Question

Let $\overrightarrow{A}=\hat{i}Acos\theta +\hat{j}Asin\theta$, be any vector. Another vector $\overrightarrow{B}$ which is normal to $\overrightarrow{A}$ is :-

• A. $\hat{i}Bcos\theta +\hat{j}Bsin\theta$
• B. $\hat{i}Bsin\theta +\hat{j}Bcos\theta$
• C. $\hat{i}Bsin\theta -\hat{j}Bcos\theta$
• D. $\hat{i}Acos\theta -\hat{j}Asin\theta$

Answer 1

The correct option is C $\hat{i}Bsin\theta -\hat{j}Bcos\theta$

$\overrightarrow{A}=Acos\theta i+Asin\theta j$
In complex number format (say) we can write this as
$A=A(cos\theta +isin\theta )$
$=A{e}^{i\left(\theta \right)}$.
Now revolving it by $\frac{\pi }{2}$ in anticlockwise sense give us
$A{e}^{i(\frac{\pi }{2}+\theta )}$
$=A(-sin\theta +icos\theta )$
Thus unit vector perpendicular to $\overrightarrow{A}$ will be of the form
$=\pm (-sin\theta i+cos\theta j)$
$=\mp (sin\theta i-cos\theta j)$
Hence the correct answer is
$Bsin\theta i-Bcos\theta j$