The angle between the line frac{x + 1}{3} = f | vedclass

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(A) The direction vector of the line is $\vec{b} = 3\hat{i} + 2\hat{j} + 4\hat{k}$.The normal vector to the plane is $\vec{n} = 2\hat{i} + 1\hat{j} -

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$\sin^{-1}\left(\frac{4}{\sqrt{406}}\right)$

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(A) The direction vector of the line is $\vec{b} = 3\hat{i} + 2\hat{j} + 4\hat{k}$.The normal vector to the plane is $\vec{n} = 2\hat{i} + 1\hat{j} - 3\hat{k}$.The angle $	heta$ between a line and a plane is given by $\sin 	heta = \left| \frac{\vec{b} \cdot \vec{n}}{|\vec{b}| |\vec{n}|} ight|$.First,calculate the dot product: $\vec{b} \cdot \vec{n} = (3)(2) + (2)(1) + (4)(-3) = 6 + 2 - 12 = -4$.Next,calculate the magnitudes: $|\vec{b}| = \sqrt{3^2 + 2^2 + 4^2} = \sqrt{9 + 4 + 16} = \sqrt{29}$ and $|\vec{n}| = \sqrt{2^2 + 1^2 + (-3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$.Thus,$\sin 	heta = \frac{|-4|}{\sqrt{29} \sqrt{14}} = \frac{4}{\sqrt{406}}$.Therefore,$	heta = \sin^{-1}\left(\frac{4}{\sqrt{406}}ight)$.Since the provided option $(b)$ is $\sin^{-1}\left(\frac{-4}{\sqrt{406}}ight)$,which is not the standard principal value for an angle between a line and a plane (as it must be non-negative),and option $(a)$ is $\sin^{-1}\left(\frac{4}{\sqrt{406}}ight)$,the correct choice is $(a)$.

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

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