Question

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

Answer 1

The correct option is B ${\cos }^{-1}\left(\frac{5}{2\sqrt{13}}\right)$

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