Question

Find the foot of perpendicular from the point (2, 3, 4) to the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}.$ Also, find the perpendicular distance from the given point to the line.

Answer 1

Let L be the foot of the perpendicular drawn from the point P (2, 3, 4) to the given line.

The coordinates of a general point on the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$ are given by
$$\begin{gathered} \frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}=\lambda \\ \mathrm{They}\mathrm{can}\mathrm{be}\mathrm{re}-\mathrm{written}\mathrm{as} \\ \frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}=\lambda \\ \Rightarrow x=-2\lambda +4 \\ y=6\lambda \\ z=-3\lambda +1 \end{gathered}$$

Let the coordinates of L be $\left(-2\lambda +4,6\lambda ,-3\lambda +1\right)$.



The direction ratios of PL are proportional to $-2\lambda +4-2,6\lambda -3,-3\lambda +1-4,\mathrm{i}.\mathrm{e}.-2\lambda +2,6\lambda -3,-3\lambda -3$.

The direction ratios of the given line are proportional to $-$2, 6, $-$3, but PL is perpendicular to the given line.

$$\begin{gathered} \therefore -2\left(-2\lambda +2\right)+6\left(6\lambda -3\right)-3\left(-3\lambda -3\right)=0 \\ \Rightarrow \lambda =\frac{13}{49} \end{gathered}$$

Substituting $\lambda =\frac{13}{49}$ in $\left(-2\lambda +4,6\lambda ,-3\lambda +1\right)$, we get the coordinates of L as $\left(\frac{170}{49},\frac{78}{49},\frac{10}{49}\right)$.

$$\begin{gathered} \therefore PL=\sqrt{{\left(\frac{170}{49}-2\right)}^2+{\left(\frac{78}{49}-3\right)}^2+{\left(\frac{10}{49}-4\right)}^2} \\ =\sqrt{\frac{44541}{2401}} \\ =\sqrt{\frac{909}{49}} \\ =\frac{3}{7}\sqrt{101} \end{gathered}$$.


Hence, the length of the perpendicular from P on PL is $\frac{3}{7}\sqrt{101}\mathrm{units}$.