Source: Extra Practice

A unit vector perpendicular to both A⃗=i^+j^\vec{A} = \hat{i} + \hat{j} and B⃗=i^−j^\vec{B} = \hat{i} - \hat{j} is:

Options

Option A is correct

k^\hat{k}

Option B

i^+j^2\frac{\hat{i} + \hat{j}}{\sqrt{2}}

Option C

i^−j^2\frac{\hat{i} - \hat{j}}{\sqrt{2}}

Option D

j^\hat{j}

Explanation

A vector perpendicular to both A⃗\vec{A} and B⃗\vec{B} is given by their cross product: C⃗=A⃗×B⃗=(i^+j^)×(i^−j^)=−i^×j^+j^×i^=−k^−k^=−2k^\vec{C} = \vec{A} \times \vec{B} = (\hat{i} + \hat{j}) \times (\hat{i} - \hat{j}) = -\hat{i} \times \hat{j} + \hat{j} \times \hat{i} = -\hat{k} - \hat{k} = -2\hat{k}. A unit vector in this direction is C⃗∣C⃗∣=−2k^2=−k^\frac{\vec{C}}{|\vec{C}|} = \frac{-2\hat{k}}{2} = -\hat{k}. Since unit vectors can be in either direction perpendicular to the plane, k^\hat{k} is also a valid unit vector perpendicular to both. Options 2 and 3 lie in the same xyxy-plane and thus cannot be perpendicular to the vectors which also lie in the xyxy-plane.