Question

Find the equation of the perpendicular drawn from the point P (2, 4, −1) to the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}.$ Also, write down the coordinates of the foot of the perpendicular from P.

Answer 1

Let L be the foot of the perpendicular drawn from the point P (2, 4, $-$1) to the given line.

The coordinates of a general point on the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$ are given by
$$\begin{gathered} \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}=\lambda \\ \Rightarrow x=\lambda -5 \\ y=4\lambda -3 \\ z=-9\lambda +6 \end{gathered}$$

Let the coordinates of L be $\left(\lambda -5,4\lambda -3,-9\lambda +6\right)$.



The direction ratios of PL are proportional to $\lambda -5-2,4\lambda -3-4,-9\lambda +6+1,\mathrm{i}.\mathrm{e}.\lambda -7,4\lambda -7,-9\lambda +7$.

The direction ratios of the given line are proportional to 1, 4, $-$9, but PL is perpendicular to the given line.

$$\begin{gathered} \therefore 1\left(\lambda -7\right)+4\left(4\lambda -7\right)-9\left(-9\lambda +7\right)=0 \\ \Rightarrow \lambda =1 \end{gathered}$$

Substituting $\lambda =1$ in $\left(\lambda -5,4\lambda -3,-9\lambda +6\right)$, we get the coordinates of L as $\left(-4,1,-3\right)$.

Equation of the line PL is

$$\begin{gathered} \frac{x-2}{-4-2}=\frac{y-4}{1-4}=\frac{z+1}{-3+1} \\ =\frac{x-2}{-6}=\frac{y-4}{-3}=\frac{z+1}{-2} \end{gathered}$$