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

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(C) Given vertices $A(4,7,8)$, $B(2,3,4)$, and $C(2,5,7)$.First, calculate the lengths of sides $AB$ and $AC$:$AB = \sqrt{(4-2)^2 + (7-3)^2 + (8-4)^2}

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

Vedclass · Answer

(C) Given vertices $A(4,7,8)$, $B(2,3,4)$, and $C(2,5,7)$.First, calculate the lengths of sides $AB$ and $AC$:$AB = \sqrt{(4-2)^2 + (7-3)^2 + (8-4)^2} = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4+16+16} = \sqrt{36} = 6$.$AC = \sqrt{(4-2)^2 + (7-5)^2 + (8-7)^2} = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3$.By the Angle Bisector Theorem, the internal bisector of angle $A$ divides the opposite side $BC$ in the ratio $AB:AC = 6:3 = 2:1$.Let $D$ be the point on $BC$ dividing it in the ratio $2:1$. Using the section formula:$D = \left( \frac{2(2) + 1(2)}{2+1}, \frac{2(5) + 1(3)}{2+1}, \frac{2(7) + 1(4)}{2+1} ight) = \left( \frac{6}{3}, \frac{13}{3}, \frac{18}{3} ight) = \left( 2, \frac{13}{3}, 6 ight)$.The length of the internal bisector is the distance $AD$:$AD = \sqrt{(4-2)^2 + (7 - \frac{13}{3})^2 + (8-6)^2} = \sqrt{2^2 + (\frac{8}{3})^2 + 2^2} = \sqrt{4 + \frac{64}{9} + 4} = \sqrt{8 + \frac{64}{9}} = \sqrt{\frac{72+64}{9}} = \sqrt{\frac{136}{9}} = \frac{\sqrt{4 	imes 34}}{3} = \frac{2}{3} \sqrt{34}$.

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

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