3.7 - Angles and perpendiculars with vectors

Two forms of scalar product

How can coordinates tell us whether two directions are perpendicular? Take $\mathbf a=(1,2,2)$ and $\mathbf c=(2,-1,0)$. Multiplying matching components and adding gives $2-2+0=0$. This number detects a right angle.

The component calculation and the geometrical angle calculation give the same number.

Why does the angle appear? Expanding $|\mathbf a-\mathbf b|^2$ by components gives $|\mathbf a|^2+|\mathbf b|^2-2\mathbf a\cdot\mathbf b$. Applying the cosine rule to the triangle formed by the same two arrows gives $|\mathbf a|^2+|\mathbf b|^2-2|\mathbf a||\mathbf b|\cos\theta$. Equate the expressions.

For nonzero vectors, a positive scalar product gives $0\le\theta<90^\circ$, zero gives $\theta=90^\circ$, and a negative scalar product gives $90^\circ<\theta\le180^\circ$. The boundary values $0^\circ$ and $180^\circ$ occur for vectors in the same and opposite directions respectively. Although $\mathbf0\cdot\mathbf b=0$, the zero vector has no direction and does not define an angle.

Reverse just the second vector. Its magnitude stays the same, but the angle between the directed arrows becomes supplementary.

Dot products also give equations for unknown components. For example, $(1,2,2)$ is perpendicular to $(2,-1,k)$ when $2-2+2k=0$, so $k=0$.

Use the sign of the dot product as a check.

Angles between lines

A line can be followed in either direction, so its direction vector may be replaced by its negative. This produces two supplementary angles. If a question asks for the acute angle between lines, choose the smaller angle; if it names an angle such as $FHG$, use the rays from the named vertex.

The smaller angle $\alpha$ between two lines, with $0\le\alpha\le90^\circ$, satisfies

where $\mathbf d$ and $\mathbf e$ are their direction vectors. The absolute value selects the smaller angle. Perpendicular lines give $90^\circ$; parallel directions give $0^\circ$. The same calculation compares the directions of skew lines without pretending they intersect.

Now put those directions inside a cuboid. The geometry will specify actual rays, so we must not automatically choose the acute angle.

[DIAGRAM: asset_name: Directed angles in a cuboid; asset_slug: cuboid-vector-angles; caption: The angle FHG follows rays from H, so the directions must point from H towards F and G; description: A perspective drawing of a cuboid has O at 0 comma 0 comma 0, A at 4 comma 0 comma 0, B at 4 comma 3 comma 0, C at 0 comma 3 comma 0, D at 0 comma 0 comma 2, E at 4 comma 0 comma 2, F at 4 comma 3 comma 2 and G at 0 comma 3 comma 2. Body diagonals OF and AG meet at H, the centre 2 comma 1.5 comma 1. The rays HF and HG are highlighted with arrows, and edge lengths 4, 3 and 2 are marked. The drawing is a perspective view, not an angle scale drawing.]