The unit vector perpendicular to each of the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\), where \(\vec{a}=\hat{\imath}+\hat{\jmath}+\hat{k}\) and \(\vec{b}=\hat{\imath}+2\hat{\jmath}+3\hat{k}\), is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-\displaystyle\frac{1}{\sqrt{6}}\hat{\imath}+\frac{2}{\sqrt{6}}\hat{\jmath}-\frac{1}{\sqrt{6}}\hat{k}\)
Let \(\vec{p} = \vec{a}+\vec{b}\) and \(\vec{q} = \vec{a}-\vec{b}\):
\(\vec{p} = 2\hat{\imath}+3\hat{\jmath}+4\hat{k}\)
\(\vec{q} = 0\hat{\imath}-\hat{\jmath}-2\hat{k}\)
The cross product \(\vec{p}\times\vec{q}\) is perpendicular to both:
\( \displaystyle \vec{p}\times\vec{q} = \begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\ 2 & 3 & 4\\ 0 & -1 & -2\end{vmatrix} \)
\( = \hat{\imath}(-6+4) - \hat{\jmath}(-4-0) + \hat{k}(-2-0) \)
\( = -2\hat{\imath}+4\hat{\jmath}-2\hat{k} \)
\( |\vec{p}\times\vec{q}| = \sqrt{4+16+4} = \sqrt{24} = 2\sqrt{6} \)
Dividing by the magnitude gives the unit vector:
\( \displaystyle \frac{-2\hat{\imath}+4\hat{\jmath}-2\hat{k}}{2\sqrt{6}} = -\frac{1}{\sqrt{6}}\hat{\imath}+\frac{2}{\sqrt{6}}\hat{\jmath}-\frac{1}{\sqrt{6}}\hat{k} \)
Shortcut: Options (2) and (3) are not even unit vectors: the squares of their components add up to \(\frac{3}{6}\) and \(\frac{9}{6}\). For Options (1) and (4), test the dot product with \(\vec{q} = -\hat{\jmath}-2\hat{k}\), using the numerators only:
Option (1): \(-2-2 = -4 \ne 0\)
Option (4): \(-2+2 = 0\)
Option (4) also gives \(-2+6-4 = 0\) with \(\vec{p}\), so it is the answer.
NCERT (2026-27 edition):
Vector (or Cross) Product of Two Vectors
