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

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(A) The angle $	heta$ between a line with direction ratios $(a, b, c)$ and a plane with normal vector $\vec{n} = (a_1, b_1, c_1)$ is given by the for

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$\sin ^{-1}\left(\frac{4}{21}\right)$

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(A) The angle $	heta$ between a line with direction ratios $(a, b, c)$ and a plane with normal vector $\vec{n} = (a_1, b_1, c_1)$ is given by the formula: $\sin 	heta = \frac{|a a_1 + b b_1 + c c_1|}{\sqrt{a^2 + b^2 + c^2} \sqrt{a_1^2 + b_1^2 + c_1^2}}$ Here,the direction ratios of the line are $(a, b, c) = (2, 1, 2)$ and the normal vector to the plane is $\vec{n} = (6, -2, -3)$. The magnitude of the direction vector of the line is $\sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$. The magnitude of the normal vector to the plane is $\sqrt{6^2 + (-2)^2 + (-3)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$. Substituting these values into the formula: $\sin 	heta = \frac{|(2)(6) + (1)(-2) + (2)(-3)|}{3 	imes 7} = \frac{|12 - 2 - 6|}{21} = \frac{4}{21}$. Therefore,$	heta = \sin ^{-1}\left(\frac{4}{21}ight)$.

The angle between the line $\frac{x-1}{2}=\frac{y+3}{1}=\frac{z+7}{2}$ and the plane $\bar{r} \cdot(6 \hat{\imath}-2 \hat{\jmath}-3 \hat{k})=5$ is

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