The angle between the plane x + y + z = 5 and | vedclass

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(D) The direction vector $\vec{v}$ of the line of intersection of the planes $3x + 4y + z - 1 = 0$ and $5x + 8y + 2z + 14 = 0$ is given by the cross p

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$\sin^{-1}\left(\sqrt{\frac{3}{17}}\right)$

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(D) The direction vector $\vec{v}$ of the line of intersection of the planes $3x + 4y + z - 1 = 0$ and $5x + 8y + 2z + 14 = 0$ is given by the cross product of their normal vectors $\vec{n_1} = (3, 4, 1)$ and $\vec{n_2} = (5, 8, 2)$.$\vec{v} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 4 & 1 \ 5 & 8 & 2 \end{vmatrix} = \hat{i}(8-8) - \hat{j}(6-5) + \hat{k}(24-20) = 0\hat{i} - 1\hat{j} + 4\hat{k}$.The normal vector to the plane $x + y + z = 5$ is $\vec{n} = (1, 1, 1)$.The angle $	heta$ between a line with direction vector $\vec{v}$ and a plane with normal vector $\vec{n}$ is given by $\sin 	heta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|}$.$\vec{v} \cdot \vec{n} = (0)(1) + (-1)(1) + (4)(1) = 3$.$|\vec{v}| = \sqrt{0^2 + (-1)^2 + 4^2} = \sqrt{17}$.$|\vec{n}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.$\sin 	heta = \frac{|3|}{\sqrt{17} \cdot \sqrt{3}} = \frac{3}{\sqrt{51}} = \frac{\sqrt{3} \cdot \sqrt{3}}{\sqrt{17} \cdot \sqrt{3}} = \sqrt{\frac{3}{17}}$.Therefore,$	heta = \sin^{-1}\left(\sqrt{\frac{3}{17}}ight)$.

The angle between the plane $x + y + z = 5$ and the line of intersection of the planes $3x + 4y + z - 1 = 0$ and $5x + 8y + 2z + 14 = 0$ is

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