If A(4, 7, 8),B(2, 3, 4),and C(2, 5, 7) are t | vedclass

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(B) Let $AD$ be the angle bisector of $\angle A$,where $D$ lies on $BC$. The point $D$ divides $BC$ in the ratio $AB : AC$.First,calculate the lengths

Vedclass · Accepted Answer

$\frac{2\sqrt{34}}{3}$

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(B) Let $AD$ be the angle bisector of $\angle A$,where $D$ lies on $BC$. The point $D$ divides $BC$ in the ratio $AB : AC$.First,calculate the lengths of sides $AB$ and $AC$:$\vec{AB} = (2-4, 3-7, 4-8) = (-2, -4, -4) \implies |\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.$\vec{AC} = (2-4, 5-7, 7-8) = (-2, -2, -1) \implies |\vec{AC}| = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.The ratio $AB : AC = 6 : 3 = 2 : 1$.Using the section formula,the coordinates of $D$ are:$D = \frac{1 \cdot B + 2 \cdot C}{1 + 2} = \frac{(2, 3, 4) + 2(2, 5, 7)}{3} = \frac{(2+4, 3+10, 4+14)}{3} = \frac{(6, 13, 18)}{3} = (2, \frac{13}{3}, 6)$.Now,the length of the angle bisector $AD$ is the distance between $A(4, 7, 8)$ and $D(2, \frac{13}{3}, 6)$:$|AD| = \sqrt{(2-4)^2 + (\frac{13}{3} - 7)^2 + (6-8)^2} = \sqrt{(-2)^2 + (\frac{13-21}{3})^2 + (-2)^2} = \sqrt{4 + (-\frac{8}{3})^2 + 4} = \sqrt{8 + \frac{64}{9}} = \sqrt{\frac{72+64}{9}} = \sqrt{\frac{136}{9}} = \frac{\sqrt{4 	imes 34}}{3} = \frac{2\sqrt{34}}{3}$.

If $A(4, 7, 8)$,$B(2, 3, 4)$,and $C(2, 5, 7)$ are the vertices of a triangle $ABC$,find the length of the angle bisector of $\angle A$.

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