If vec{OA} = 2hat{i} + 2hat{j} + hat{k},vec{O | vedclass

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(B) Given vectors $\vec{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{OB} = 2\hat{i} + 4\hat{j} + 4\hat{k}$.First,calculate the magnitudes:$|\vec{OA}

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$136$

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(B) Given vectors $\vec{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{OB} = 2\hat{i} + 4\hat{j} + 4\hat{k}$.First,calculate the magnitudes:$|\vec{OA}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = 3$.$|\vec{OB}| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = 6$.Let $\vec{a} = \frac{\vec{OA}}{|\vec{OA}|} = \frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{1}{3}\hat{k}$ and $\vec{b} = \frac{\vec{OB}}{|\vec{OB}|} = \frac{2}{6}\hat{i} + \frac{4}{6}\hat{j} + \frac{4}{6}\hat{k} = \frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$.The internal angle bisector vector $\vec{OP}$ is along the direction of $(\vec{a} + \vec{b})$.$\vec{a} + \vec{b} = (\frac{2}{3} + \frac{1}{3})\hat{i} + (\frac{2}{3} + \frac{2}{3})\hat{j} + (\frac{1}{3} + \frac{2}{3})\hat{k} = 1\hat{i} + \frac{4}{3}\hat{j} + 1\hat{k}$.The length of the angle bisector $k$ is the magnitude of the vector $\vec{OP} = \lambda(\vec{a} + \vec{b})$. Using the section formula,the bisector vector is $\vec{OP} = \frac{|\vec{OB}|\vec{OA} + |\vec{OA}|\vec{OB}}{|\vec{OA}| + |\vec{OB}|} = \frac{6\vec{OA} + 3\vec{OB}}{9} = \frac{2\vec{OA} + \vec{OB}}{3}$.$\vec{OP} = \frac{2(2\hat{i} + 2\hat{j} + \hat{k}) + (2\hat{i} + 4\hat{j} + 4\hat{k})}{3} = \frac{6\hat{i} + 8\hat{j} + 6\hat{k}}{3} = 2\hat{i} + \frac{8}{3}\hat{j} + 2\hat{k}$.$k^2 = |\vec{OP}|^2 = 2^2 + (\frac{8}{3})^2 + 2^2 = 4 + \frac{64}{9} + 4 = 8 + \frac{64}{9} = \frac{72 + 64}{9} = \frac{136}{9}$.Therefore,$9k^2 = 136$.

If $\vec{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$,$\vec{OB} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ and the length of the internal bisector of $\angle BOA$ of triangle $AOB$ is $k$,then $9k^2 =$

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