Question
Using vector method, find the incenter of the triangle whose vertices are $\mathrm{A}(0,3,0)$, $\mathrm{B}(0,0,4)$ and $\mathrm{C}(0,3,4)$.

Answer

Let $\bar{a}=3 \hat{j}, \bar{b}=4 \hat{k}$ and $\bar{c}=3 \hat{j}+4 \hat{k}$
$
\begin{array}{ll}
\therefore & \overline{\mathrm{AB}}=\bar{b}-\bar{a}=-3 \hat{j}+4 \hat{k}, \overline{\mathrm{AC}}=\bar{c}-\bar{a}=4 \hat{k}, \overline{\mathrm{BC}}=\bar{c}-\bar{b}=3 \hat{j} \\
\therefore & |\overline{\mathrm{AB}}|=5,|\overline{\mathrm{AC}}|=4,|\overline{\mathrm{BC}}|=3
\end{array}
$
If $\mathrm{H}(\bar{h})$ is the incenter of triangle $\mathrm{ABC}$ then,
$
\begin{aligned}
& \therefore & \bar{h} & =\frac{|\overline{\mathrm{BC}}| \bar{a}+|\overline{\mathrm{AC}}| \bar{b}+|\overline{\mathrm{AB}}| \bar{c}}{|\overline{\mathrm{BC}}|+|\overline{\mathrm{AC}}|+|\overline{\mathrm{AB}}|} \\
& \therefore & \bar{h} & =\frac{3(3 j)+4(4 k)+5(3 j+4 k)}{3+4+5} \\
& & & =\frac{9 j+16 k+15 j+20 k}{12} \\
& \therefore & \bar{h} & =\frac{24 \hat{j}+36 \hat{k}}{12} \\
& \therefore & \bar{h} & =2 \hat{j}+3 \hat{k} \\
& \text { And } & \mathrm{H} & \equiv(0,2,3)
\end{aligned}
$
Note : In $\triangle \mathrm{ABC}$,
1) P.V. of Centroid is given by $\frac{\bar{a}+\bar{b}+\bar{c}}{3}$
2) P.V. of Incentre is given by $\frac{|\overline{\mathrm{AB}}| \bar{c}+|\overline{\mathrm{BC}}| \bar{a}+|\overline{\mathrm{AC}}| \bar{b}}{|\overline{\mathrm{AB}}|+|\overline{\mathrm{BC}}|+|\overline{\mathrm{AC}}|}$
3) P.V. of Orthocentre is given by $\frac{\tan \mathrm{A} \bar{a}+\tan \mathrm{B} \bar{b}+\tan \mathrm{C} \bar{c}}{\tan \mathrm{A}+\tan \mathrm{B}+\tan \mathrm{C}}$ (Verify)

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