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(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively. Without loss of generality,let $\vec{a} = \vec{0}$.Giv

Vedclass · Accepted Answer

$164$

Vedclass · Answer

(A) Let the position vectors of vertices $A, B, C$ be $\vec{a}, \vec{b}, \vec{c}$ respectively. Without loss of generality,let $\vec{a} = \vec{0}$.Given $\vec{AB} = \vec{b} - \vec{a} = 2\hat{i}-\hat{j}+\hat{k}$,so $\vec{b} = 2\hat{i}-\hat{j}+\hat{k}$.Given $\vec{CA} = \vec{a} - \vec{c} = \hat{i}-3\hat{j}-5\hat{k}$,so $\vec{c} = -(\hat{i}-3\hat{j}-5\hat{k}) = -\hat{i}+3\hat{j}+5\hat{k}$.The centroid $G$ has position vector $\vec{g} = \frac{\vec{a}+\vec{b}+\vec{c}}{3} = \frac{\vec{0} + (2\hat{i}-\hat{j}+\hat{k}) + (-\hat{i}+3\hat{j}+5\hat{k})}{3} = \frac{\hat{i}+2\hat{j}+6\hat{k}}{3}$.Now,$\overrightarrow{AG} = \vec{g} - \vec{a} = \frac{1}{3}(\hat{i}+2\hat{j}+6\hat{k})$. Thus,$|\overrightarrow{AG}|^2 = \frac{1}{9}(1^2+2^2+6^2) = \frac{41}{9}$.$\overrightarrow{BG} = \vec{g} - \vec{b} = (\frac{1}{3}-2)\hat{i} + (\frac{2}{3}+1)\hat{j} + (2-1)\hat{k} = -\frac{5}{3}\hat{i} + \frac{5}{3}\hat{j} + \hat{k}$. Thus,$|\overrightarrow{BG}|^2 = \frac{25}{9} + \frac{25}{9} + 1 = \frac{59}{9}$.$\overrightarrow{CG} = \vec{g} - \vec{c} = (\frac{1}{3}+1)\hat{i} + (\frac{2}{3}-3)\hat{j} + (2-5)\hat{k} = \frac{4}{3}\hat{i} - \frac{7}{3}\hat{j} - 3\hat{k}$. Thus,$|\overrightarrow{CG}|^2 = \frac{16}{9} + \frac{49}{9} + 9 = \frac{65+81}{9} = \frac{146}{9}$.Finally,$6(|\overrightarrow{AG}|^2+|\overrightarrow{BG}|^2+|\overrightarrow{CG}|^2) = 6(\frac{41+59+146}{9}) = 6(\frac{246}{9}) = 6 	imes \frac{82}{3} = 2 	imes 82 = 164$.

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