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

Answer 1

Answer: (B)

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

Answer 2

$L_1$ is parallel to $\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\\ 2&-2&3\\\ 1&-1&1\end{vmatrix}=\hat{i}+\hat{j}$
$L_2$ is parallel to $\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\\ 1&2&-1\\\ 3&-1&2\end{vmatrix}=3\hat{i}-5\hat{j}-7\hat{k}$
Also, $L_2$ passes through $\left(\frac{5}{7}, \frac{8}{7},0\right)$
So, required plane is $\begin{vmatrix}x-\frac{5}{7}&y-\frac{8}{7}&z\\\ 1&1&0\\\ 3&-5&-7\end{vmatrix}=0$
$\Rightarrow 7x - 7y + 8z + 3 = 0$
Now, perpendicular distance $=\frac{3}{\sqrt{162}}=\frac{1}{3\sqrt{2}}$