If \(\hat{a}\) is a unit vector perpendicular to both the vectors \(\vec{b}=\hat{\jmath}+2\hat{k}\) and \(\vec{c}= \hat{\imath}+ 2\hat{\jmath}\), then \(\hat{a}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{-4\hat{\imath}+2\hat{\jmath}-\hat{k}}{\sqrt{21}}\)
The cross product \(\vec{b} \times \vec{c}\) is perpendicular to both \(\vec{b}\) and \(\vec{c}\):
\(\displaystyle \vec{b} \times \vec{c} = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\0&1&2\\1&2&0\end{vmatrix}\)
\(= \hat{\imath}(0 - 4) - \hat{\jmath}(0 - 2) + \hat{k}(0 - 1)\)
\(= -4\hat{\imath} + 2\hat{\jmath} - \hat{k}\)
\(|\vec{b} \times \vec{c}| = \sqrt{16 + 4 + 1} = \sqrt{21}\)
Dividing by the magnitude gives the unit vectors
\(\displaystyle \hat{a} = \pm\frac{-4\hat{\imath} + 2\hat{\jmath} - \hat{k}}{\sqrt{21}}\)
Only the + sign appears among the options, in Option (3).
Shortcut: The dot product of the correct option’s numerator with \(\vec{b} = (0, 1, 2)\) and with \(\vec{c} = (1, 2, 0)\) must be 0:
Option (1): with \(\vec{b}\), \(2 + 2 = 4 \ne 0\)
Option (2): with \(\vec{c}\), \(4 + 4 = 8 \ne 0\)
Option (3): with \(\vec{b}\), \(2 - 2 = 0\); with \(\vec{c}\), \(-4 + 4 = 0\)
Option (4): with \(\vec{b}\), \(-2 - 2 = -4 \ne 0\)