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(B) Let the position vector of vertex $C$ be $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$.Given,centroid $G = \hat{i} = 1\hat{i} + 0\hat{j} + 0\hat{k}$.

Vedclass · Accepted Answer

$74$

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(B) Let the position vector of vertex $C$ be $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$.Given,centroid $G = \hat{i} = 1\hat{i} + 0\hat{j} + 0\hat{k}$.We know that $G = \frac{\vec{A} + \vec{B} + \vec{C}}{3}$,where $\vec{A} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{B} = 2\hat{i} + 4\hat{j} - 4\hat{k}$.So,$3(1\hat{i} + 0\hat{j} + 0\hat{k}) = (2\hat{i} + \hat{j} + \hat{k}) + (2\hat{i} + 4\hat{j} - 4\hat{k}) + (x\hat{i} + y\hat{j} + z\hat{k})$.$3\hat{i} = (4+x)\hat{i} + (5+y)\hat{j} + (-3+z)\hat{k}$.Comparing coefficients: $4+x = 3 \Rightarrow x = -1$; $5+y = 0 \Rightarrow y = -5$; $-3+z = 0 \Rightarrow z = 3$.Thus,$\vec{C} = -\hat{i} - 5\hat{j} + 3\hat{k}$.Now,$AG^2 = |\vec{G} - \vec{A}|^2 = |(1-2)\hat{i} + (0-1)\hat{j} + (0-1)\hat{k}|^2 = |-1\hat{i} - 1\hat{j} - 1\hat{k}|^2 = 1+1+1 = 3$.$BG^2 = |\vec{G} - \vec{B}|^2 = |(1-2)\hat{i} + (0-4)\hat{j} + (0+4)\hat{k}|^2 = |-1\hat{i} - 4\hat{j} + 4\hat{k}|^2 = 1+16+16 = 33$.$CG^2 = |\vec{G} - \vec{C}|^2 = |(1+1)\hat{i} + (0+5)\hat{j} + (0-3)\hat{k}|^2 = |2\hat{i} + 5\hat{j} - 3\hat{k}|^2 = 4+25+9 = 38$.Therefore,$AG^2 + BG^2 + CG^2 = 3 + 33 + 38 = 74$.

If $\hat{i}$ is the position vector of the centroid $G$ of triangle $ABC$ and $2\hat{i}+\hat{j}+\hat{k}$ and $2\hat{i}+4\hat{j}-4\hat{k}$ are respectively the position vectors of its vertices $A$ and $B$,then $AG^2+BG^2+CG^2=$

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