Scalar and Vector Products Notes — JC1 H2 Math

Scalar product (component form)

a·b = a₁b₁ + a₂b₂ + a₃b₃

Applies to a = a₁i + a₂j + a₃k and b = b₁i + b₂j + b₃k. Dotting a vector with itself gives a·a = |a|².

Vector product (component form)

$\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} \times \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} = \begin{pmatrix} a_2b_3 - a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_1b_2 - a_2b_1 \end{pmatrix}$

This is not given in MF26, so memorise it. The middle component is −(a₁b₃ − a₃b₁). The result is perpendicular to both a and b.

Angle between two vectors

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

Requires non-zero vectors measured tail-to-tail, with 0 ≤ θ ≤ π. For the acute angle between two lines, use |d₁·d₂| in the numerator.

Length of projection of a onto b

$|\mathbf{a}\cdot\hat{\mathbf{b}}| = \frac{|\mathbf{a}\cdot\mathbf{b}|}{|\mathbf{b}|}$

b̂ = b ÷ |b| must be a unit vector. a·b̂ is the signed length. The projection vector (a·b̂)b̂ gives the foot of the perpendicular.

Perpendicular distance from a point to a line

$|\mathbf{a}\times\hat{\mathbf{b}}| = \frac{|\mathbf{a}\times\mathbf{b}|}{|\mathbf{b}|}$

Gives the distance from the point with position vector a to the line through the origin in the direction of b. Check with Pythagoras: (a·b̂)² + |a × b̂|² = |a|².

Area of triangle

$\text{Area} = \frac{1}{2}|\mathbf{a}\times\mathbf{b}|$

a and b are adjacent sides drawn tail-to-tail, and the parallelogram they span has area |a × b|. Use the full vectors and never divide by |b|. For triangle ABC, cross two sides from the same vertex, such as AB and AC.