Distance of a point from a line

Theorem 6.5 (distance of a point from a line). Given a line $\bar{r} = \bar{a} + \lambda\bar{b}$ and any point $P$ with position vector $\bar{\alpha}$, we want a formula for the perpendicular distance of $P$ from the line, without having to construct the actual foot of the perpendicular by hand every time. Let $A$ be the known base point of the line (position vector $\bar{a}$), and let $M$ be the foot of the perpendicular dropped from $P$ onto the line — so $M$ is the point on the line closest to $P$, and $PM$ is perpendicular to the line. Because $M$ lies on the line, the segment $AM$ is exactly the projection of $\overline{AP}$ onto the line's direction; and since the line runs parallel to $\bar{b}$, this is the same as the projection of $\overline{AP}$ onto $\bar{b}$ itself. Using the standard formula for the scalar projection of one vector onto another, $AM = \frac{\overline{AP}\cdot\bar{b}}{|\bar{b}|} = \frac{(\bar{\alpha}-\bar{a})\cdot\bar{b}}{|\bar{b}|}.$ Now triangle $AMP$ is right-angled at $M$ (since $PM$ is perpendicular to the line, and $AM$ lies along the line), so Pythagoras' theorem gives $PM^2 = AP^2 - AM^2$. Since $AP = |\bar{\alpha}-\bar{a}|$, this becomes $PM^2 = |\bar{\alpha}-\bar{a}|^2 - \left[\frac{(\bar{\alpha}-\bar{a})\cdot\bar{b}}{|\bar{b}|}\right]^2,$ and taking the square root gives the required distance: $PM = \sqrt{|\bar{\alpha}-\bar{a}|^2 - \left[\frac{(\bar{\alpha}-\bar{a})\cdot\bar{b}}{|\bar{b}|}\right]^2}.$ In words: the squared distance from A to P splits, via Pythagoras, into a part running along the line (the squared projection, which measures how far along the line the foot of the perpendicular sits) and a part running perpendicular to the line (the actual distance we want) — so subtracting the along-line part from the full squared distance $AP^2$ isolates exactly the perpendicular part. This one formula works whether the point $P$ is given directly with a position vector, or the line is given in Cartesian symmetric form (in which case it is easiest to first find the actual coordinates of the foot of the perpendicular M by demanding that $PM$ be perpendicular to the line's direction, as the second worked example below does).

Worked examples.

Ex.(12). Find the length of the perpendicular from $P(3,2,1)$ to the line $\bar{r} = (7\hat{i}+7\hat{j}+6\hat{k}) + \lambda(-2\hat{i}+2\hat{j}+3\hat{k})$. This length is exactly the distance of P from the line, so Theorem 6.5 applies directly with $\bar{\alpha} = 3\hat{i}+2\hat{j}+\hat{k}$, $\bar{a} = 7\hat{i}+7\hat{j}+6\hat{k}$, and $\bar{b} = -2\hat{i}+2\hat{j}+3\hat{k}$. First, $\bar{\alpha}-\bar{a} = (3\hat{i}+2\hat{j}+\hat{k}) - (7\hat{i}+7\hat{j}+6\hat{k}) = -4\hat{i}-5\hat{j}-5\hat{k}$, so $|\bar{\alpha}-\bar{a}|^2 = (-4)^2+(-5)^2+(-5)^2 = 16+25+25 = 66.$ Next, the dot product with $\bar{b}$: $(\bar{\alpha}-\bar{a})\cdot\bar{b} = (-4)(-2) + (-5)(2) + (-5)(3) = 8 - 10 - 15 = -17$, and $|\bar{b}| = \sqrt{(-2)^2+2^2+3^2} = \sqrt{17}$. Substituting into Theorem 6.5's formula: $PM = \sqrt{66 - \left(\frac{-17}{\sqrt{17}}\right)^2} = \sqrt{66 - 17} = \sqrt{49} = 7\ \text{units}.$ (Here $\left(\frac{-17}{\sqrt{17}}\right)^2 = \frac{289}{17} = 17$, which is why the projection term simplifies so cleanly to $17$.) …