A unit vector perpendicular to the vectors \(\hat{\imath} - \hat{\jmath}\) and \(\hat{\imath} + \hat{\jmath}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\hat{k}\)
Let \(\vec{a} = \hat{\imath} - \hat{\jmath}\) and \(\vec{b} = \hat{\imath} + \hat{\jmath}\). The cross product \(\vec{a} \times \vec{b}\) is perpendicular to both:
\(\displaystyle \vec{a} \times \vec{b} = \begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\1 & -1 & 0\\1 & 1 & 0\end{vmatrix}\)
\(= \hat{\imath}(0 - 0) - \hat{\jmath}(0 - 0) + \hat{k}(1 + 1) = 2\hat{k}\)
Dividing by its magnitude 2 gives the unit vector \(\hat{k}\). (The other unit vector, \(-\hat{k}\), is not among the options.)
Shortcut: Both given vectors have no \(\hat{k}\) component, so \(\hat{k}\) is perpendicular to both: \(\hat{k}\cdot(\hat{\imath} \mp \hat{\jmath}) = 0\). Options (2), (3) and (4) are parallel to one of the given vectors, so they cannot be perpendicular to it.