The position vectors of the points A and B wi | vedclass

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(B) Given the position vectors of $A$ and $B$ as $\vec{a} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + 4\hat{j} + 4\hat{k}$.First,calcul

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$\frac{\sqrt{136}}{3}$

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(B) Given the position vectors of $A$ and $B$ as $\vec{a} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + 4\hat{j} + 4\hat{k}$.First,calculate the lengths of the sides $OA$ and $OB$:$OA = |\vec{a}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.$OB = |\vec{b}| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.According to the angle bisector theorem,the internal bisector $OD$ of $\angle BOA$ divides the side $AB$ in the ratio of the adjacent sides,which is $OA : OB = 3 : 6 = 1 : 2$.Using the section formula,the position vector of point $D$ is given by:$\vec{d} = \frac{2\vec{a} + 1\vec{b}}{1 + 2} = \frac{2(2\hat{i} + 2\hat{j} + \hat{k}) + (2\hat{i} + 4\hat{j} + 4\hat{k})}{3} = \frac{(4+2)\hat{i} + (4+4)\hat{j} + (2+4)\hat{k}}{3} = \frac{6\hat{i} + 8\hat{j} + 6\hat{k}}{3} = 2\hat{i} + \frac{8}{3}\hat{j} + 2\hat{k}$.The length of the internal bisector $OD$ is the magnitude of $\vec{d}$:$|OD| = \sqrt{2^2 + (\frac{8}{3})^2 + 2^2} = \sqrt{4 + \frac{64}{9} + 4} = \sqrt{8 + \frac{64}{9}} = \sqrt{\frac{72 + 64}{9}} = \sqrt{\frac{136}{9}} = \frac{\sqrt{136}}{3}$.

The position vectors of the points $A$ and $B$ with respect to $O$ are $2 \hat{i}+2 \hat{j}+\hat{k}$ and $2 \hat{i}+4 \hat{j}+4 \hat{k}$. The length of the internal bisector of $\angle BOA$ of $\triangle AOB$ is:

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