Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let $\bar{a}=2\hat{j}+\hat{k},\hat{b}=-2\hat{i},\hat{c}=-2\hat{i}+2\hat{k}$ $\therefore \overline{\mathrm{AB}}=-2\hat{\mathrm{i}}-2\hat{\mathrm{j}}-\hat{\mathrm{k}}$, $\overline{\mathrm{BC}}=2\hat{\mathrm{k}}$ $\begin{array}{rl}& \overline{\mathrm{AC}}=-2\hat{\mathrm{i}}-2\hat{\mathrm{j}}+\hat{\mathrm{k}}\\ & \Rightarrow |\overline{\mathrm{AB}}|=3,|\overline{\mathrm{BC}}|=2,|\overline{\mathrm{AC}}|=3\end{array}$ Incentre of $△\mathrm{ABC}$ is given by $\frac{|\overline{\mathrm{AB}}|\overline{\mathrm{c}}+|\overline{\mathrm{BC}}|\overline{\mathrm{a}}+|\overline{\mathrm{AC}}|\overline{\mathrm{b}}}{|\overline{\mathrm{AB}}|+|\overline{\mathrm{BC}}|+|\overline{\mathrm{AC}}|}$ $\begin{array}{rl}& =\frac{3(-2\hat{i}+2\hat{k})+2(2\hat{j}+\hat{k})+3(-2\hat{i})}{3+2+3}\\ & =\frac{-12\hat{i}+4\hat{j}+8\hat{k}}{8}\\ & =-\frac{3}{2}\hat{i}+\frac{1}{2}\hat{j}+\hat{k}\end{array}$