The points A(3, 2, 0),B(5, 3, 2),and C(-9, 6, | vedclass

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(B) Let the vertices be $A(3, 2, 0)$,$B(5, 3, 2)$,and $C(-9, 6, -3)$.First,calculate the lengths of sides $AB$ and $AC$:$AB = \sqrt{(5-3)^2 + (3-2)^2

Vedclass · Accepted Answer

$\left( \frac{19}{8}, \frac{57}{16}, \frac{17}{16} \right)$

Vedclass · Answer

(B) Let the vertices be $A(3, 2, 0)$,$B(5, 3, 2)$,and $C(-9, 6, -3)$.First,calculate the lengths of sides $AB$ and $AC$:$AB = \sqrt{(5-3)^2 + (3-2)^2 + (2-0)^2} = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.$AC = \sqrt{(-9-3)^2 + (6-2)^2 + (-3-0)^2} = \sqrt{(-12)^2 + 4^2 + (-3)^2} = \sqrt{144 + 16 + 9} = \sqrt{169} = 13$.According to the Angle Bisector Theorem,the bisector $AD$ divides the side $BC$ in the ratio $AB : AC = 3 : 13$.Using the section formula,the coordinates of $D$ are given by:$D = \left( \frac{3(-9) + 13(5)}{3+13}, \frac{3(6) + 13(3)}{3+13}, \frac{3(-3) + 13(2)}{3+13} \right)$$D = \left( \frac{-27 + 65}{16}, \frac{18 + 39}{16}, \frac{-9 + 26}{16} \right)$$D = \left( \frac{38}{16}, \frac{57}{16}, \frac{17}{16} \right) = \left( \frac{19}{8}, \frac{57}{16}, \frac{17}{16} \right)$.Thus,option $B$ is correct.

The points $A(3, 2, 0)$,$B(5, 3, 2)$,and $C(-9, 6, -3)$ are the vertices of a triangle $ABC$. If $AD$ is the internal bisector of $\angle BAC$ which meets $BC$ at $D$,then the coordinates of $D$ are:

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