Question

If $A(4,7,8),B(2,3,4)$ and $C(2,5,7)$ are the vertices of $△ABC$, than the length of the internal bisector of the angle $A$ is

• A. $\frac{1}{2}\sqrt{34}$
• B. $\frac{1}{3}\sqrt{34}$
• C. $\frac{2}{3}\sqrt{34}$
• D. $\frac{3}{8}\sqrt{17}$

Answer 1

Answer: (C)

Given that, points $A(4,7,8),B(2,3,4)$ and $C(2,5,7)$
Length
$AB=\sqrt{{\left(x_A-x_B\right)}^2+{\left(y_A-y_B\right)}^2+{\left(z_A-z_B\right)}^2}$
$=\sqrt{(4-2{)}^2+(7-3{)}^2+(8-4{)}^2}$
$=\sqrt{(2{)}^2+(4{)}^2+(4{)}^2}=\sqrt{4+16+16}$
$AB=6$
Length
$AC=\sqrt{{\left(x_A-x_C\right)}^2+{\left(y_A-y_C\right)}^2+{\left(z_A-z_C\right)}^2}$
$=\sqrt{(4-2{)}^2+(7-5{)}^2+(8-7{)}^2}$
$=\sqrt{(2{)}^2+(2{)}^2+(1{)}^2}=\sqrt{4+4+1}$
$AC=3\Rightarrow AB:AC=6:3=2:1$
Now, let say the internal bisector of the angle $A$ meets the side $BC$ 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
$=\sqrt{(4-2{)}^2+{\left(7-\frac{13}{3}\right)}^2+(8-6{)}^2}$
$=\sqrt{(2{)}^2+{\left(\frac{8}{3}\right)}^2+(2{)}^2}$
$=\sqrt{4+\frac{64}{9}+4}=\sqrt{\frac{136}{9}}=\frac{2}{3}\sqrt{34}$