If the coordinates of the vertices of a trian | vedclass

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(A) Since $AD$ is the angle bisector of $\angle A$,by the Angle Bisector Theorem,we have $\frac{AB}{AC} = \frac{BD}{CD}$.First,calculate the lengths o

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$\left(\frac{13}{3}, \frac{10}{3}, 4\right)$

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(A) Since $AD$ is the angle bisector of $\angle A$,by the Angle Bisector Theorem,we have $\frac{AB}{AC} = \frac{BD}{CD}$.First,calculate the lengths of sides $AB$ and $AC$:$AB = \sqrt{(5-7)^2 + (4-6)^2 + (6-4)^2} = \sqrt{(-2)^2 + (-2)^2 + 2^2} = \sqrt{4+4+4} = \sqrt{12} = 2\sqrt{3}$.$AC = \sqrt{(3-7)^2 + (2-6)^2 + (0-4)^2} = \sqrt{(-4)^2 + (-4)^2 + (-4)^2} = \sqrt{16+16+16} = \sqrt{48} = 4\sqrt{3}$.Thus,the ratio $\frac{BD}{CD} = \frac{AB}{AC} = \frac{2\sqrt{3}}{4\sqrt{3}} = \frac{1}{2}$.Point $D$ divides the line segment $BC$ internally in the ratio $1:2$. Using the section formula,the coordinates of $D$ are:$D = \left(\frac{1(3) + 2(5)}{1+2}, \frac{1(2) + 2(4)}{1+2}, \frac{1(0) + 2(6)}{1+2}\right)$$D = \left(\frac{3+10}{3}, \frac{2+8}{3}, \frac{0+12}{3}\right) = \left(\frac{13}{3}, \frac{10}{3}, 4\right)$.

If the coordinates of the vertices of a $\triangle ABC$ are $A(7,6,4)$,$B(5,4,6)$,$C(3,2,0)$ and the bisector of $\angle BAC$ meets the side $BC$ at $D$,then the coordinates of $D$ are

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