In the triangle with vertices A(3,2,0),B(5,3, | vedclass

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Question

(B) The angle bisector theorem states that the bisector of $\angle BAC$ divides the opposite side $BC$ in the ratio of the sides adjacent to the angle

Vedclass · Accepted Answer

$\left(\frac{38}{16}, \frac{57}{16}, \frac{17}{16}\right)$

Vedclass · Answer

(B) The angle bisector theorem states that the bisector of $\angle BAC$ divides the opposite side $BC$ in the ratio of the sides adjacent to the angle,i.e.,$BD/DC = AB/AC$. First,calculate the lengths of sides $AB$ and $AC$: $AB = \sqrt{(5-3)^2 + (3-2)^2 + (2-0)^2} = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$. $AC = \sqrt{(-9-3)^2 + (6-2)^2 + (-3-0)^2} = \sqrt{(-12)^2 + 4^2 + (-3)^2} = \sqrt{144 + 16 + 9} = \sqrt{169} = 13$. Thus,the ratio $BD:DC = AB:AC = 3:13$. Using the section formula,the coordinates of $D$ are given by: $D = \left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n}\right)$ where $m=3, n=13$,$B(5,3,2)$,and $C(-9,6,-3)$. $x = \frac{3(-9) + 13(5)}{3+13} = \frac{-27 + 65}{16} = \frac{38}{16}$. $y = \frac{3(6) + 13(3)}{3+13} = \frac{18 + 39}{16} = \frac{57}{16}$. $z = \frac{3(-3) + 13(2)}{3+13} = \frac{-9 + 26}{16} = \frac{17}{16}$. Therefore,the coordinates of $D$ are $\left(\frac{38}{16}, \frac{57}{16}, \frac{17}{16}\right)$.

In the triangle with vertices $A(3,2,0)$,$B(5,3,2)$,and $C(-9,6,-3)$,the bisector of $\angle BAC$ meets $BC$ at $D$. The coordinates of $D$ are

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