The length of the internal bisector of angle | vedclass

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(C) Let $AD$ be the internal bisector of $\angle A$ meeting $BC$ at $D$.According to the Angle Bisector Theorem,$D$ divides the side $BC$ in the ratio

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $AD$ be the internal bisector of $\angle A$ meeting $BC$ at $D$.According to the Angle Bisector Theorem,$D$ divides the side $BC$ in the ratio of the adjacent sides $AB:AC$.First,calculate the lengths of sides $AB$ and $AC$:$AB = \sqrt{(2-4)^2 + (3-7)^2 + (4-8)^2} = \sqrt{(-2)^2 + (-4)^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.$AC = \sqrt{(2-4)^2 + (5-7)^2 + (7-8)^2} = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.The ratio $AB:AC = 6:3 = 2:1$.Using the section formula,the coordinates of $D$ are:$D = \left( \frac{2(2) + 1(2)}{2+1}, \frac{2(5) + 1(3)}{2+1}, \frac{2(7) + 1(4)}{2+1} ight) = \left( \frac{4+2}{3}, \frac{10+3}{3}, \frac{14+4}{3} ight) = \left( 2, \frac{13}{3}, 6 ight)$.Now,calculate the length of $AD$:$AD = \sqrt{(2-4)^2 + (\frac{13}{3} - 7)^2 + (6-8)^2} = \sqrt{(-2)^2 + (-\frac{8}{3})^2 + (-2)^2} = \sqrt{4 + \frac{64}{9} + 4} = \sqrt{8 + \frac{64}{9}} = \sqrt{\frac{72+64}{9}} = \sqrt{\frac{136}{9}} = \frac{\sqrt{4 	imes 34}}{3} = \frac{2\sqrt{34}}{3}$.

The length of the internal bisector of angle $A$ in $\triangle ABC$ with vertices $A(4,7,8)$,$B(2,3,4)$ and $C(2,5,7)$ is

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