Let vec{OA} = -4hat{i} + 3hat{k} and vec{OB} | vedclass

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(A) The vector $\vec{OD}$ that bisects $\angle AOB$ is in the direction of the sum of the unit vectors along $\vec{OA}$ and $\vec{OB}$.First,find the

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$\pm(\hat{i} + \hat{j} + 2\hat{k})$

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(A) The vector $\vec{OD}$ that bisects $\angle AOB$ is in the direction of the sum of the unit vectors along $\vec{OA}$ and $\vec{OB}$.First,find the unit vectors $\hat{a}$ and $\hat{b}$:$|\vec{OA}| = \sqrt{(-4)^2 + 0^2 + 3^2} = \sqrt{16 + 9} = 5$$\hat{a} = \frac{\vec{OA}}{|\vec{OA}|} = \frac{-4\hat{i} + 3\hat{k}}{5}$$|\vec{OB}| = \sqrt{14^2 + 2^2 + (-5)^2} = \sqrt{196 + 4 + 25} = \sqrt{225} = 15$$\hat{b} = \frac{\vec{OB}}{|\vec{OB}|} = \frac{14\hat{i} + 2\hat{j} - 5\hat{k}}{15}$The direction of the bisector is $\vec{v} = \hat{a} + \hat{b} = \frac{-12\hat{i} + 9\hat{k} + 14\hat{i} + 2\hat{j} - 5\hat{k}}{15} = \frac{2\hat{i} + 2\hat{j} + 4\hat{k}}{15} = \frac{2}{15}(\hat{i} + \hat{j} + 2\hat{k})$.Since $\vec{OD}$ is along this direction,$\vec{OD} = \lambda(\hat{i} + \hat{j} + 2\hat{k})$.Given $|\vec{OD}| = \sqrt{6}$,we have $|\lambda| \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6} \implies |\lambda| \sqrt{6} = \sqrt{6} \implies |\lambda| = 1$.Thus,$\vec{OD} = \pm(\hat{i} + \hat{j} + 2\hat{k})$.

Let $\vec{OA} = -4\hat{i} + 3\hat{k}$ and $\vec{OB} = 14\hat{i} + 2\hat{j} - 5\hat{k}$. If $\vec{OD}$ bisects $\angle AOB$ and $|\vec{OD}| = \sqrt{6}$,then $\vec{OD} =$

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