if A(5,4,6)B(1,-1,3)and C(4,3,2) are the vertices of a triangle and the internal bisector of <BAC meets BC at D find the coordinates of D. Also find the coordinates of D. also find the length of AD
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$$\begin{gathered} \mathrm{Let} ∆\mathrm{ABC} \mathrm{be} \mathrm{a} \mathrm{triangle} \mathrm{and} \mathrm{AD}\hspace{0.17em}\mathrm{is} \mathrm{the} \mathrm{angle} \mathrm{bisector}. \\ \mathrm{Applying} \mathrm{angle} \mathrm{bisector} \mathrm{we} \mathrm{get}: \\ \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{CD}} \\ \mathrm{A}\left(5,4,6\right), \mathrm{B}\left(1,-1,3\right) \mathrm{and} \mathrm{C}\left(4,3,2\right) \\ \mathrm{Applying} \mathrm{section} \mathrm{formula} \mathrm{we} \mathrm{get} \\ \mathrm{AB}=\sqrt{{\left(1-5\right)}^2+{\left(-1-4\right)}^2+{\left(3-6\right)}^2}=\sqrt{16+25+9}=\sqrt{50}=5\sqrt{2} \mathrm{units} \\ \mathrm{AC}=\sqrt{{\left(4-5\right)}^2+{\left(3-4\right)}^2+{\left(2-6\right)}^2}=\sqrt{1+1+16}=3\sqrt{2} \mathrm{units} \\ \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{5\sqrt{2}}{3\sqrt{2}} \\ \Rightarrow \frac{\mathrm{BD}}{\mathrm{CD}}=\frac{5}{3} \\ \mathrm{Thus}, \mathrm{point} \mathrm{D}\hspace{0.17em}\mathrm{divides} \mathrm{the} \mathrm{line} \mathrm{segment} \mathrm{joining} \mathrm{the} \mathrm{points} \mathrm{B}\left(1,-1,3\right) \mathrm{and} \mathrm{C}\left(4,3,2\right) \\ \mathrm{in} \mathrm{the} \mathrm{ratio} 5:3 \\ \mathrm{Let} \mathrm{the} \mathrm{coordinates} \mathrm{of} \mathrm{D} \mathrm{be} \mathrm{D}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right) \\ \mathrm{Hence} \mathrm{by} \mathrm{section} \mathrm{formula} \mathrm{we} \mathrm{get}; \\ \mathrm{x}=\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{\mathrm{m}+\mathrm{n}}=\frac{5\times 4+3\times 1}{5+3}=\frac{23}{8} \\ \mathrm{y}=\frac{{\mathrm{my}}_2+{\mathrm{ny}}_1}{\mathrm{m}+\mathrm{n}}=\frac{5\times 3-3\times 1}{5+3}=\frac{12}{8}=\frac{3}{2} \\ \mathrm{z}=\frac{{\mathrm{mz}}_2+{\mathrm{nz}}_1}{\mathrm{m}+\mathrm{n}}=\frac{5\times 2+3\times 3}{5+3}=\frac{19}{8} \\ \mathbf{Hence}\mathbf{ }\mathbf{Coordinates}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{D}\mathbf{\hspace{0.17em}}\mathbf{are}\mathbf{ }\mathbf{D}\left(\frac{\mathbf{23}}{\mathbf{8}}\mathbf{,}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{,}\frac{\mathbf{19}}{\mathbf{8}}\right) \\ \mathrm{Length} \mathrm{of} \mathrm{AD}\hspace{0.17em}\mathrm{will} \mathrm{be} \mathrm{given} \mathrm{by} \\ \mathrm{AD}=\sqrt{\left(5-\frac{23}{8}\right)+{\left(4-\frac{3}{2}\right)}^2+{\left(6-\frac{19}{8}\right)}^2} \\ =\sqrt{\left(\frac{40-23}{8}\right)+{\left(\frac{8-3}{2}\right)}^2+{\left(\frac{48-19}{8}\right)}^2} \\ =\sqrt{{\left(\frac{17}{8}\right)}^2+{\left(\frac{5}{2}\right)}^2+{\left(\frac{29}{8}\right)}^2} \\ =\frac{1}{8}\sqrt{289+400+841} \\ =\frac{1}{8}\sqrt{1530} \\ \mathbf{Hence}\mathbf{ }\mathbf{AD}\mathbf{=}\frac{\mathbf{1}}{\mathbf{8}}\sqrt{\mathbf{1530}}\mathbf{ }\mathbf{units} \end{gathered}$$

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