Let vec{a} = 3hat{i} + 2hat{j} + 2hat{k} and | vedclass

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(A) First,calculate $\vec{a} + \vec{b} = (3+1)\hat{i} + (2+2)\hat{j} + (2-2)\hat{k} = 4\hat{i} + 4\hat{j}$.Next,calculate $\vec{a} - \vec{b} = (3-1)\h

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$4(2\hat{i} - 2\hat{j} - \hat{k})$

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(A) First,calculate $\vec{a} + \vec{b} = (3+1)\hat{i} + (2+2)\hat{j} + (2-2)\hat{k} = 4\hat{i} + 4\hat{j}$.Next,calculate $\vec{a} - \vec{b} = (3-1)\hat{i} + (2-2)\hat{j} + (2-(-2))\hat{k} = 2\hat{i} + 4\hat{k}$.$A$ vector perpendicular to both $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ is given by their cross product:$\vec{v} = (\vec{a} + \vec{b}) 	imes (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 4 & 0 \ 2 & 0 & 4 \end{vmatrix} = \hat{i}(16 - 0) - \hat{j}(16 - 0) + \hat{k}(0 - 8) = 16\hat{i} - 16\hat{j} - 8\hat{k}$.Simplify the vector: $8(2\hat{i} - 2\hat{j} - \hat{k})$.The magnitude of this vector is $8 \sqrt{2^2 + (-2)^2 + (-1)^2} = 8 \sqrt{4+4+1} = 8 	imes 3 = 24$.We need a vector with magnitude $12$,which is half of the magnitude of $\vec{v}$.Thus,the required vector is $\pm \frac{12}{24} 	imes 8(2\hat{i} - 2\hat{j} - \hat{k}) = \pm 4(2\hat{i} - 2\hat{j} - \hat{k})$.Comparing with the options,$4(2\hat{i} - 2\hat{j} - \hat{k})$ is the correct choice.

Let $\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}$ be two vectors. If a vector perpendicular to both the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ has the magnitude $12$,then one such vector is

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