Question

Given the vertices of triangle by position vectors $\hat{i}+\hat{j}+\hat{k},\hat{i}+\hat{k}and\hat{j}+\hat{k}$ the centroid and Incentre of the triangle will be given by

• A. $\frac{2\hat{i}+2\hat{j}+3\hat{k}}{3},\frac{(\sqrt{2}+1)\hat{i}+(\sqrt{2}+1)\hat{j}+(\sqrt{2}+2)\hat{k}}{2+\sqrt{2}}$
• B. $\frac{2\hat{i}+2\hat{j}+3\hat{k}}{3},\frac{(\sqrt{2}+1)\hat{i}+(\sqrt{2}+2)\hat{j}+(\sqrt{2}+1)\hat{k}}{2+\sqrt{2}}$
• C. $\frac{2\hat{i}+2\hat{j}+3\hat{k}}{3},\frac{(\sqrt{2}+1)\hat{i}+(\sqrt{2}+2)\hat{j}+(\sqrt{2}+2)\hat{k}}{2+\sqrt{2}}$
• D. $\frac{2\hat{i}+2\hat{j}+3\hat{k}}{3},\frac{(\sqrt{2}+2)\hat{i}+(\sqrt{2}+1)\hat{j}+(\sqrt{2}+2)\hat{k}}{2+\sqrt{2}}$

Answer 1

The correct option is A $\frac{2\hat{i}+2\hat{j}+3\hat{k}}{3},\frac{(\sqrt{2}+1)\hat{i}+(\sqrt{2}+1)\hat{j}+(\sqrt{2}+2)\hat{k}}{2+\sqrt{2}}$

Given position vecto $\overline{p_1},\overline{p_2}and\overline{p_3}$. We know centroid is given by $G=\frac{p_1+p_2+p_3}{3}$
$I=\frac{|\overline{p_2}-\overline{p_3}|\overline{p_1}+|\overline{p_1}-\overline{p_3}|\overline{p_2}+|\overline{p_1}-\overline{p_2}|\overline{p_3}}{|\overline{p_2}-\overline{p_3}|+|\overline{p_2}-\overline{p_1}|+|\overline{p_3}-\overline{p_1}|}$
Therefor, $G=\frac{(\hat{i}+\hat{j}+\hat{k})+(\hat{i}+\hat{k})+(\hat{j}+\hat{k})}{3}$
$=\frac{2\hat{i}+2\hat{j}+3\hat{k}}{3}$
$|\overline{p_2}-\overline{p_3}|=|\hat{i}+\hat{k}-\hat{j}-\hat{k}|=|\hat{i}-\hat{l}|=\sqrt{2}|\overline{p_2}-\overline{p_1}|=|\hat{i}+\hat{k}-\hat{i}-\hat{j}-\hat{k}|=|-\hat{j}|=1|\overline{p_3}-\overline{p_1}|=|\hat{j}+\hat{k}-\hat{i}-\hat{j}-\hat{k}|=|-\hat{i}|=1$
$\therefore I=\frac{\sqrt{2}(\hat{i}+\hat{j}+\hat{k})+1(\hat{i}+\hat{k})+(\hat{j}+\hat{k})}{\sqrt{2}+1+1}$
$=\frac{(\sqrt{2}+1)\hat{i}+(\sqrt{2+1)}\hat{j}+(\sqrt{2}+2)\hat{k}}{2+\sqrt{2}}$
Correct option is A.