Question

Find the equation of the plane if the foot of the perpendicular from origin to the plane is $(2,3,-5)$

• A. $2x+3y+5y=38$
• B. $2x+3y-5y=38$
• C. $2x-3y-5y=38$
• D. None of these

Answer 1

Answer: (B)

$2x+3y-5y=38$

General equation of plan $\Rightarrow ax+by+c=d$

Given $(2,3,-5)$ is on the plane.

$2a+3b-5c=d----\left(1\right)$

Directional ratios of the perpendicular to the plane

$\Rightarrow (a,b,c)$

And the perpendicular passes through the origin.

$\therefore$ equation of line $\Rightarrow \frac{x-2}{a}=\frac{y-3}{b}=\frac{z+5}{c}=k$

The line passes through origin $\Rightarrow x=ak+2$

$y=bk+3$

$z=ck-5$

$\Rightarrow ak+2=0$

$\Rightarrow a=-2/k$

$bk+3=0\Rightarrow b=-3/k$

$ck-5=0\Rightarrow c=5/k$

$\therefore$ equation of plane

$\Rightarrow \frac{-2}{k}\left(x\right)-\frac{3}{k}y+\frac{5}{k}z=d$

$-2x-3y+5z=dk$

It passes through $(2,3,-5)$

$-2\left(2\right)-3\left(3\right)+5(-5)=dk$

$-38=dk$

$\therefore$ equation of plane $\Rightarrow 5z+38=2x+3y$

$\Rightarrow 2x+3y-5z=38$