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A unit vector perpendicular to both the vectors $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\hat{\mathbf{j}}+\hat{\mathbf{k}}$ is
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Verified Answer
The correct answer is:
$\frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{3}}$
Let $\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\overrightarrow{\mathbf{b}}=\hat{\mathbf{j}}+\hat{\mathbf{k}}$
Now, $\quad \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right|$ $=\hat{\mathbf{i}}(1-0)-\hat{\mathbf{j}}(1-0)+\hat{\mathbf{k}}(1-0)$ $=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$
and $\quad|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|=\sqrt{1^{2}+(-1)^{2}+1^{2}}=\sqrt{3}$
$\therefore$ Required unit vector $=\frac{\overrightarrow{\mathbf{a} \times \overrightarrow{\mathbf{b}}}}{\mid \overrightarrow{\mathbf{a} \times \overrightarrow{\mathbf{b}} \mid}}$
$=\frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{3}}$
Now, $\quad \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right|$ $=\hat{\mathbf{i}}(1-0)-\hat{\mathbf{j}}(1-0)+\hat{\mathbf{k}}(1-0)$ $=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$
and $\quad|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|=\sqrt{1^{2}+(-1)^{2}+1^{2}}=\sqrt{3}$
$\therefore$ Required unit vector $=\frac{\overrightarrow{\mathbf{a} \times \overrightarrow{\mathbf{b}}}}{\mid \overrightarrow{\mathbf{a} \times \overrightarrow{\mathbf{b}} \mid}}$
$=\frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{3}}$
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