The four points A(2,-1,3), B(4,-2,1), C(4,5,- | vedclass

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(B) Let the given points be $A(2,-1,3), B(4,-2,1), C(4,5,-7)$ and $D(2,6,-5)$.First, we find the mid-points of the diagonals $AC$ and $BD$:Mid-point o

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Parallelogram

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(B) Let the given points be $A(2,-1,3), B(4,-2,1), C(4,5,-7)$ and $D(2,6,-5)$.First, we find the mid-points of the diagonals $AC$ and $BD$:Mid-point of $AC = \left(\frac{2+4}{2}, \frac{-1+5}{2}, \frac{3-7}{2}ight) = (3, 2, -2)$.Mid-point of $BD = \left(\frac{4+2}{2}, \frac{-2+6}{2}, \frac{1-5}{2}ight) = (3, 2, -2)$.Since the mid-points of the diagonals are the same, the quadrilateral $ABCD$ is a parallelogram.Next, we check the side lengths:$AB = \sqrt{(4-2)^2 + (-2 - (-1))^2 + (1-3)^2} = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.$BC = \sqrt{(4-4)^2 + (5 - (-2))^2 + (-7-1)^2} = \sqrt{0^2 + 7^2 + (-8)^2} = \sqrt{0+49+64} = \sqrt{113}$.Since $AB 
eq BC$, it is not a rhombus or a square.Now, check if it is a rectangle by calculating the dot product of adjacent sides:$\vec{AB} = (4-2)\hat{i} + (-2+1)\hat{j} + (1-3)\hat{k} = 2\hat{i} - \hat{j} - 2\hat{k}$.$\vec{BC} = (4-4)\hat{i} + (5+2)\hat{j} + (-7-1)\hat{k} = 0\hat{i} + 7\hat{j} - 8\hat{k}$.$\vec{AB} \cdot \vec{BC} = (2)(0) + (-1)(7) + (-2)(-8) = 0 - 7 + 16 = 9 
eq 0$.Since the dot product is not zero, the sides are not perpendicular.Therefore, $ABCD$ is a parallelogram.

The four points $A(2,-1,3), B(4,-2,1), C(4,5,-7)$ and $D(2,6,-5)$ form a:

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