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(A) Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.First,calculate $\vec{a} + \vec{b} = (1+1)\hat{i} + (1

Vedclass · Accepted Answer

$\frac{-1}{\sqrt{6}}\hat{i} + \frac{2}{\sqrt{6}}\hat{j} - \frac{1}{\sqrt{6}}\hat{k}$

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(A) Given $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.First,calculate $\vec{a} + \vec{b} = (1+1)\hat{i} + (1+2)\hat{j} + (1+3)\hat{k} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.Next,calculate $\vec{a} - \vec{b} = (1-1)\hat{i} + (1-2)\hat{j} + (1-3)\hat{k} = 0\hat{i} - \hat{j} - 2\hat{k}$.$A$ vector perpendicular to both $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ is their cross product $\vec{c} = (\vec{a} + \vec{b}) 	imes (\vec{a} - \vec{b})$.$\vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 0 & -1 & -2 \end{vmatrix} = \hat{i}(-6 - (-4)) - \hat{j}(-4 - 0) + \hat{k}(-2 - 0) = -2\hat{i} + 4\hat{j} - 2\hat{k}$.The magnitude is $|\vec{c}| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}$.The required unit vector is $\pm \frac{\vec{c}}{|\vec{c}|} = \pm \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}} = \pm (\frac{-1}{\sqrt{6}}\hat{i} + \frac{2}{\sqrt{6}}\hat{j} - \frac{1}{\sqrt{6}}\hat{k})$.Comparing with the options,option $A$ is correct.

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

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