Question

Find the coordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, −1, 3) and C(2, −3, −1). [NCERT EXEMPLAR]

Answer 1

The Cartesian equation of the line joining points B(0, −1, 3) and C(2, −3, −1) is

$$\begin{gathered} \frac{x-0}{2-0}=\frac{y-\left(-1\right)}{-3-\left(-1\right)}=\frac{z-3}{-1-3} \\ \mathrm{Or}\frac{x}{2}=\frac{y+1}{-2}=\frac{\mathrm{z}-3}{-4} \end{gathered}$$

Let L be the foot of the perpendicular drawn from the point A(1, 8, 4) to the line $\frac{x}{2}=\frac{y+1}{-2}=\frac{\mathrm{z}-3}{-4}$.

The coordinates of general point on the line $\frac{x}{2}=\frac{y+1}{-2}=\frac{\mathrm{z}-3}{-4}$ are given by

$$\begin{gathered} \frac{x}{2}=\frac{y+1}{-2}=\frac{\mathrm{z}-3}{-4}=\lambda \\ \mathrm{Or}x=2\lambda ,y=-2\lambda -1,z=-4\lambda +3 \end{gathered}$$

Let the coordinates of L be $\left(2\lambda ,-2\lambda -1,-4\lambda +3\right)$. Therefore, the direction ratios of AL are proportional to

$2\lambda -1,-2\lambda -1-8,-4\lambda +3-4$ or $2\lambda -1,-2\lambda -9,-4\lambda -1$

Direction ratios of the given line are proportional to 2, −2, −4.

But, AL is perpendicular to the given line.

$$\begin{gathered} \therefore 2\times \left(2\lambda -1\right)+\left(-2\right)\times \left(-2\lambda -9\right)+\left(-4\right)\times \left(-4\lambda -1\right)=0 \\ \Rightarrow 4\lambda -2+4\lambda +18+16\lambda +4=0 \\ \Rightarrow 24\lambda +20=0 \\ \Rightarrow \lambda =-\frac{5}{6} \end{gathered}$$

Putting $\lambda =-\frac{5}{6}$ in $\left(2\lambda ,-2\lambda -1,-4\lambda +3\right)$, we get

$\left(2\times \left(-\frac{5}{6}\right),-2\times \left(-\frac{5}{6}\right)-1,-4\times \left(-\frac{5}{6}\right)+3\right)=\left(-\frac{5}{3},\frac{2}{3},\frac{19}{3}\right)$

Thus, the required coordinates of the foot of the perpendicular are $\left(-\frac{5}{3},\frac{2}{3},\frac{19}{3}\right)$.