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

Answer 1

$L_1$ is line of the intersection of the plane

$2x-2y+3z-2=0$ and $x-y+z+1=0$ and $L_2$ is line

of intersection of the plane $x+2y-z-3=0$ and $3x-y+2z-1=0$

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

$L_2$ 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)

so, equation of plane is

$\left|\begin{array}{ccc}x-\frac{5}{7}& y-\frac{8}{7}& z\\ 1& 1& 0\\ 3& -5& -7\end{array}\right|=0$

$7x-7y+8z+3=0$

Now, perpendicular distance from origin is

$\left|\frac{3}{\sqrt{7^2+7^2+8^2}}\right|=\frac{3}{\sqrt{162}}=\frac{1}{3\sqrt{2}}$