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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 4.78 min answer

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  1. What this dot point is asking
  2. Vectors in three dimensions
  3. The dot product
  4. Finding the angle between vectors
  5. Perpendicularity and projection

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 i,j,k\mathbf{i}, \mathbf{j}, \mathbf{k}:

a=a1i+a2j+a3k=(a1a2a3).\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
∣a∣=a12+a22+a32.|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}.

A unit vector in the direction of a\mathbf{a} is a^=1∣a∣a\hat{\mathbf{a}} = \dfrac{1}{|\mathbf{a}|}\mathbf{a}, which has length 11.

The dot product

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

Finding the angle between vectors

Rearranging gives cos⁡θ=a⋅b∣a∣∣b∣\cos\theta = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}.

Perpendicularity and projection

The scalar projection of a\mathbf{a} onto b\mathbf{b} is a⋅b∣b∣\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}, the signed length of the shadow of a\mathbf{a} along b\mathbf{b}. The vector projection is (a⋅b∣b∣2)b\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.

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.

2018 SACE Stage 22 marksThe points are A(1, 2, -3), B(5, 3, -2) and C(6, 7, -3). Find the dot product AB . AC.
Show worked answer →

First find the two vectors from A.
AB = B - A = (5-1, 3-2, -2-(-3)) = (4, 1, 1).
AC = C - A = (6-1, 7-2, -3-(-3)) = (5, 5, 0). [1 mark]

The dot product multiplies matching components and adds:
AB . AC = (4)(5) + (1)(5) + (1)(0) = 20 + 5 + 0 = 25. [1 mark]

2018 SACE Stage 21 marksUsing A(1, 2, -3), B(5, 3, -2), C(6, 7, -3), find cos(angle BAC).
Show worked answer →

The angle BAC is the angle between AB and AC at A. Use cos(angle) = (AB . AC) / (|AB| |AC|).

From the dot product, AB . AC = 25.
|AB| = sqrt(4^2 + 1^2 + 1^2) = sqrt(18) = 3 sqrt(2).
|AC| = sqrt(5^2 + 5^2 + 0^2) = sqrt(50) = 5 sqrt(2). [implicit]

So cos(angle BAC) = 25 / (3 sqrt(2) x 5 sqrt(2)) = 25 / (15 x 2) = 25/30 = 5/6. [1 mark]

2017 SACE Stage 22 marksFigure 1 shows rhombus OPQR with OP = p and OR = r. Show that OQ . PR = |r|^2 - |p|^2, then hence prove that the diagonals of the rhombus OPQR are perpendicular, giving reasons.
Show worked answer →

The diagonals are OQ and PR. In the rhombus OQ = OP + OR = p + r, and PR = OR - OP = r - p.

Compute the dot product, expanding like a binomial:
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 . r - p . p = |r|^2 - |p|^2. [1 mark]

In a rhombus all sides are equal, so |p| = |r| (OP and OR are adjacent sides). Therefore OQ . PR = |r|^2 - |p|^2 = 0. A zero dot product of two non-zero vectors means they are perpendicular, so the diagonals OQ and PR are perpendicular. [1 mark]