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(A) Given $\bar{a}=\hat{i}+\hat{j}+\hat{k}$ and $\bar{b}=3 \hat{i}-2 \hat{j}+5 \hat{k}$.First,calculate the sum and difference of the vectors:$\bar{a}

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$\frac{-14 \hat{i}+4 \hat{j}+10 \hat{k}}{\sqrt{312}}$

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(A) Given $\bar{a}=\hat{i}+\hat{j}+\hat{k}$ and $\bar{b}=3 \hat{i}-2 \hat{j}+5 \hat{k}$.First,calculate the sum and difference of the vectors:$\bar{a}+\bar{b} = (\hat{i}+\hat{j}+\hat{k})+(3 \hat{i}-2 \hat{j}+5 \hat{k}) = 4 \hat{i}-\hat{j}+6 \hat{k}$$\bar{a}-\bar{b} = (\hat{i}+\hat{j}+\hat{k})-(3 \hat{i}-2 \hat{j}+5 \hat{k}) = -2 \hat{i}+3 \hat{j}-4 \hat{k}$The vector perpendicular to both $(\bar{a}+\bar{b})$ and $(\bar{a}-\bar{b})$ is given by their cross product:$\vec{v} = (\bar{a}+\bar{b}) 	imes (\bar{a}-\bar{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & -1 & 6 \ -2 & 3 & -4 \end{vmatrix}$$\vec{v} = \hat{i}((-1)(-4) - (6)(3)) - \hat{j}((4)(-4) - (6)(-2)) + \hat{k}((4)(3) - (-1)(-2))$$\vec{v} = \hat{i}(4 - 18) - \hat{j}(-16 + 12) + \hat{k}(12 - 2) = -14 \hat{i} + 4 \hat{j} + 10 \hat{k}$The magnitude of this vector is $|\vec{v}| = \sqrt{(-14)^2 + 4^2 + 10^2} = \sqrt{196 + 16 + 100} = \sqrt{312}$.The required unit vector is $\frac{\vec{v}}{|\vec{v}|} = \frac{-14 \hat{i}+4 \hat{j}+10 \hat{k}}{\sqrt{312}}$.

The unit vector perpendicular to each of the vectors $\bar{a}+\bar{b}$ and $\bar{a}-\bar{b}$,where $\bar{a}=\hat{i}+\hat{j}+\hat{k}$ and $\bar{b}=3 \hat{i}-2 \hat{j}+5 \hat{k}$ is

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