Let \(\vec{a}=\hat{\imath}+\hat{\jmath}-\hat{k}\), \(\vec{b}=\hat{\imath}-\hat{\jmath}+\hat{k}\) and \(\vec{c}=2\hat{\imath}+3\hat{\jmath}-\hat{k}\). If \(\hat{n}\) is a unit vector such that \(\vec{a}\cdot\hat{n} = 0\) and \(\vec{b}\cdot\hat{n} = 0\), then \(|\vec{c}\cdot\hat{n}|\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\sqrt{2}\)
A vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) is along their cross product:
\( \displaystyle \vec{a}\times\vec{b} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 1 & -1 \\ 1 & -1 & 1 \end{vmatrix} \)
\( = \hat{\imath}(1-1)-\hat{\jmath}(1+1)+\hat{k}(-1-1)\)
\( = -2\hat{\jmath}-2\hat{k}\)
Dividing by its magnitude \(2\sqrt{2}\), the unit vector is
\( \displaystyle \hat{n} = \pm\frac{1}{\sqrt{2}}(\hat{\jmath}+\hat{k}) \)
Then
\( \displaystyle |\vec{c}\cdot\hat{n}| = \frac{|3-1|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \)