The angle between the line frac{x - 1}{-2} = | vedclass

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(A) The direction vector of the line is $\vec{b} = -2\hat{i} + \hat{j} + 2\hat{k}$.The normal vector to the plane is $\vec{n} = 3\hat{i} + 2\hat{j} +

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$\sin^{-1}\left(\frac{8}{21}\right)$

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(A) The direction vector of the line is $\vec{b} = -2\hat{i} + \hat{j} + 2\hat{k}$.The normal vector to the plane is $\vec{n} = 3\hat{i} + 2\hat{j} + 6\hat{k}$.The angle $	heta$ between a line and a plane is given by $\sin 	heta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.Calculate the dot product: $\vec{b} \cdot \vec{n} = (-2)(3) + (1)(2) + (2)(6) = -6 + 2 + 12 = 8$.Calculate the magnitudes: $|\vec{b}| = \sqrt{(-2)^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.$|\vec{n}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.Thus,$\sin 	heta = \frac{|8|}{3 	imes 7} = \frac{8}{21}$.Therefore,$	heta = \sin^{-1}\left(\frac{8}{21}ight)$.

The angle between the line $\frac{x - 1}{-2} = \frac{y - 2}{1} = \frac{z + 1}{2}$ and the plane $3x + 2y + 6z = 1$ is:

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