A tetrahedron has vertices at P(2,1,3),Q(-1,1 | vedclass

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(D) The normal vector to face $OPQ$ is given by the cross product of vectors $\vec{OP}$ and $\vec{OQ}$.$\vec{n_1} = \vec{OP} 	imes \vec{OQ} = \begin{

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$\cos ^{-1}\left(\frac{25}{\sqrt{59} \sqrt{35}}\right)$

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(D) The normal vector to face $OPQ$ is given by the cross product of vectors $\vec{OP}$ and $\vec{OQ}$.$\vec{n_1} = \vec{OP} 	imes \vec{OQ} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & 3 \ -1 & 1 & 2 \end{vmatrix} = \hat{i}(2-3) - \hat{j}(4+3) + \hat{k}(2+1) = -\hat{i} - 7\hat{j} + 3\hat{k}$.The normal vector to face $PQR$ is given by the cross product of vectors $\vec{PQ}$ and $\vec{PR}$.$\vec{PQ} = (-1-2, 1-1, 2-3) = (-3, 0, -1)$ and $\vec{PR} = (1-2, 2-1, 1-3) = (-1, 1, -2)$.$\vec{n_2} = \vec{PQ} 	imes \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & 0 & -1 \ -1 & 1 & -2 \end{vmatrix} = \hat{i}(0+1) - \hat{j}(6-1) + \hat{k}(-3-0) = \hat{i} - 5\hat{j} - 3\hat{k}$.The angle $	heta$ between the faces is the angle between their normal vectors:$\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} = \frac{|(-1)(1) + (-7)(-5) + (3)(-3)|}{\sqrt{(-1)^2+(-7)^2+3^2} \sqrt{1^2+(-5)^2+(-3)^2}}$$\cos 	heta = \frac{|-1 + 35 - 9|}{\sqrt{1+49+9} \sqrt{1+25+9}} = \frac{25}{\sqrt{59} \sqrt{35}}$.Therefore,$	heta = \cos ^{-1}\left(\frac{25}{\sqrt{59} \sqrt{35}}ight)$.

$A$ tetrahedron has vertices at $P(2,1,3)$,$Q(-1,1,2)$,$R(1,2,1)$ and $O(0,0,0)$. The angle between the faces $OPQ$ and $PQR$ is

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