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

Question

Acute angle between the line $ \frac{x-5}{2}=\frac{y+1}{-1}=\frac{z+4}{1} $ and the plane $ 3 x-4 y-z+5=0 $ is, $ \sin ^{-1}\left(\frac{9}{\sqrt{364}}\right) $, $ \sin ^{-1}\left(\frac{9}{2 \sqrt{13}}\right) $, $ \cos ^{-1}\left(\frac{9}{\sqrt{364}}\right) $, $ \cos ^{-1}\left(\frac{5}{2 \sqrt{13}}\right) $

MARKS App · Accepted Answer

$ L: \frac{(x-5)}{2}=\frac{(y+1)}{-1}=\frac{(z+4)}{1} $ Dr's of line: $ 2,-1,1 $ P: $ 3 x-4 y-z+5=0 $ 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| $ $$ \begin{array}{l} \sin \theta=\frac{9}{\sqrt{156}} \ \theta=\sin ^{-1}\left(\frac{9}{\sqrt{156}}\right) \ \theta=\sin ^{-1}\left(\frac{9}{2 \sqrt{39}}\right) \ \theta=\cos ^{-1}\left(\frac{5}{2 \sqrt{13}}\right) \end{array} $$

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