Question

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

• A. ${\sin }^{-1}\left(\frac{-4}{\sqrt{406}}\right)$
• B. ${\cos }^{-1}\left(\frac{-4}{\sqrt{406}}\right)$
• C. ${\sin }^{-1}\left(\frac{-2}{\sqrt{406}}\right)$
• D. none of these.

Answer 1

Answer: (D)

The d.n, of the line are $<3,2,4>$ and attitude numbers of the plane are $<2,1,-3>$.
Hence, the angle between the line and the plane is given by
$\sin \theta =\frac{|3\times 2+2\times 1+4\times \left(-3\right)|}{\sqrt{3^2+2^2+4^2}\sqrt{2^2+1^2+3^2}}$
i.e., by $\sin \theta =\frac{4}{\sqrt{29}\sqrt{14}}=\frac{4}{\sqrt{406}}$