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