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Mathematics Question on Three Dimensional Geometry

Acute angel between the line x−52=y+1−1=z+41\frac{x-5}{2} = \frac{y+1}{-1} = \frac{z+4}{1} and the plane 3x - 4y - z + 5 = 0 is

A

cos⁡−1(5213)\cos^{-1} \left(\frac{5}{2\sqrt{13}}\right)

B

cos⁡−1(9364)\cos^{-1} \left(\frac{9}{\sqrt{364}}\right)

C

sin⁡−1(523)\sin^{-1} \left(\frac{5}{2\sqrt{3}}\right)

D

sin⁡−1(9364)\sin^{-1} \left(\frac{9}{\sqrt{364}}\right)

Answer

cos⁡−1(5213)\cos^{-1} \left(\frac{5}{2\sqrt{13}}\right)

Explanation

Solution

dr�s of line 2, -1, 1
dr�s of normal to plane -3, 4, 1
sinθ=∣−6−4+14+1+19+16+1∣=∣−9156∣=9156sin \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}}
θ=sin−1(9156)=sin−1(9239)=cos−1(5213)\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)