Let \(\vec{a} =\hat{\imath}+\hat{\jmath},\ \vec{b}=\hat{\imath}-\hat{\jmath}\) and \(\vec{c} =\hat{\imath}+\hat{\jmath} +\hat{k}\). If \(\hat{m}\) is a unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\), then \(|\vec{c}.\hat{m}|\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1
A vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) is their cross product:
\( \displaystyle \vec{a}\times\vec{b} = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\1&1&0\\1&-1&0\end{vmatrix} \)
\(= \hat{\imath}(0 - 0) - \hat{\jmath}(0 - 0) + \hat{k}(-1 - 1) = -2\hat{k}\)
Dividing by its magnitude 2 gives the unit vector:
\(\hat{m} = \pm\hat{k}\)
Then
\(\vec{c}\cdot\hat{m} = \pm(\hat{\imath}+\hat{\jmath}+\hat{k})\cdot\hat{k} = \pm 1\)
\(|\vec{c}\cdot\hat{m}| = 1\)
Shortcut: \(\vec{a}\) and \(\vec{b}\) have no \(\hat{k}\) component, so both lie in the xy-plane, and the unit vector perpendicular to both is \(\pm\hat{k}\). Its dot product with \(\vec{c}\) is the \(\hat{k}\) component of \(\vec{c}\), which is 1 in absolute value.