Given triangle ABC such that A = 2hat{i} - ha | vedclass

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(B) The vertices of $	riangle ABC$ are $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(3, -4, -4)$.First,we calculate the lengths of the sides using the distance

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$A$ right-angled triangle

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(B) The vertices of $	riangle ABC$ are $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(3, -4, -4)$.First,we calculate the lengths of the sides using the distance formula:$AB = \sqrt{(1-2)^2 + (-3 - (-1))^2 + (-5-1)^2} = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}$.$BC = \sqrt{(3-1)^2 + (-4 - (-3))^2 + (-4 - (-5))^2} = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.$CA = \sqrt{(2-3)^2 + (-1 - (-4))^2 + (1 - (-4))^2} = \sqrt{(-1)^2 + 3^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$.Now,check for the Pythagorean theorem:$BC^2 + CA^2 = 6 + 35 = 41 = AB^2$.Since $AB^2 = BC^2 + CA^2$,the triangle satisfies the condition for a right-angled triangle.Thus,$	riangle ABC$ is a right-angled triangle.

Given $\triangle ABC$ such that $A = 2\hat{i} - \hat{j} + \hat{k}$,$B = \hat{i} - 3\hat{j} - 5\hat{k}$,and $C = 3\hat{i} - 4\hat{j} - 4\hat{k}$,then $\triangle ABC$ is:

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