If A(3,2,3), B(1,4,6), and C(7,4,5) are the t | vedclass

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(A) In a parallelogram $ABCD$, the diagonals $AC$ and $BD$ bisect each other at point $M$.$M$ is the midpoint of $AC$, so $M = \left(\frac{3+7}{2}, \f

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$\cos ^{-1}\left(\frac{16}{\sqrt{357}}\right)$

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(A) In a parallelogram $ABCD$, the diagonals $AC$ and $BD$ bisect each other at point $M$.$M$ is the midpoint of $AC$, so $M = \left(\frac{3+7}{2}, \frac{2+4}{2}, \frac{3+5}{2}ight) = (5, 3, 4)$.Since $M$ is also the midpoint of $BD$, let $D = (x, y, z)$. Then $\left(\frac{x+1}{2}, \frac{y+4}{2}, \frac{z+6}{2}ight) = (5, 3, 4)$.Solving for $x, y, z$: $x+1=10 \Rightarrow x=9$; $y+4=6 \Rightarrow y=2$; $z+6=8 \Rightarrow z=2$. Thus, $D = (9, 2, 2)$.The vector $\vec{DC} = (7-9, 4-2, 5-2) = (-2, 2, 3)$. The direction ratios are $a_1 = -2, b_1 = 2, c_1 = 3$.The vector $\vec{DB} = (1-9, 4-2, 6-2) = (-8, 2, 4)$. The direction ratios are $a_2 = -8, b_2 = 2, c_2 = 4$.The angle $	heta$ between $\vec{DC}$ and $\vec{DB}$ is given by $\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.$\cos 	heta = \frac{|(-2)(-8) + (2)(2) + (3)(4)|}{\sqrt{(-2)^2 + 2^2 + 3^2} \sqrt{(-8)^2 + 2^2 + 4^2}} = \frac{|16 + 4 + 12|}{\sqrt{4+4+9} \sqrt{64+4+16}} = \frac{32}{\sqrt{17} \sqrt{84}} = \frac{32}{\sqrt{17} \cdot 2\sqrt{21}} = \frac{16}{\sqrt{357}}$.Therefore, $	heta = \cos ^{-1}\left(\frac{16}{\sqrt{357}}ight)$.

If $A(3,2,3)$, $B(1,4,6)$, and $C(7,4,5)$ are the three vertices of a parallelogram $ABCD$, then the angle between its diagonal through $D$ and the side $DC$ is

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