If the coordinates of the points A, B, C are | vedclass

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(D) The direction ratios of line $AB$ are $(2 - (-1), 3 - 3, 5 - 2) = (3, 0, 3)$.The direction ratios of line $AC$ are $(3 - (-1), 5 - 3, -2 - 2) = (4

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$90$

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(D) The direction ratios of line $AB$ are $(2 - (-1), 3 - 3, 5 - 2) = (3, 0, 3)$.The direction ratios of line $AC$ are $(3 - (-1), 5 - 3, -2 - 2) = (4, 2, -4)$.Let $	heta$ be the angle between $AB$ and $AC$. The formula for the cosine of the angle between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is $\cos 	heta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.Substituting the values: $\cos A = \frac{|(3)(4) + (0)(2) + (3)(-4)|}{\sqrt{3^2 + 0^2 + 3^2} \sqrt{4^2 + 2^2 + (-4)^2}}$.$\cos A = \frac{|12 + 0 - 12|}{\sqrt{18} \sqrt{16 + 4 + 16}} = \frac{0}{\sqrt{18} \sqrt{36}} = 0$.Since $\cos A = 0$,we have $A = 90^o$.

If the coordinates of the points $A, B, C$ are $(-1, 3, 2)$,$(2, 3, 5)$,and $(3, 5, -2)$ respectively,then $\angle A = \dots^o$.

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