Let overrightarrow{OA}=hat{i}+2 hat{j}-2 hat{ | vedclass

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(B) The magnitudes of the vectors are $|\overrightarrow{OA}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1+4+4} = 3$ and $|\overrightarrow{OB}| = \sqrt{(-2)^2

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$\hat{i}+5 \hat{j}+4 \hat{k}$

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(B) The magnitudes of the vectors are $|\overrightarrow{OA}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1+4+4} = 3$ and $|\overrightarrow{OB}| = \sqrt{(-2)^2 + (-3)^2 + 6^2} = \sqrt{4+9+36} = 7$. The unit vectors along $\overrightarrow{OA}$ and $\overrightarrow{OB}$ are $\hat{a} = \frac{1}{3}(\hat{i} + 2\hat{j} - 2\hat{k})$ and $\hat{b} = \frac{1}{7}(-2\hat{i} - 3\hat{j} + 6\hat{k})$. The vector along the angle bisector is given by $\lambda(\hat{a} + \hat{b}) = \lambda \left( \frac{\hat{i} + 2\hat{j} - 2\hat{k}}{3} + \frac{-2\hat{i} - 3\hat{j} + 6\hat{k}}{7} \right)$. Simplifying the expression: $\lambda \left( \frac{7\hat{i} + 14\hat{j} - 14\hat{k} - 6\hat{i} - 9\hat{j} + 18\hat{k}}{21} \right) = \frac{\lambda}{21}(\hat{i} + 5\hat{j} + 4\hat{k})$. Given $|\overrightarrow{OC}| = \sqrt{42}$,we have $\frac{|\lambda|}{21} \sqrt{1^2 + 5^2 + 4^2} = \sqrt{42}$. $\frac{|\lambda|}{21} \sqrt{1 + 25 + 16} = \sqrt{42} \implies \frac{|\lambda|}{21} \sqrt{42} = \sqrt{42} \implies |\lambda| = 21$. Taking $\lambda = 21$,we get $\overrightarrow{OC} = \hat{i} + 5\hat{j} + 4\hat{k}$.

Let $\overrightarrow{OA}=\hat{i}+2 \hat{j}-2 \hat{k}$ and $\overrightarrow{OB}=-2 \hat{i}-3 \hat{j}+6 \hat{k}$ be the position vectors of two points $A$ and $B$. If $C$ is a point on the bisector of $\angle AOB$ and $OC=\sqrt{42}$,then $\overrightarrow{OC}=$

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