Find a unit vector perpendicular to both \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) (A) $\left[\left(\mathrm{A}^{\rightarrow} \cdot \mathrm{B}^{\rightarrow}\right) / \mathrm{AB}\right]$ (B) $\left[\left(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{-}\right) /(\mathrm{AB} \sin \theta)\right]$ (C) $\left[\left(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\right) /(\mathrm{AB} \cos \theta)\right]$ (D) $\left[\left(\mathrm{A}^{\rightarrow} \cdot \mathrm{B}^{\rightarrow}\right) /(\mathrm{AB} \sin \theta)\right]$

First, find the cross product of the two given vectors A and B as it will give us the vector perpendicular to both A and B. C = \(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\)

Next, find the magnitude of the cross product vector which will be required for converting the vector into a unit vector. |\(\mathrm{C}^{\rightarrow} \)| = |\(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\)|

Now, find the unit vector of the cross product (vector C) by dividing it by its magnitude. Unit vector = \(\frac{\mathrm{C}^{\rightarrow}}{|\mathrm{C}^{\rightarrow}|}\) = \(\frac{\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}}{|\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}|}\) Another way to express the denominator is by using the magnitudes of vectors A and B and the sine of the angle between them, \(\theta\). |\(\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}\)| = \(\mathrm{AB} \sin\theta\) Therefore, Unit vector = \(\frac{\mathrm{A}^{\rightarrow} \times \mathrm{B}^{\rightarrow}}{\mathrm{AB} \sin\theta}\) Comparing this result to the given options, we get the correct answer as (B).