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. ${\cos }^{-1}\left(\frac{3}{\sqrt{17}}\right)$
• B. ${\cos }^{-1}\left(\sqrt{\frac{3}{17}}\right)$
• C. ${\sin }^{-1}\left(\frac{3}{\sqrt{17}}\right)$
• D. ${\sin }^{-1}\left(\sqrt{\frac{3}{17}}\right)$

Answer 1

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

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

So, the direction ratios of the line can be chosen as $-\hat{j}+4\hat{k}$

So, the angle between the plane $x+y+z+5=0$ and the line obtained is given by ${\sin }^{-1}\frac{-1+4}{\sqrt{17}\sqrt{3}}={\sin }^{-1}\sqrt{\frac{3}{17}}$

So, option D is the correct answer.