Find the angle between the line frac{x+1}{2}= | vedclass

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(A) The direction vector of the line is $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$.The normal vector to the plane is $\vec{n} = 10\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 $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$.The normal vector to the plane is $\vec{n} = 10\hat{i} + 2\hat{j} - 11\hat{k}$.Let $\phi$ be the angle between the line and the plane. The formula for the angle is given by $\sin \phi = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.First,calculate the dot product: $\vec{b} \cdot \vec{n} = (2)(10) + (3)(2) + (6)(-11) = 20 + 6 - 66 = -40$.Next,calculate the magnitudes: $|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.$|\vec{n}| = \sqrt{10^2 + 2^2 + (-11)^2} = \sqrt{100 + 4 + 121} = \sqrt{225} = 15$.Thus,$\sin \phi = \frac{|-40|}{7 	imes 15} = \frac{40}{105} = \frac{8}{21}$.Therefore,$\phi = \sin^{-1}\left(\frac{8}{21}ight)$.

Find the angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane $10x+2y-11z=3$.

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