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

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$\pm \frac{2}{3}\hat{i} \mp \frac{2}{3}\hat{j} \mp \frac{1}{3}\hat{k}$

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(A) Given $\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}$.First,calculate the sum and difference of the vectors:$\vec{a}+\vec{b} = (3+1)\hat{i} + (2+2)\hat{j} + (2-2)\hat{k} = 4\hat{i} + 4\hat{j}$$\vec{a}-\vec{b} = (3-1)\hat{i} + (2-2)\hat{j} + (2-(-2))\hat{k} = 2\hat{i} + 4\hat{k}$Now,find the cross product $(\vec{a}+\vec{b}) 	imes (\vec{a}-\vec{b})$:$(\vec{a}+\vec{b}) 	imes (\vec{a}-\vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 4 & 0 \ 2 & 0 & 4 \end{vmatrix}$$= \hat{i}(16-0) - \hat{j}(16-0) + \hat{k}(0-8) = 16\hat{i} - 16\hat{j} - 8\hat{k}$Calculate the magnitude of the cross product:$|(\vec{a}+\vec{b}) 	imes (\vec{a}-\vec{b})| = \sqrt{16^2 + (-16)^2 + (-8)^2} = \sqrt{256 + 256 + 64} = \sqrt{576} = 24$The unit vector perpendicular to both is given by:$\pm \frac{(\vec{a}+\vec{b}) 	imes (\vec{a}-\vec{b})}{|(\vec{a}+\vec{b}) 	imes (\vec{a}-\vec{b})|} = \pm \frac{16\hat{i} - 16\hat{j} - 8\hat{k}}{24}$$= \pm \frac{2\hat{i} - 2\hat{j} - \hat{k}}{3} = \pm \frac{2}{3}\hat{i} \mp \frac{2}{3}\hat{j} \mp \frac{1}{3}\hat{k}$

Find a unit vector perpendicular to each of the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b},$ where $\vec{a}=3 \hat{i}+2 \hat{j}+2 \hat{k}$ and $\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}.$

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