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Q.
If $A(4,7,8), B(2,3,4)$ and $C(2,5,7)$ are the vertices of $\triangle ABC$, than the length of the internal bisector of the angle $A$ is
TS EAMCET 2020
Solution:
Given that, points $A(4,7,8), B(2,3,4)$ and $C(2,5,7)$
Length
$A B=\sqrt{\left(x_{A}-x_{B}\right)^{2}+\left(y_{A}-y_{B}\right)^{2}+\left(z_{A}-z_{B}\right)^{2}}$
$=\sqrt{(4-2)^{2}+(7-3)^{2}+(8-4)^{2}}$
$=\sqrt{(2)^{2}+(4)^{2}+(4)^{2}}=\sqrt{4+16+16}$
$A B=6$
Length
$A C=\sqrt{\left(x_{A}-x_{C}\right)^{2}+\left(y_{A}-y_{C}\right)^{2}+\left(z_{A}-z_{C}\right)^{2}}$
$=\sqrt{(4-2)^{2}+(7-5)^{2}+(8-7)^{2}}$
$=\sqrt{(2)^{2}+(2)^{2}+(1)^{2}}=\sqrt{4+4+1}$
$A C=3 \Rightarrow A B: A C=6: 3=2: 1$
Now, let say the internal bisector of the angle $A$ meets the side $B C$ that point
$D\left(x_{D}, y_{D}, z_{D}\right)$
$\therefore x$-coordinate of point $D$,
$x_{D}=\frac{m x_{C}+n x_{B}}{m +n}=\frac{2 \times 2+1 \times 2}{2+1}$
$\Rightarrow x_{D}=2$
$y$-coordinate of point $D$,
$y_{D}=\frac{m y_{C}+n y_{B}}{m +n}=\frac{2 \times 5+1 \times 3}{2+1}$
$=\frac{13}{3}$
$z$-coordinate of point $D, z$-coordinate of point $D$,
$z_{D}=\frac{m z_{C}+n z_{B}}{m +n}=\frac{2 \times 7+1 \times 4}{2+1}$
$=\frac{18}{3}=6$
$\because$ Point $D\left(2, \frac{13}{3}, 6\right)$
So, length
$=\sqrt{(4-2)^{2}+\left(7-\frac{13}{3}\right)^{2}+(8-6)^{2}}$
$=\sqrt{(2)^{2}+\left(\frac{8}{3}\right)^{2}+(2)^{2}}$
$=\sqrt{4+\frac{64}{9}+4}=\sqrt{\frac{136}{9}}=\frac{2}{3} \sqrt{34}$