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 form the plane, containing the lines $L_1$ and $L_2$ is :

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

Answer 1

The correct option is B $\frac{1}{4\sqrt{2}}$

Given that,

$L_1$ is interaction of phases $2x-2y+3z-2=0\to P_{1}x-y+z+1=0\to P_2$

$L_2$ is interaction of phases $x+2y-z-3=0\to P_{3}3x-y+2z-1=0\to P_4$

Normal to plane 1 is $\overline{n_1}=2\hat{i}-2\hat{j}+3\hat{k}$

Normal to plane 2 is $\overline{n_2}=\hat{i}-\hat{j}+\hat{k}$

Normal to plane 3 is $\overline{n_3}=\hat{i}+2\hat{j}-\hat{k}$

Normal to plane 4 is $\overline{n_4}=3\hat{i}-\hat{j}+2\hat{k}$

$L_1$ will lie along $\overline{n_1}\times \overline{n_2}=\left|\begin{array}{ccc}i& j& k\\ 2& -2& 3\\ 1& -1& 1\end{array}\right|=\hat{i}+\hat{j}$

$L_2$ will lie along $\overline{n_3}\times \overline{n_4}=\left|\begin{array}{ccc}i& j& k\\ 1& 2& -1\\ 3& -1& 2\end{array}\right|=3\hat{i}-5\hat{j}-7\hat{k}$

Normal to the plane containing $L_1$ and $L_2=(\hat{i}+\hat{j})\times (3\hat{i}-5\hat{j}-7\hat{k})=\left|\begin{array}{ccc}i& j& k\\ 1& 1& 0\\ 3& -5& -7\end{array}\right|=-7\hat{i}+7\hat{j}-8\hat{k}$

equation of plane having $L_1$ & $L_2=-7x+7y-8z+d=0$ ---eq.1

By checking $P_1$ and $P_2$ , by putting x = 0, then y = 5 & z = 4 satisfy both the planes.

So, as the point ( 0 ,5 , 4) lies o both planes $P_1,P_2$ , it also lies on plane containing $L_1$ and $L_2$.

By putting (0 , 5 , 4) in eq.1 , we get d = -3 $

$\therefore$ Equation is -7x + 7y - 8z - 3 = 0

$\therefore$ Distance from origin $=\frac{d}{\sqrt{a^2+b^2+c^2}}=\frac{+3}{\sqrt{7^2+7^2+8^2}}=\frac{+3}{\sqrt{162}}=\frac{1}{3\sqrt{2}}$