The foot of the perpendicular drawn from the origin to the plane $x+y+3 z-4=0$ is

The foot of the perpendicular drawn from the origin to the plane $x+y+3 z-4=0$ is
  1. $\left(\frac{2}{11}, \frac{2}{11}, \frac{9}{11}\right)$
  2. $\left(\frac{4}{11}, \frac{4}{11}, \frac{12}{11}\right)$
  3. $\left(\frac{1}{7}, \frac{1}{7}, \frac{6}{7}\right)$
  4. $\left(\frac{1}{5}, \frac{1}{5}, \frac{3}{5}\right)$

Solution

d.r. of $\perp$ er drawn from origin to the given plane are $1,1,3$. Hence equation of $\perp e r$ line to the plane and passing through origin is $\frac{x}{1}=\frac{y}{1}=\frac{z}{3}=K$ ... say Now let foot of the $\perp$ er be $P(K, K, 3 K)$ Also this point P lies on given plane $\therefore \mathrm{K}+\mathrm{K}+9 \mathrm{~K}-4=0 \Rightarrow 11 \mathrm{~K}=4 \Rightarrow \mathrm{K}=\frac{4}{11}$ Hence $P \equiv\left(\frac{4}{11}, \frac{4}{11}, \frac{12}{11}\right)$

Asked in: MHT CET 2020 (13 Oct Shift 2)

Practice more Line and Plane questions on Aicharya