Let the points on the plane P be equidistant | vedclass

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(C) The plane $P$ is the perpendicular bisector plane of the line segment joining $A(-4, 2, 1)$ and $B(2, -2, 3)$.The normal vector $\vec{n}_1$ of pla

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$\frac{\pi}{3}$

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(C) The plane $P$ is the perpendicular bisector plane of the line segment joining $A(-4, 2, 1)$ and $B(2, -2, 3)$.The normal vector $\vec{n}_1$ of plane $P$ is the vector $\vec{AB} = (2 - (-4))\hat{i} + (-2 - 2)\hat{j} + (3 - 1)\hat{k} = 6\hat{i} - 4\hat{j} + 2\hat{k}$.We can take the normal vector as $\vec{n}_1 = 3\hat{i} - 2\hat{j} + \hat{k}$.The midpoint of $AB$ is $M = \left(\frac{-4+2}{2}, \frac{2-2}{2}, \frac{1+3}{2}ight) = (-1, 0, 2)$.The equation of plane $P$ is $3(x + 1) - 2(y - 0) + 1(z - 2) = 0$,which simplifies to $3x - 2y + z + 1 = 0$.The second plane is $P': 2x + y + 3z - 1 = 0$,with normal vector $\vec{n}_2 = 2\hat{i} + \hat{j} + 3\hat{k}$.The angle $	heta$ between the two planes is given by $\cos 	heta = \left| \frac{\vec{n}_1 \cdot \vec{n}_2}{|\vec{n}_1| |\vec{n}_2|} ight|$.$\vec{n}_1 \cdot \vec{n}_2 = (3)(2) + (-2)(1) + (1)(3) = 6 - 2 + 3 = 7$.$|\vec{n}_1| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$.$|\vec{n}_2| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$.$\cos 	heta = \left| \frac{7}{\sqrt{14} \cdot \sqrt{14}} ight| = \frac{7}{14} = \frac{1}{2}$.Therefore,$	heta = \cos^{-1}\left(\frac{1}{2}ight) = \frac{\pi}{3}$.

Let the points on the plane $P$ be equidistant from the points $A(-4, 2, 1)$ and $B(2, -2, 3)$. Then the acute angle between the plane $P$ and the plane $2x + y + 3z = 1$ is

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