A(-1, 2, -3), B(5, 0, -6), C(0, 4, -1) are th | vedclass

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(A) Given vertices $A(-1, 2, -3), B(5, 0, -6), C(0, 4, -1)$.First,calculate the lengths of sides $AB$ and $AC$:$AB = \sqrt{(-1-5)^2 + (2-0)^2 + (-3+6)

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$\frac{25}{\sqrt{714}}, \frac{8}{\sqrt{714}}, \frac{5}{\sqrt{714}}$

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(A) Given vertices $A(-1, 2, -3), B(5, 0, -6), C(0, 4, -1)$.First,calculate the lengths of sides $AB$ and $AC$:$AB = \sqrt{(-1-5)^2 + (2-0)^2 + (-3+6)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.$AC = \sqrt{(-1-0)^2 + (2-4)^2 + (-3+1)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.By the Angle Bisector Theorem,the internal bisector of $\angle BAC$ divides $BC$ in the ratio $AB:AC = 7:3$.Let $D$ be the point on $BC$ dividing it in ratio $7:3$. Using the section formula:$D = \left( \frac{7(0) + 3(5)}{7+3}, \frac{7(4) + 3(0)}{7+3}, \frac{7(-1) + 3(-6)}{7+3} \right) = \left( \frac{15}{10}, \frac{28}{10}, \frac{-25}{10} \right) = \left( \frac{3}{2}, \frac{14}{5}, -\frac{5}{2} \right)$.The vector $\vec{AD} = D - A = \left( \frac{3}{2} - (-1), \frac{14}{5} - 2, -\frac{5}{2} - (-3) \right) = \left( \frac{5}{2}, \frac{4}{5}, \frac{1}{2} \right)$.To find the direction cosines,normalize $\vec{AD}$:$|\vec{AD}| = \sqrt{(\frac{5}{2})^2 + (\frac{4}{5})^2 + (\frac{1}{2})^2} = \sqrt{\frac{25}{4} + \frac{16}{25} + \frac{1}{4}} = \sqrt{\frac{26}{4} + \frac{16}{25}} = \sqrt{\frac{13}{2} + \frac{16}{25}} = \sqrt{\frac{325 + 32}{50}} = \sqrt{\frac{357}{50}} = \sqrt{\frac{714}{100}} = \frac{\sqrt{714}}{10}$.Direction cosines are $\frac{5/2}{\sqrt{714}/10}, \frac{4/5}{\sqrt{714}/10}, \frac{1/2}{\sqrt{714}/10} = \frac{25}{\sqrt{714}}, \frac{8}{\sqrt{714}}, \frac{5}{\sqrt{714}}$.

$A(-1, 2, -3), B(5, 0, -6), C(0, 4, -1)$ are the vertices of a triangle $ABC$. The direction cosines of the internal bisector of $\angle BAC$ are

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