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

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

Answer 1

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



Vectors along the given lines $L_1,L_2$ are
$\widehat{n_1}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -2& 3\\ 1& -1& 1\end{array}\right|=\hat{i}+\hat{j}$
and $\widehat{n_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& -1\\ 3& -1& 2\end{array}\right|=\widehat{3i}-5\hat{j}-7\hat{k}$

Putting $y=0$ in $1^{st}$ given two equation of planes, we get
$2x+3z=2$ and $2x+2z=-2$
$\Rightarrow z=4$
$\therefore$ Point on the plane is $(-5,0,4)$ and normal vector of required plane is
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 0\\ 3& -5& -7\end{array}\right|=-7\hat{i}+7\hat{j}-8\hat{k}$

Hence, equation of plane is $-7x+7y-8z-3=0$
$\therefore$ Perpendicular distance is $\frac{3}{\sqrt{162}}=\frac{1}{3\sqrt{2}}$