Q.Find the co-ordinates of the foot of the perpendicular drawn from the point $2\hat{i} - \hat{j} + 5\hat{k}$ to the line $\vec{r} = (11\hat{i} - 2\hat{j} - 8\hat{k}) + \lambda(10\hat{i} - 4\hat{j} - 11\hat{k})$. Also find the length of the perpendicular.

The perpendicular distance of a point $P(\bar{\alpha})$ from a line $\bar{r} = \bar{a}+\lambda\bar{b}$ can be found without ever locating the actual foot of the perpendicular, by combining a projection with the Pythagorean theorem. If $M$ is the foot of the perpendicular from $P$ onto the line, then $AM$ — the distance along the line from the line's base point $A$ to $M$ — is exactly the scalar projection of $\overline{AP} = \bar{\alpha}-\bar{a}$ onto the line's direction $\bar{b}$, namely $AM = \dfrac{(\bar{\alpha}-\bar{a})\cdot\bar{b}}{|\bar{b}|}$. Since triangle $AMP$ is right-angled at $M$, Pythagoras' theorem gives $PM^2 = AP^2 - AM^2$, so the required perpendicular distance is $PM = \sqrt{|\bar{\alpha}-\bar{a}|^2 - \left[\frac{(\bar{\alpha}-\bar{a})\cdot\bar{b}}{|\bar{b}|}\right]^2}.$ This single formula applies directly whenever the line is already given in vector form. When a line is instead given in Cartesian symmetric form, it is often just as quick, and sometimes quicker, to parametrize the foot of the perpendicular $M$ using the line's own parameter, then solve for that parameter by demanding that $\overline{PM}$ be perpendicular to the line's direction ratios (their dot product must vanish) — after which the ordinary distance formula between the two known points $P$ and $M$ gives the answer directly. Both routes compute the same underlying geometric quantity and must always agree.