The acute angle between the line frac{x-5}{2} | vedclass

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(D) The direction ratios of the line are $\vec{b} = (2, -1, 1)$.The normal vector to the plane $3x - 4y - z + 5 = 0$ is $\vec{n} = (3, -4, -1)$.The an

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

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(D) The direction ratios of the line are $\vec{b} = (2, -1, 1)$.The normal vector to the plane $3x - 4y - z + 5 = 0$ is $\vec{n} = (3, -4, -1)$.The angle $	heta$ between a line with direction vector $\vec{b}$ and a plane with normal $\vec{n}$ is given by $\sin 	heta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.Calculating the dot product: $\vec{b} \cdot \vec{n} = (2)(3) + (-1)(-4) + (1)(-1) = 6 + 4 - 1 = 9$.Calculating the magnitudes: $|\vec{b}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4+1+1} = \sqrt{6}$ and $|\vec{n}| = \sqrt{3^2 + (-4)^2 + (-1)^2} = \sqrt{9+16+1} = \sqrt{26}$.Thus,$\sin 	heta = \frac{|9|}{\sqrt{6} \cdot \sqrt{26}} = \frac{9}{\sqrt{156}} = \frac{9}{\sqrt{4 	imes 39}} = \frac{9}{2\sqrt{39}}$.Since $\sin^2 	heta + \cos^2 	heta = 1$,$\cos 	heta = \sqrt{1 - \sin^2 	heta} = \sqrt{1 - \frac{81}{156}} = \sqrt{\frac{75}{156}} = \sqrt{\frac{25}{52}} = \frac{5}{2\sqrt{13}}$.Therefore,$	heta = \cos^{-1}\left(\frac{5}{2\sqrt{13}}ight)$.

The 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

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