A unit vector perpendicular to both the vectors \(\hat{\imath}+2\hat{\jmath}-\hat{k}\) and \(\hat{\imath}-\hat{\jmath}+\hat{k}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle \pm\frac{1}{\sqrt{14}}(\hat{\imath}-2\hat{\jmath}-3\hat{k}) \)
Let \(\vec{a}=\hat{\imath}+2\hat{\jmath}-\hat{k}\) and \(\vec{b}=\hat{\imath}-\hat{\jmath}+\hat{k}\). The cross product \(\vec{a}\times\vec{b}\) is perpendicular to both.
\( \displaystyle \vec{a}\times\vec{b} = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & 2 & -1 \\ 1 & -1 & 1 \end{vmatrix} \)
\( = \hat{\imath}(2-1)-\hat{\jmath}(1+1)+\hat{k}(-1-2)\)
\( = \hat{\imath}-2\hat{\jmath}-3\hat{k}\)
\(|\vec{a}\times\vec{b}| = \sqrt{1+4+9} = \sqrt{14}\)
The unit vectors perpendicular to both are \(\displaystyle\pm\frac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}\):
\( \displaystyle \pm\frac{1}{\sqrt{14}}(\hat{\imath}-2\hat{\jmath}-3\hat{k}) \)