Let \(\vec a =\hat{\imath} +\hat{\jmath}+\hat{k}\) and \(\vec b = \hat{\imath}+2\hat{\jmath}+3\hat{k}\), then a unit vector perpendicular to both vectors \((\vec a +\vec b)\) and \((\vec a -\vec b)\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{\sqrt{6}}(-\hat{\imath}+2\hat{\jmath}-\hat{k})\)
\(\vec a + \vec b = 2\hat{\imath} + 3\hat{\jmath} + 4\hat{k}\)
\(\vec a - \vec b = -\hat{\jmath} - 2\hat{k}\)
The cross product of two vectors is perpendicular to both:
\( (\vec a + \vec b) \times (\vec a - \vec b) = \begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\2&3&4\\0&-1&-2\end{vmatrix} \)
\( = \hat{\imath}(-6 + 4) - \hat{\jmath}(-4 - 0) + \hat{k}(-2 - 0) \)
\( = -2\hat{\imath} + 4\hat{\jmath} - 2\hat{k} \)
Its magnitude is \(\sqrt{4 + 16 + 4} = 2\sqrt{6}\). Dividing by it gives the unit vector:
\( \displaystyle \frac{-2\hat{\imath} + 4\hat{\jmath} - 2\hat{k}}{2\sqrt{6}} = \frac{1}{\sqrt{6}}(-\hat{\imath} + 2\hat{\jmath} - \hat{k}) \)
Shortcut: Take the dot product of each option with \(\vec a + \vec b = (2, 3, 4)\) and \(\vec a - \vec b = (0, -1, -2)\). Option (1): \(-2 + 6 - 4 = 0\) and \(-2 + 2 = 0\). The others fail the first test: Option (2) gives \(2 - 6 - 4 = -8\), Option (3) gives \(2 + 3 + 4 = 9\), Option (4) gives \(2 + 6 + 4 = 12\).
NCERT (2026-27 edition):
Vector (or Cross) Product of Two Vectors
