Question

Find the foot of the perpendicular from (1, 2, −3) to the line $\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}.$

Answer 1

Let L be the foot of the perpendicular drawn from the point P (1, 2, $-$3) to the given line.

The coordinates of a general point on the line $\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}$ are given by
$$\begin{gathered} \frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}=\lambda \\ \Rightarrow x=2\lambda -1 \\ y=-2\lambda +3 \\ z=-\lambda \end{gathered}$$

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



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

The direction ratios of the given line are proportional to 2, $-$2, $-$1, but PL is perpendicular to the given line.

$$\begin{gathered} \therefore 2\left(2\lambda -2\right)-2\left(-2\lambda +1\right)-1\left(-\lambda +3\right)=0 \\ \Rightarrow \lambda =1 \end{gathered}$$

Substituting $\lambda =1$ in $\left(2\lambda -1,-2\lambda +3,-\lambda \right)$, we get the coordinates of L as $\left(1,1,-1\right)$.