Physics - Vectors Question with Solution | TestHub

Given vectors are $\vec{A}=2\hat{i}+\hat{j}$ and $\vec{B}=\hat{i}-\hat{j}$.

A vector perpendicular to both $\vec{A}$ and $\vec{B}$ is given by their cross product $\vec{A}\times \vec{B}$.

$\vec{A}\times \vec{B}=(2\hat{i}+\hat{j})\times (\hat{i}-\hat{j})$

$\vec{A}\times \vec{B}=2(\hat{i}\times \hat{i})-2(\hat{i}\times \hat{j})+(\hat{j}\times \hat{i})-(\hat{j}\times \hat{j})$

Since $\hat{i}\times \hat{i}=0$, $\hat{j}\times \hat{j}=0$, $\hat{i}\times \hat{j}=\hat{k}$, and $\hat{j}\times \hat{i}=-\hat{k}$:

$\vec{A}\times \vec{B}=0-2\hat{k}-\hat{k}-0=-3\hat{k}$

The magnitude of this vector is $|\vec{A}\times \vec{B}|=|-3\hat{k}|=3$.

The unit vector perpendicular to both $\vec{A}$ and $\vec{B}$ is:

$\hat{n}=\frac{\vec{A}\times \vec{B}}{|\vec{A}\times \vec{B}|}=\frac{-3\hat{k}}{3}=-\hat{k}$

Answer: $-\hat{k}$