Let the three sides of a triangle $A B C$ be given by the vectors $2 \hat{i}-\hat{j}+\hat{k}, \hat{i

Question

Let the three sides of a triangle $A B C$ be given by the vectors $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$. Let $G$ be the centroid of the triangle $A B C$. Then $6\left(|\overrightarrow{A G}|^{2}+|\overrightarrow{B G}|^{2}+|\overrightarrow{C G}|^{2}ight)$ is equal to $\_\_\_\_$ .\

Professor PC Thomas Classes · Answer

(164): We have in $	riangle A B C$,\ $\overrightarrow{A B}+\overrightarrow{C A}=\overrightarrow{B C}$\ Let PV of $\vec{A}$ be $\overrightarrow{0}$ then $\overrightarrow{A B}=\vec{B}-\vec{A}$\ ⇒ P.V. of $\vec{B}=2 \hat{i}-\hat{j}+\hat{k}$\ $\overrightarrow{C A}=\vec{A}-\vec{C}$\ P.V of $\vec{C}=-\hat{i}+3 \hat{j}+5 \hat{k}$ Now, P.V. of $\vec{G}=\frac{\vec{A}+\vec{B}+\vec{C}}{3}=\frac{1}{3}(\hat{i}+2 \hat{j}+6 \hat{k})$\ Then $\overrightarrow{A G}=\frac{1}{3}(\hat{i}+2 \hat{j}+6 \hat{k})$\ $\Rightarrow|\overrightarrow{A G}|^{2}=\frac{41}{9}$\ $\overrightarrow{B G}=\left(\frac{1}{3}-2ight) \hat{i}+\left(\frac{2}{3}+1ight) \hat{j}+(2-1) \hat{k}$\ $\Rightarrow|\overrightarrow{B G}|^{2}=\frac{59}{9}$\ $\Rightarrow \overrightarrow{C G}=\left(\frac{1}{3}+1ight) \hat{i}+\left(\frac{2}{3}-3ight) \hat{j}+(2-5) \hat{k}$\ $\Rightarrow|\overrightarrow{C G}|^{2}=\frac{146}{9}$\ $	herefore 6\left(|\overrightarrow{A G}|^{2}+|\overrightarrow{B G}|^{2}+|\overrightarrow{C G}|^{2}ight)$\ $=6\left(\frac{41}{9}+\frac{59}{9}+\frac{146}{9}ight)=164$\