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

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

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

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(B) The direction vector of the line is $\vec{b} = 3\hat{i} + 4\hat{j} + 2\hat{k}$ and the normal vector to the plane is $\vec{n} = 2\hat{i} - 3\hat{j} + 1\hat{k}$.The angle $	heta$ between a line and a plane is given by $\sin 	heta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.$\vec{b} \cdot \vec{n} = (3)(2) + (4)(-3) + (2)(1) = 6 - 12 + 2 = -4$.$|\vec{b}| = \sqrt{3^2 + 4^2 + 2^2} = \sqrt{9 + 16 + 4} = \sqrt{29}$.$|\vec{n}| = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$.Therefore,$\sin 	heta = \frac{|-4|}{\sqrt{29} \sqrt{14}} = \frac{4}{\sqrt{406}}$.Thus,$	heta = \sin^{-1} \left( \frac{4}{\sqrt{406}} ight)$.

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

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