Question

Let $\overrightarrow{A}=\hat{i}A\cos \theta +\hat{j}A\sin \theta$ be any vector. Another vector $\overrightarrow{B}$ which is normal to$\overrightarrow{A}$ is

• A. $\hat{i}B\cos \theta -\hat{j}B\sin \theta$
• B. $\hat{i}B\sin \theta +\hat{j}B\cos \theta$
• C. $\hat{i}B\cos \theta +\hat{j}B\sin \theta$
• D. $\hat{i}B\sin \theta -\hat{j}B\cos \theta$

Answer 1

The correct option is D $\hat{i}B\sin \theta -\hat{j}B\cos \theta$

Given,
$\overrightarrow{A}=\hat{i}A\cos \theta +\hat{j}A\sin \theta$

Let the vector $\overrightarrow{B}=x\hat{i}+y\hat{i}\dots \left(i\right)$
And $\overrightarrow{B}$ is normal to $\overrightarrow{A}$
$\Rightarrow \overrightarrow{A}\cdot \overrightarrow{B}=0$

So, $Ax\cos \theta +Ay\sin \theta =0\dots \left(ii\right)$

$x\cos \theta =-y\sin \theta \dots \left(iii\right)$

$\frac{x}{y}=\frac{-\sin \theta }{\cos \theta }$

$\frac{x^2}{y^2}+1=\frac{{\sin }^2\theta }{{\cos }^2\theta }+1$

$\frac{x^2+y^2}{y^2}=\frac{1}{{\cos }^2\theta }$

$\therefore y=\pm B\cos \theta$

Now for $y=B\cos \theta$, from equation $\left(iii\right)$,
$\Rightarrow x=-B\sin \theta$

Substituting $x$ and $y$ in equation $\left(i\right)$, we get,
$\Rightarrow \overrightarrow{B}=-\hat{i}B\sin \theta +\hat{j}B\cos \theta$

Now for $y=-B\cos \theta$, from equation $(iii$),
$x=B\sin \theta$

Substituting $x$ and $y$ in equation ($i$),
we get,
$\overrightarrow{B}=\hat{i}B\sin \theta -\hat{j}B\cos \theta$

Final answer: (c).