Question

Find the foot of the perpendicular from (0, 2, 7) on the line $\frac{x+2}{-1}=\frac{y-1}{3}=\frac{z-3}{-2}.$

Answer 1

Let L be the foot of the perpendicular drawn from the point P (0, 2, 7) to the given line.

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

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



The direction ratios of PL are proportional to $-\lambda -2-0,3\lambda +1-2,-2\lambda +3-7,\mathrm{i}.\mathrm{e}.-\lambda -2,3\lambda -1,-2\lambda -4$.

The direction ratios of the given line are proportional to $-$1, 3, $-$2, but PL is perpendicular to the given line.

$$\begin{gathered} \therefore -1\left(-\lambda -2\right)+3\left(3\lambda -1\right)-2\left(-2\lambda -4\right)=0 \\ \Rightarrow \lambda =-\frac{1}{2} \end{gathered}$$
Substituting $\lambda =-\frac{1}{2}$ in $\left(-\lambda -2,3\lambda +1,-2\lambda +3\right)$, we get the coordinates of L as $\left(-\frac{3}{2},-\frac{1}{2},4\right)$.

Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.