Question

The points $A(3,2,0)$, $B(5,3,2)$ and $C(0,2,4)$ are the vertices of a triangle. Find the distance of the point $A$ from the point in which the bisector of $??AC$ meets $[BC]$.

• A. $8\sqrt{510}$
• B. $\sqrt{510}$
• C. $\frac{1}{8}\sqrt{510}$
• D. None of these

Answer 1

The Correct Option is C

Let $D$ be the points at which bisector of $??AC$ meets $[BC]$, then $D$ divides $[BC]$ internally in the ratio $c : b$ where $c = |AB|$ and $b = |AC|$.

Now $c = \left|AB\right| = \sqrt{\left(5-3\right)^{2}+\left(3-2\right)^{2}+\left(2-0\right)^{2}} = 3$ units and $b = \left|AC\right| = \sqrt{\left(0-3\right)^{2}+\left(2-2\right)^{2}+\left(4-0\right)^{2}}=\sqrt{25}=5$ units $\therefore D$ divides $\left[BC\right]$ in the ratio $3 : 5$ Hence, $D \equiv \left(\frac{3\times0+5\times 5}{3+5}, \frac{3\times2+5\times3}{3+5}, \frac{3\times4+5\times2}{3+5}\right)$, i.e., $D \equiv \left(\frac{25}{8}, \frac{21}{8}, \frac{22}{8}\right)$. Now, $\left|AD\right|= \sqrt{\left(\frac{25}{8}-3\right)^{2}+\left(\frac{21}{8}-2\right)^{2}+\left(\frac{22}{8}-0\right)^{2}}$ $= \sqrt{\frac{1}{64}+\frac{25}{64}+\frac{484}{64}}$ $=\sqrt{\frac{510}{64}}=\frac{1}{8}\sqrt{510}$