If A(4,3,5),B(0,-2,2),and C(3,2,1) are three | vedclass

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Question

(A) 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

$\left(\frac{15}{8}, \frac{4}{8}, \frac{11}{8}\right)$

Vedclass · Answer

(A) 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.,$BD:CD = AB:AC$.First,calculate the lengths of sides $AB$ and $AC$:$AB = \sqrt{(0-4)^2 + (-2-3)^2 + (2-5)^2} = \sqrt{(-4)^2 + (-5)^2 + (-3)^2} = \sqrt{16 + 25 + 9} = \sqrt{50} = 5\sqrt{2}$.$AC = \sqrt{(3-4)^2 + (2-3)^2 + (1-5)^2} = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$.Thus,the ratio $BD:CD = AB:AC = 5\sqrt{2} : 3\sqrt{2} = 5:3$.Point $D$ divides $BC$ internally in the ratio $5:3$. Using the section formula,the coordinates of $D$ are:$D = \left(\frac{5(3) + 3(0)}{5+3}, \frac{5(2) + 3(-2)}{5+3}, \frac{5(1) + 3(2)}{5+3}\right)$$D = \left(\frac{15+0}{8}, \frac{10-6}{8}, \frac{5+6}{8}\right) = \left(\frac{15}{8}, \frac{4}{8}, \frac{11}{8}\right)$.

If $A(4,3,5)$,$B(0,-2,2)$,and $C(3,2,1)$ are three points,then the coordinates of the point $D$ where the bisector of $\angle BAC$ meets the side $BC$ are:

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