If three consecutive vertices of a parallelog | vedclass

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(B) Given,$A(4,3,5)$,$B(0,6,0)$,$C(-8,1,4)$ and $D$ are the vertices of a parallelogram.Let $D$ be the point $(x, y, z)$.Since the diagonals of a para

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$\cos ^{-1}\left(\frac{55}{\sqrt{149} \sqrt{161}}\right)$

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(B) Given,$A(4,3,5)$,$B(0,6,0)$,$C(-8,1,4)$ and $D$ are the vertices of a parallelogram.Let $D$ be the point $(x, y, z)$.Since the diagonals of a parallelogram bisect each other,the mid-point of $AC$ is equal to the mid-point of $BD$.$\left(\frac{4+(-8)}{2}, \frac{3+1}{2}, \frac{5+4}{2}ight) = \left(\frac{x+0}{2}, \frac{y+6}{2}, \frac{z+0}{2}ight)$$\left(-2, 2, \frac{9}{2}ight) = \left(\frac{x}{2}, \frac{y+6}{2}, \frac{z}{2}ight)$Equating the coordinates,we get $x = -4$,$y = -2$,$z = 9$.Thus,$D$ is $(-4, -2, 9)$.Now,the vector $\vec{AC} = (-8-4)\hat{i} + (1-3)\hat{j} + (4-5)\hat{k} = -12\hat{i} - 2\hat{j} - \hat{k}$.The vector $\vec{BD} = (-4-0)\hat{i} + (-2-6)\hat{j} + (9-0)\hat{k} = -4\hat{i} - 8\hat{j} + 9\hat{k}$.Let $	heta$ be the angle between $\vec{AC}$ and $\vec{BD}$.$\cos 	heta = \frac{|\vec{AC} \cdot \vec{BD}|}{|\vec{AC}| |\vec{BD}|} = \frac{|(-12)(-4) + (-2)(-8) + (-1)(9)|}{\sqrt{(-12)^2 + (-2)^2 + (-1)^2} \sqrt{(-4)^2 + (-8)^2 + 9^2}}$$\cos 	heta = \frac{|48 + 16 - 9|}{\sqrt{144 + 4 + 1} \sqrt{16 + 64 + 81}} = \frac{55}{\sqrt{149} \sqrt{161}}$Therefore,$	heta = \cos ^{-1}\left(\frac{55}{\sqrt{149} \sqrt{161}}ight)$.

If three consecutive vertices of a parallelogram are $A(4,3,5)$,$B(0,6,0)$,$C(-8,1,4)$ and $D$ is the fourth vertex,then the angle between $AC$ and $BD$ is

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