The co-ordinates of the foot of the perpendicular drawn from the origin to the plane $2 x+6 y-3 z=63$ are

The co-ordinates of the foot of the perpendicular drawn from the origin to the plane $2 x+6 y-3 z=63$ are
  1. $(4,2,-4)$
  2. $\left(\frac{18}{7}, \frac{54}{7}, \frac{-27}{7}\right)$
  3. $\left(\frac{2}{7}, \frac{6}{7}, \frac{-3}{7}\right)$
  4. $\left(\frac{9}{7}, \frac{6}{7}, \frac{-3}{7}\right)$

Solution

The co-ordinates of foot of perpendicular can be obtained by $\begin{aligned} & \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=\frac{-\left(a_1+b y_1+c_1-d\right)}{a^2+b^2+c^2} \\ & \Rightarrow \frac{x-0}{2}=\frac{y-0}{6}=\frac{z-0}{-3}=\frac{-(2 \times 0+6 \times 0-3 \times 0-63)}{2^2+6^2+(-3)^2} \\ & \Rightarrow \frac{x}{2}=\frac{y}{6}=\frac{z}{-3}=\frac{63}{49} \\ & \Rightarrow x=\frac{18}{7}, y=\frac{54}{7}, z=\frac{-27}{7} \\ & \Rightarrow\left(\frac{18}{7}, \frac{54}{7}, \frac{-27}{7}\right)\end{aligned}$

Asked in: MHT CET 2022 (05 Aug Shift 2)

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