Q.The length of the perpendicular from (1, 6, 3) to the line $\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3}$ is
(A) 3
(B) $\sqrt{11}$
(C) $\sqrt{13}$
(D) 5

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 q …