Question

If the projection of the line $\frac{x}{2}=\frac{y-1}{2}=\frac{z-1}{1}$ on a plane $P$ is $\frac{x}{1}=\frac{y-1}{1}=\frac{z-1}{-1} $ Then the distance of plane $P$ from origin is

• A. $\sqrt{3}$
• B. $\sqrt{\frac{3}{2}}$
• C. $\sqrt{6}$
• D. $\frac{2}{\sqrt{3}}$

Answer 1

Answer: (B)

Projection of line $\frac{x}{2}=\frac{y-1}{2}=\frac{z-1}{1}$ on a plane $P$ is
$\frac{x}{1}=\frac{y-1}{1}=\frac{z-1}{-1}$
$\therefore $ Plane through these lines is perpendicular to the plane $P$.
Normal to the plane determined by the given lines is
$\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ 2&2&1\\ 1&1&-1\end{vmatrix}=3 \hat{i}+3 \hat{j} $
$\therefore $ Direction ratios normal to the required plane is
$\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ -3&3&0\\ 1&1&-1\end{vmatrix}=-3 \hat{i}-3 \hat{j}-6 \hat{k}$
$\therefore $ Equation of the plane is $x+y+2 z-3=0$ as it passes through (0,1,1)
Its distance from origin is $\sqrt{\frac{3}{2}}$