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(C) The direction ratios of the line are $\vec{v} = (2, 5, 1)$.The normal vector to the plane $8x + 2y - z = 4$ is $\vec{n} = (8, 2, -1)$.The angle $\

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$\sin ^{-1}\left(\frac{25}{\sqrt{2070}}\right)$

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(C) The direction ratios of the line are $\vec{v} = (2, 5, 1)$.The normal vector to the plane $8x + 2y - z = 4$ is $\vec{n} = (8, 2, -1)$.The angle $	heta$ between a line with direction vector $\vec{v}$ and a plane with normal $\vec{n}$ is given by $\sin 	heta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|}$.Calculating the dot product: $\vec{v} \cdot \vec{n} = (2)(8) + (5)(2) + (1)(-1) = 16 + 10 - 1 = 25$.Calculating the magnitudes: $|\vec{v}| = \sqrt{2^2 + 5^2 + 1^2} = \sqrt{4 + 25 + 1} = \sqrt{30}$ and $|\vec{n}| = \sqrt{8^2 + 2^2 + (-1)^2} = \sqrt{64 + 4 + 1} = \sqrt{69}$.Thus,$\sin 	heta = \frac{25}{\sqrt{30} \sqrt{69}} = \frac{25}{\sqrt{2070}}$.Therefore,$	heta = \sin ^{-1}\left(\frac{25}{\sqrt{2070}}ight)$.

The angle between the line with the direction ratios $(2, 5, 1)$ and the plane $8x + 2y - z = 4$ is

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