbf{a}\cdot(\mathbf{b}+\mathbf{c}) = \mathbf{a}\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{c}$ , so an expression like $(\mathbf{p}+\mathbf{r})\cdot(\mathbf{r}-\mathbf{p})$ expands exactly like a binomial product. It also satisfies $\mathbf{a}\cdot\mathbf{a} = |\mathbf{a}|^2$ , which is the bridge between the algebra of dot products and the geometry of lengths. These properties are what let you prove geometric results - such as the perpendicularity of a rhombus's diagonals - purely by expanding dot products, without coordinates.

Finding a value for perpendicularity

For what value of $t$ are $\mathbf{u} = 3\mathbf{i} + t\mathbf{j} - \mathbf{k}$ and $\mathbf{v} = 2\mathbf{i} + \mathbf{j} + 4\mathbf{k}$ perpendicular?

Step 1: Set up the perpendicularity condition

Two nonzero vectors are perpendicular if and only if their dot product equals zero. This follows directly from the formula $\mathbf{u}\cdot\mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\theta$ , since $\cos 90^{\circ} = 0$ . So we require $\mathbf{u}\cdot\mathbf{v} = 0$ .

Step 2: Compute the dot product in terms of $t$ and solve

Multiplying corresponding components and setting the result equal to zero gives an equation in $t$ .

\mathbf{u}\cdot\mathbf{v} = (3)(2) + (t)(1) + (-1)(4) = 6 + t - 4 = t + 2.

Perpendicular means

t + 2 = 0

, so

t = -2

Final answer: $t = -2$

Exam-style practice questions

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

SACE 20223 marksCalculator-assumed. The points are

A(1, 2, -3)

B(5, 3, -2)

and

C(6, 7, -3)

. Find

\vec{AB} \cdot \vec{AC}

and hence

\cos(\angle BAC)

Show worked answer →

First the two vectors from $A$ : $\vec{AB} = (4, 1, 1)$ and $\vec{AC} = (5, 5, 0)$ .

Dot product: $\vec{AB} \cdot \vec{AC} = (4)(5) + (1)(5) + (1)(0) = 25$ . [1 mark]

Magnitudes: $|\vec{AB}| = \sqrt{4^2 + 1^2 + 1^2} = \sqrt{18} = 3\sqrt{2}$ and $|\vec{AC}| = \sqrt{5^2 + 5^2} = \sqrt{50} = 5\sqrt{2}$ . [1 mark]

So $\cos(\angle BAC) = \dfrac{25}{3\sqrt{2}\times 5\sqrt{2}} = \dfrac{25}{30} = \dfrac{5}{6}$ . [1 mark]

SACE 20212 marksCalculator-free. The rhombus

OPQR

has

\vec{OP} = \mathbf{p}

and

\vec{OR} = \mathbf{r}

. Show that

\vec{OQ} \cdot \vec{PR} = |\mathbf{r}|^2 - |\mathbf{p}|^2

, then prove the diagonals are perpendicular.

Show worked answer →

The diagonals are $\vec{OQ} = \mathbf{p} + \mathbf{r}$ and $\vec{PR} = \mathbf{r} - \mathbf{p}$ .

Expand: $\vec{OQ} \cdot \vec{PR} = (\mathbf{p} + \mathbf{r}) \cdot (\mathbf{r} - \mathbf{p}) = \mathbf{p}\cdot\mathbf{r} - \mathbf{p}\cdot\mathbf{p} + \mathbf{r}\cdot\mathbf{r} - \mathbf{r}\cdot\mathbf{p}$ . Since $\mathbf{p}\cdot\mathbf{r} = \mathbf{r}\cdot\mathbf{p}$ , the middle terms cancel, leaving $|\mathbf{r}|^2 - |\mathbf{p}|^2$ . [1 mark]

In a rhombus all sides are equal, so $|\mathbf{p}| = |\mathbf{r}|$ , giving $\vec{OQ} \cdot \vec{PR} = 0$ . A zero dot product of two nonzero vectors means they are perpendicular, so the diagonals are perpendicular. [1 mark]

SACE 20232 marksCalculator-free. Find the value of

t

for which

\mathbf{u} = 3\mathbf{i} + t\mathbf{j} - \mathbf{k}

and

\mathbf{v} = 2\mathbf{i} + \mathbf{j} + 4\mathbf{k}

are perpendicular.

Show worked answer →

Set the dot product to zero:

\mathbf{u}\cdot\mathbf{v} = (3)(2) + (t)(1) + (-1)(4) = 6 + t - 4 = t + 2.

Perpendicular means $t + 2 = 0$ , so $t = -2$ .

Marks: one for the dot product expression, one for solving $t = -2$ .