Question

Vector $\underset{A}{\overset{}{\to }}$ has a magnitude of 6 units and is in the direction of $+xaxis.$ Vector $\underset{B}{\overset{}{\to }}$ has a magnitude of 4units lies in the x-y plane making an angle of ${30}^o$ with $+x-axis$ . Find the vector product $\underset{A}{\overset{}{\to }}\times \underset{B}{\overset{}{\to }}$

Answer 1

Converting the vectors into catesium from

$\overrightarrow{A}=|\overrightarrow{A}|\hat{i}=6\hat{i}$

$\overrightarrow{B}=|\overrightarrow{B}|[\cos \theta \hat{i}+\sin \theta \hat{j}]=4[\cos {30}^o\hat{i}+\sin {30}^o]$

$\overrightarrow{B}=2.(3\hat{i}+2\hat{j}$

So, $\overrightarrow{A}\times \overrightarrow{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 6& 0& 0\\ 2\sqrt{3}& 2& 0\end{array}\right|$

$=(0.0-2.0)\hat{i}+(2\sqrt{3}.0-6.0)\hat{j}+(6.2-2\sqrt{3}.0)\hat{k}$

$=12\hat{k}$

The vector $\overrightarrow{A}\times \overrightarrow{B}$ is $12$ magnitude along the normal to the $x-y$ plane.