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 $\frac{1}{3\sqrt{2}}$

Normal to plane$=2x-2y+3z-2=0$ is $\overrightarrow{n_1}=2\hat{i}-2\hat{j}-3\hat{k}$

Normal to plane$=x-y+z+1=0$ is $\overrightarrow{n_2}=\hat{i}-\hat{j}-\hat{k}$

Normal to plane$=x+2y-z-3=0$ is $\overrightarrow{n_3}=\hat{i}+2\hat{j}-\hat{k}$

Normal to plane$=3x-y+2z-1=0$ is $\overrightarrow{n_4}=3\hat{i}-\hat{j}+2\hat{k}$

So line of intersection of planes 1 and 2 is along $\overrightarrow{n_1}\times \overrightarrow{n_2}=\hat{i}+\hat{j}:\overrightarrow{L_1}$

similarly the line of intersection lof planes 3 and 4 is along $\overrightarrow{n_3}\times \overrightarrow{n_4}=3\hat{i}-5\hat{j}-7\hat{k}:\overrightarrow{L_2}$

The normal to the plane containing $\overrightarrow{L_1}$ and $\overrightarrow{L_2}$ is along $(\overrightarrow{i}+\overrightarrow{j})\times (3\hat{i}-5\hat{j}-7\hat{k})=-7\hat{i}+7\hat{j}-3\hat{k}$

By inspection $\overrightarrow{L_1}$ passes through$(0,5,4)$

so the equation of the required plane is$=-7x+7y-8z-3=0$

and its distance from origin is $\frac{3}{(7^2+7^2+8^2{)}^{\frac{1}{2}}}=\frac{3}{\sqrt{162}}=\frac{1}{3\sqrt{2}}.$