Mathematics - Vector Question with Solution | TestHub

By given data
$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{CB}}$
Let pv of $\overrightarrow{\mathrm{A}}$ are $\overrightarrow{\mathrm{O}}$ then
$\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{B}}-\overrightarrow{\mathrm{A}}$
i.e. $pv$ of $\overrightarrow{\mathrm{B}}-=2\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{C}}$
i.e. pv of $\overrightarrow{\mathrm{C}}=-(\hat{\mathrm{i}}-3\hat{\mathrm{j}}-5\hat{\mathrm{k}})$
Now pv of centroid
$\begin{array}{rl}& (\overrightarrow{\mathrm{G}})-=\frac{\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}}{3}=\frac{\overrightarrow{0}+(2,-1,1)+(-1,3,5)}{3}\\ & \overrightarrow{\mathrm{G}}=\frac{1}{3}(\hat{\mathrm{i}}+2\hat{\mathrm{j}}+6\hat{\mathrm{k}})\end{array}$
Now $\overrightarrow{\mathrm{AG}}=\frac{1}{3}(\hat{\mathrm{i}}+2\hat{\mathrm{j}}+6\hat{\mathrm{k}})$
$\Rightarrow |\overrightarrow{\mathrm{AG}}{|}^2=\frac{1}{9}\times 41$
$\overrightarrow{\mathrm{BG}}=\left(\frac{1}{3}-2\right)\hat{\mathrm{i}}+\left(\frac{2}{3}+1\right)\hat{\mathrm{j}}+(2-1)\hat{\mathrm{k}}$
$\Rightarrow |\overrightarrow{\mathrm{BG}}{|}^2=\frac{59}{9}$
$\begin{array}{rl}& \overrightarrow{\mathrm{CG}}=\left(\frac{1}{3}+1\right)\hat{\mathrm{i}}+\left(\frac{2}{3}-3\right)\hat{\mathrm{j}}+(2-5)\hat{\mathrm{k}}\\ & \Rightarrow |\overrightarrow{\mathrm{CG}}{|}^2=\frac{146}{9}\end{array}$
Now
$\begin{array}{rl}& 6\left[|\overrightarrow{\mathrm{AG}}{|}^2+|\overrightarrow{\mathrm{BG}}{|}^2+|\overrightarrow{\mathrm{CG}}{|}^2\right]=6\times \left[\frac{41}{9}+\frac{59}{9}+\frac{146}{9}\right]\\ & =6\times \frac{246}{9}=164\end{array}$