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The unit vector perpendicular to the vectors $6 \hat{i}+2 \hat{j}+3 \hat{k}$ and $3 \hat{i}-6 \hat{j}-2 \hat{k}$ is $-$
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The correct answer is:
$\frac{2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{7}$
Unit vector perpendicular to both the given vectors is, $\frac{(6 \hat{i}+2 \hat{j}+3 \hat{k}) \times(3 \hat{i}-6 \hat{j}-2 \hat{k})}{|(6 \hat{i}+2 \hat{j}+3 \hat{k}) \times(3 \hat{i}-6 \hat{j}-2 \hat{k})|}=\frac{2 \hat{i}+3 \hat{j}-6 \hat{k}}{7}$
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