Question

Acute angel between the line $\frac{x-5}{2} = \frac{y+1}{-1} = \frac{z+4}{1} $ and the plane 3x - 4y - z + 5 = 0 is

• A. $\cos^{-1} \left(\frac{5}{2\sqrt{13}}\right)$
• B. $\cos^{-1} \left(\frac{9}{\sqrt{364}}\right)$
• C. $\sin^{-1} \left(\frac{5}{2\sqrt{3}}\right)$
• D. $\sin^{-1} \left(\frac{9}{\sqrt{364}}\right)$

Answer 1

Answer: (A)

dr’s of line 2, -1, 1
dr’s of normal to plane -3, 4, 1
$sin \theta=\left|\frac{-6-4+1}{\sqrt{4+1+1}\sqrt{9+16+1}}\right|=\left|\frac{-9}{\sqrt{156}}\right|=\frac{9}{\sqrt{156}}$
$\theta=sin^{-1}\left(\frac{9}{\sqrt{156}}\right)=sin^{-1}\left(\frac{9}{2\sqrt{39}}\right)=cos^{-1}\left(\frac{5}{2\sqrt{13}}\right)$