If A(1, 2, -3),B(2, 3, -1),and C(3, 1, 1) are | vedclass

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(B) First,calculate the lengths of the sides of $	riangle ABC$: $c = AB = \sqrt{(2-1)^2 + (3-2)^2 + (-1 - (-3))^2} = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6

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$\frac{3 \sqrt{3}}{\sqrt{7}}$

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(B) First,calculate the lengths of the sides of $	riangle ABC$: $c = AB = \sqrt{(2-1)^2 + (3-2)^2 + (-1 - (-3))^2} = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}$ $a = BC = \sqrt{(3-2)^2 + (1-3)^2 + (1 - (-1))^2} = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3$ $b = AC = \sqrt{(3-1)^2 + (1-2)^2 + (1 - (-3))^2} = \sqrt{2^2 + (-1)^2 + 4^2} = \sqrt{21}$ Using the cosine rule: $\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{21 + 6 - 9}{2 	imes \sqrt{21} 	imes \sqrt{6}} = \frac{18}{2 \sqrt{126}} = \frac{9}{3 \sqrt{14}} = \frac{3}{\sqrt{14}}$ $\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{9 + 6 - 21}{2 	imes 3 	imes \sqrt{6}} = \frac{-6}{6 \sqrt{6}} = -\frac{1}{\sqrt{6}}$ Therefore,$\left|\frac{\cos A}{\cos B}ight| = \left|\frac{3}{\sqrt{14}} \div \left(-\frac{1}{\sqrt{6}}ight)ight| = \left| -\frac{3 \sqrt{6}}{\sqrt{14}} ight| = \frac{3 \sqrt{3 	imes 2}}{\sqrt{7 	imes 2}} = \frac{3 \sqrt{3}}{\sqrt{7}}$.

If $A(1, 2, -3)$,$B(2, 3, -1)$,and $C(3, 1, 1)$ are the vertices of $\triangle ABC$,then $\left|\frac{\cos A}{\cos B}\right| = $

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