Distance from a Point to a Line

Distance from point P (position vector p) to the line r = a + λb. The cross product |b × (a−p)| gives the area of the parallelogram with sides b and (a−p); dividing by |b| gives the perpendicular height.

Let the line be $\mathbf{r} = \mathbf{a}+\lambda\mathbf{b}$ and the point be $P$ with position vector $\mathbf{p}$.

The vector from the base point $A$ (position vector $\mathbf{a}$) to $P$ is $\overrightarrow{AP} = \mathbf{p}-\mathbf{a}$.

The area of the parallelogram formed by $\mathbf{b}$ and $(\mathbf{p}-\mathbf{a})$ is $|\mathbf{b}\times(\mathbf{p}-\mathbf{a})|$.

This area also equals $|\mathbf{b}| \cdot d$ where $d$ is the perpendicular distance from $P$ to the line.

Therefore:

$$d = \frac{|\mathbf{b}\times(\mathbf{p}-\mathbf{a})|}{|\mathbf{b}|}$$

Cartesian method (alternative): Find the foot of perpendicular $F$ from $P$ to the line. The foot is at parameter $\lambda_0 = \frac{(\mathbf{p}-\mathbf{a})\cdot\mathbf{b}}{|\mathbf{b}|^2}$. Then $d = |PF| = |\mathbf{p}-(\mathbf{a}+\lambda_0\mathbf{b})|$.

When the point lies on the line: $\mathbf{p}-\mathbf{a}$ is parallel to $\mathbf{b}$, so $\mathbf{b}\times(\mathbf{p}-\mathbf{a}) = \mathbf{0}$, giving $d = 0$.