Let P(3, 2, 3),Q(4, 6, 2),and R(7, 3, 2) be t | vedclass

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(D) The vertices are $P(3, 2, 3)$,$Q(4, 6, 2)$,and $R(7, 3, 2)$.To find $\angle QPR$,we need the direction ratios of vectors $\vec{PQ}$ and $\vec{PR}$

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$\frac{\pi}{3}$

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(D) The vertices are $P(3, 2, 3)$,$Q(4, 6, 2)$,and $R(7, 3, 2)$.To find $\angle QPR$,we need the direction ratios of vectors $\vec{PQ}$ and $\vec{PR}$.$\vec{PQ} = (4-3, 6-2, 2-3) = (1, 4, -1)$.$\vec{PR} = (7-3, 3-2, 2-3) = (4, 1, -1)$.Let $	heta = \angle QPR$. The cosine of the angle between two vectors $\vec{a}$ and $\vec{b}$ is given by $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.$\vec{PQ} \cdot \vec{PR} = (1)(4) + (4)(1) + (-1)(-1) = 4 + 4 + 1 = 9$.$|\vec{PQ}| = \sqrt{1^2 + 4^2 + (-1)^2} = \sqrt{1 + 16 + 1} = \sqrt{18}$.$|\vec{PR}| = \sqrt{4^2 + 1^2 + (-1)^2} = \sqrt{16 + 1 + 1} = \sqrt{18}$.Therefore,$\cos 	heta = \frac{9}{\sqrt{18} \cdot \sqrt{18}} = \frac{9}{18} = \frac{1}{2}$.Since $\cos 	heta = \frac{1}{2}$,we have $	heta = \frac{\pi}{3}$.

Let $P(3, 2, 3)$,$Q(4, 6, 2)$,and $R(7, 3, 2)$ be the vertices of $\triangle PQR$. Then,the angle $\angle QPR$ is

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