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(D) Let $\overrightarrow{a} = \hat{i} + \hat{j}$ and $\overrightarrow{b} = \hat{j} + \hat{k}$.The vector perpendicular to both $\overrightarrow{a}$ an

Vedclass · Accepted Answer

$\frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}$

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(D) Let $\overrightarrow{a} = \hat{i} + \hat{j}$ and $\overrightarrow{b} = \hat{j} + \hat{k}$.The vector perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by the cross product $\overrightarrow{a} 	imes \overrightarrow{b}$.$\overrightarrow{a} 	imes \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 0 \ 0 & 1 & 1 \end{vmatrix} = \hat{i}(1 - 0) - \hat{j}(1 - 0) + \hat{k}(1 - 0) = \hat{i} - \hat{j} + \hat{k}$.The magnitude of this vector is $|\overrightarrow{a} 	imes \overrightarrow{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.The required unit vector is $\pm \frac{\overrightarrow{a} 	imes \overrightarrow{b}}{|\overrightarrow{a} 	imes \overrightarrow{b}|} = \pm \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}$.Comparing this with the given options,the correct option is $\frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}$.

$A$ unit vector perpendicular to both the vectors $\hat{i}+\hat{j}$ and $\hat{j}+\hat{k}$ is

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