Question
Using vector method, find the incentre of the triangle whose vertices are $P(0,4,0), Q(0,0,3)$ and $R(0,4,3)$.
✓
Answer
Let $\bar{p}, \bar{q}, \bar{r}$be the position vectors of vertices P, Q, R of Δ PQR respectively
$\bar{p}=4 \hat{j}, \bar{q}=3 \widehat{k}, \bar{r}=4 \hat{j}+3 \widehat{k}$
$\overline{P Q}=\bar{q}-\bar{p}=3 \widehat{k}-4 \hat{j}=-4 \hat{j}+3 \widehat{k}$
$\overline{Q R}=\bar{r}-\bar{q}=4 \hat{j}+3 \widehat{k}-3 \widehat{k}=4 \hat{j}$
$\overline{R P}=\bar{p}-\bar{r}=4 \hat{j}-4 \hat{j}-3 \widehat{k}=-3 \widehat{k}$
Let x, y, z be the lengths of opposites of vertices P,Q,R respectively.
$x=|\overline{Q R}|=4$
$y=|\overline{R P}|=3$
$z=|\overline{P Q}|=\sqrt{16+9}=\sqrt{25}=5$
If $H (\bar{h})$ is the incentre of $\Delta PQR$ then
$\bar{h}=\frac{x \bar{p}+y \bar{q}+z \bar{r}}{x+y+z}$
$=\frac{4(4 \hat{j})+3(3 \widehat{k})+5(4 \hat{j}+3 \widehat{k})}{4+3+5}$
$=\frac{16 \hat{j}+9 \widehat{k}+20 \hat{j}+15 \widehat{k}}{12}$
$=\frac{36 \hat{j}+24 \widehat{k}}{12}=3 \hat{j}+2 \widehat{k}$
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