The points A(2, -1, 4),B(1, 0, -1),C(1, 2, 3) | vedclass

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(D) Given points are $A(2, -1, 4)$,$B(1, 0, -1)$,$C(1, 2, 3)$,and $D(2, 1, 8)$.First,we calculate the lengths of the sides using the distance formula

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parallelogram

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(D) Given points are $A(2, -1, 4)$,$B(1, 0, -1)$,$C(1, 2, 3)$,and $D(2, 1, 8)$.First,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.$AB = \sqrt{(1-2)^2 + (0-(-1))^2 + (-1-4)^2} = \sqrt{(-1)^2 + 1^2 + (-5)^2} = \sqrt{1+1+25} = \sqrt{27}$.$BC = \sqrt{(1-1)^2 + (2-0)^2 + (3-(-1))^2} = \sqrt{0^2 + 2^2 + 4^2} = \sqrt{0+4+16} = \sqrt{20}$.$CD = \sqrt{(2-1)^2 + (1-2)^2 + (8-3)^2} = \sqrt{1^2 + (-1)^2 + 5^2} = \sqrt{1+1+25} = \sqrt{27}$.$DA = \sqrt{(2-2)^2 + (-1-1)^2 + (4-8)^2} = \sqrt{0^2 + (-2)^2 + (-4)^2} = \sqrt{0+4+16} = \sqrt{20}$.Since $AB = CD = \sqrt{27}$ and $BC = DA = \sqrt{20}$,the opposite sides are equal.Now,we check the diagonals $AC$ and $BD$.$AC = \sqrt{(1-2)^2 + (2-(-1))^2 + (3-4)^2} = \sqrt{(-1)^2 + 3^2 + (-1)^2} = \sqrt{1+9+1} = \sqrt{11}$.$BD = \sqrt{(2-1)^2 + (1-0)^2 + (8-(-1))^2} = \sqrt{1^2 + 1^2 + 9^2} = \sqrt{1+1+81} = \sqrt{83}$.Since $AC 
eq BD$,the diagonals are not equal.Therefore,the points $A, B, C$,and $D$ form a parallelogram.

The points $A(2, -1, 4)$,$B(1, 0, -1)$,$C(1, 2, 3)$,and $D(2, 1, 8)$ form a

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