The length of perpendicular from the point (1, 0, 1) to the plane x - y + z = 4 is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\frac{2}{\sqrt{3}}$
$\vec n=\hat i-\hat j+\hat k$
let a point be fool of perpendicular on planes from (1, 0, 1)
so eq. of line along $\vec n$ containing (1, 0, 1)
$\frac{x-1}{1}=\frac{y}{-1}=\frac{z-1}{1}=λ$
$x=λ+1,y=-λ,z=λ+1$
so $λ+1+λ+λ+1=4$ (in eq. of plane)
so $3λ=2$
$λ=\frac{2}{3}$
$x=\frac{5}{3},y=-\frac{2}{3},z=\frac{5}{3}$
So distance = $\sqrt{(1-\frac{5}{3})^2+(0+\frac{2}{3})^2+(1-\frac{5}{3})^2}$
$=\frac{2}{\sqrt{3}}$