Question

An angle between the plane, $x+y+z=5$ and the line of intersection of the planes,$3x+4y+z-1=0$ and $5x+8y+2z+14=0$, is

• A. ${\cos }^{-1}\frac{3}{\sqrt{17}}$
• B. ${\sin }^{-1}\frac{3}{\sqrt{17}}$
• C. ${\cos }^{-1}\sqrt{\frac{3}{17}}$
• D. ${\sin }^{-1}\sqrt{\frac{3}{17}}$

Answer 1

The correct option is D ${\sin }^{-1}\sqrt{\frac{3}{17}}$

Normal to $3x+4y+z=1$ is $3\hat{i}+4\hat{j}+\hat{k}$.
Normal to $5x+8y+2z=-14$ is $5\hat{i}+8\hat{j}+2\hat{k}$.
The line at which these planes intersect is perpendicular to both normals, hence its direction ratios are directly proportional to the cross product vector of the normals.
DR's of line$L_1=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 4& 1\\ 5& 8& 2\end{array}\right|=(8-8)\hat{i}-(6-5)\hat{j}+(24-20)\hat{k}=-\hat{j}+4\hat{k}$
Angle between $L_1$ and plane $x+y+z-5=0$
$\sin \theta =\frac{0-1+4}{\sqrt{0+1+16}\sqrt{1+1+1}}=\frac{3}{\sqrt{17}\sqrt{3}}=\sqrt{\frac{3}{17}}$
Required angle $={\sin }^{-1}\sqrt{\frac{3}{17}}$