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How do vectors extend to three dimensions, and how do we measure distances and angles between them?

Three-dimensional vectors in component form, the magnitude and distance between points in space, the scalar (dot) product and the angle between two vectors, and the condition for perpendicular vectors.

A CCEA A-Level Mathematics answer on three-dimensional vectors in component form, the magnitude of a vector and the distance between points in space, the scalar product, the angle between two vectors, and the condition for two vectors to be perpendicular.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

CCEA wants you to work with three-dimensional vectors in component form, find the magnitude of a vector and the distance between two points in space, calculate the scalar (dot) product, use it to find the angle between two vectors, and recognise the condition for two vectors to be perpendicular. This extends the two-dimensional vectors of AS 1 into space.

The answer

Three-dimensional vectors

Magnitude and distance in space

The scalar (dot) product

Angle and perpendicularity

Rearranging the scalar product gives the angle between two vectors:

cos⁡θ=a⋅b∣a∣ ∣b∣.\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|}.

Two non-zero vectors are perpendicular exactly when a⋅b=0\mathbf{a} \cdot \mathbf{b} = 0, since cos⁡90∘=0\cos 90^\circ = 0. This is the quickest test for a right angle in space.

Worked example: testing for a right angle

Examples in context

Example 1. Work done by a force. The work done by a force F\mathbf{F} over a displacement d\mathbf{d} is the scalar product F⋅d\mathbf{F} \cdot \mathbf{d}, which picks out the component of force along the motion. The dot product is the physical meaning of "force times distance in the direction of travel".

Example 2. The angle of a roof truss. The angle between two girders meeting at a joint, each described by a three-dimensional vector, is found from the scalar-product formula. Engineers use exactly this calculation to set angles in space frames.

Try this

Q1. Find the magnitude of v=2i−j+2k\mathbf{v} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}. [2 marks]

  • Cue. 4+1+4=3\sqrt{4 + 1 + 4} = 3.

Q2. Find a⋅b\mathbf{a} \cdot \mathbf{b} for a=i+2j+3k\mathbf{a} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} and b=4i−j+2k\mathbf{b} = 4\mathbf{i} - \mathbf{j} + 2\mathbf{k}. [2 marks]

  • Cue. 4−2+6=84 - 2 + 6 = 8.

Q3. What does a⋅b=0\mathbf{a} \cdot \mathbf{b} = 0 tell you about two non-zero vectors? [1 mark]

  • Cue. They are perpendicular.

Exam-style practice questions

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

CCEA 20206 marksThe vectors are a=2i+3j−k\mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k} and b=i−2j+2k\mathbf{b} = \mathbf{i} - 2\mathbf{j} + 2\mathbf{k}. Find the angle between them.
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The scalar product is a⋅b=(2)(1)+(3)(−2)+(−1)(2)=2−6−2=−6\mathbf{a} \cdot \mathbf{b} = (2)(1) + (3)(-2) + (-1)(2) = 2 - 6 - 2 = -6.

The magnitudes are ∣a∣=4+9+1=14|\mathbf{a}| = \sqrt{4 + 9 + 1} = \sqrt{14} and ∣b∣=1+4+4=9=3|\mathbf{b}| = \sqrt{1 + 4 + 4} = \sqrt{9} = 3.

Then cos⁡θ=a⋅b∣a∣∣b∣=−6314=−214=−0.5345\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \frac{-6}{3\sqrt{14}} = \frac{-2}{\sqrt{14}} = -0.5345.

So θ=cos⁡−1(−0.5345)=122.3∘\theta = \cos^{-1}(-0.5345) = 122.3^\circ.

Markers reward the scalar product, both magnitudes, the cosine formula, and the angle.

CCEA 20195 marksThe points A(1,0,2)A(1, 0, 2) and B(4,−2,6)B(4, -2, 6) are given. Find the vector AB→\overrightarrow{AB} and the distance ABAB.
Show worked answer →

The displacement is AB→=b−a=(4−1)i+(−2−0)j+(6−2)k=3i−2j+4k\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (4 - 1)\mathbf{i} + (-2 - 0)\mathbf{j} + (6 - 2)\mathbf{k} = 3\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}.

The distance is ∣AB→∣=32+(−2)2+42=9+4+16=29=5.39|\overrightarrow{AB}| = \sqrt{3^2 + (-2)^2 + 4^2} = \sqrt{9 + 4 + 16} = \sqrt{29} = 5.39 (to three significant figures).

Markers reward the displacement vector, applying the three-dimensional Pythagoras, and the distance.

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