Question

The distance of the plane passing through the point $P(1,1,1)$ and perpendicular to the line $\frac{x-1}{3}=\frac{y-1}{0}=\frac{z-1}{4}$ from the origin is

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{7}{5}$
• D. $1$

Answer 1

Answer: (C)

$\frac{7}{5}$

Equation of the plane passing through $(1,1,1)$ is given by,

$A(x-1)+B(y-1)+C(z-1)=0$ ...$\left(1\right)$

Since the line is perpendicular to the plane $\left(1\right)$

$\therefore 3(x-1)+0(y-1)+4(z-1)=0$

$3x+4z-7=0$

Hence, distance from $(0,0,0)$ to the plane $\left(1\right)$ is,

$d=\left|\frac{-7}{\sqrt{3^2+4^2}}\right|=\frac{7}{5}$