Let O be the origin and the position vectors | vedclass

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(B) Let $\vec{a} = 2 \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{b} = 2 \hat{i} + 4 \hat{j} + 4 \hat{k}$.Lengths are $|\vec{a}| = \sqrt{2^2 + 2^2 + 1^2}

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$\frac{2}{3} \sqrt{34}$

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(B) Let $\vec{a} = 2 \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{b} = 2 \hat{i} + 4 \hat{j} + 4 \hat{k}$.Lengths are $|\vec{a}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4+4+1} = 3$ and $|\vec{b}| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4+16+16} = 6$.The angle bisector $OC$ divides $AB$ in the ratio $|\vec{a}| : |\vec{b}| = 3 : 6 = 1 : 2$.Using the section formula,the position vector of $C$ is $\vec{c} = \frac{1(\vec{b}) + 2(\vec{a})}{1+2} = \frac{(2 \hat{i} + 4 \hat{j} + 4 \hat{k}) + 2(2 \hat{i} + 2 \hat{j} + \hat{k})}{3} = \frac{6 \hat{i} + 8 \hat{j} + 6 \hat{k}}{3} = 2 \hat{i} + \frac{8}{3} \hat{j} + 2 \hat{k}$.The length $OC = |\vec{c}| = \sqrt{2^2 + (\frac{8}{3})^2 + 2^2} = \sqrt{4 + \frac{64}{9} + 4} = \sqrt{8 + \frac{64}{9}} = \sqrt{\frac{72+64}{9}} = \sqrt{\frac{136}{9}} = \frac{\sqrt{136}}{3} = \frac{2 \sqrt{34}}{3}$.

Let $O$ be the origin and the position vectors of $A$ and $B$ be $2 \hat{i}+2 \hat{j}+\hat{k}$ and $2 \hat{i}+4 \hat{j}+4 \hat{k}$ respectively. If the internal bisector of $\angle AOB$ meets the line $AB$ at $C$,then the length of $OC$ is

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