Question

If $L_1$ is the line of intersection of the planes $2x-2y+3z-2=0,x-y+z+1=0$ and $L_2$ is the line of intersection of the planes , $x+2x-z-3=0,3x-y+2z-1=0$ then the distance of the origin from the plane containing the lines $L_1$ and $L_2,$ is ?

• A. $\frac{1}{4\sqrt{2}}$
• B. $\frac{1}{3\sqrt{2}}$
• C. $\frac{1}{2\sqrt{2}}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is B $\frac{1}{3\sqrt{2}}$

(B) $L_1$ is line of intersection of plane

$L_1$ is parallel to $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -2& 3\\ 1& -1& 1\end{array}\right|=\hat{i}+\hat{j}$

$L_2$ is parallel to $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -1\\ 3& -1& 2\end{array}\right|=3\hat{i}-5\hat{j}-7\hat{k}$

Also, $L_2$ passes through $(\frac{5}{7},\frac{8}{7},0)$ $[$ put $z=0$ in last two planes$]$

$\Rightarrow$ Equation of plane$\frac{3}{\sqrt{7^2+7^2+8^2}}=\frac{1}{3\sqrt{2}}$.