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WASpecialist MathematicsSyllabus dot point

How does the dot product measure the angle between two three-dimensional vectors and project one onto another?

Compute the scalar (dot) product of vectors in three dimensions and use it for angles, perpendicularity and projections

WACE Specialist Unit 3 scalar product: the component and geometric definitions of the dot product, the angle formula, the perpendicularity test, and scalar and vector projections, with a worked example in three dimensions.

Generated by Claude Opus 4.76 min answer

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  1. What this dot point is asking
  2. Two definitions, one product
  3. The angle between two vectors
  4. Perpendicularity test
  5. Projections

What this dot point is asking

SCSA wants you to compute the dot product from components, use it to find angles and test perpendicularity, and resolve one vector along another by projection.

Two definitions, one product

The scalar product has a component definition and a geometric definition that agree:

The component form is the computational tool; the geometric form abcosθ|\mathbf{a}||\mathbf{b}|\cos\theta explains why it measures angle. The product is commutative and distributes over addition, and aa=a2\mathbf{a}\cdot\mathbf{a} = |\mathbf{a}|^2.

The angle between two vectors

Rearranging the geometric form gives

cosθ=abab.\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 result lies in [0,π][0, \pi].

Perpendicularity test

This is the quickest way to check or impose a right angle, for instance to find a value of a parameter that makes two vectors perpendicular.

Projections

The scalar projection of a\mathbf{a} onto b\mathbf{b} is abb\tfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}, the signed length of the shadow of a\mathbf{a} along b\mathbf{b}. The vector projection multiplies this by the unit vector b^\hat{\mathbf{b}}:

projba=abb2b.\text{proj}_{\mathbf{b}}\,\mathbf{a} = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\,\mathbf{b}.

This resolves a\mathbf{a} into a component along b\mathbf{b} and a remaining component perpendicular to it.