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$\quad$ Let $\overrightarrow{\mathrm{P}}=\hat{\mathrm{I}} \mathrm{P} \sin \theta-\hat{\mathrm{P}} \cos \theta$, be any vector. Another vector $\overrightarrow{\mathrm{Q}}$ which is perpendicular
to $\overrightarrow{\mathrm{P}}$ is
Options:
to $\overrightarrow{\mathrm{P}}$ is
Solution:
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Verified Answer
The correct answer is:
$(\hat{\mathrm{I}} \mathrm{Q} \cos \theta+\hat{\mathrm{j}} \mathrm{Q} \sin \theta)$
$\vec{P}=P \sin \theta \hat{i}-P \cos \theta \hat{j}$
also, $\vec{Q}=Q \cos \theta\hat{i}+Q \sin \theta\hat{j}$
$\therefore \vec{P} .\vec{Q}=(P \sin \theta \hat{i}-P \cos \theta \hat{j}) \cdot(Q \cos \theta \hat{i}+Qsin\theta \hat{j})$
$=P Q \sin \theta \cos \theta-P \cos \theta- Q\sin \theta$
$=0$
also, $\vec{Q}=Q \cos \theta\hat{i}+Q \sin \theta\hat{j}$
$\therefore \vec{P} .\vec{Q}=(P \sin \theta \hat{i}-P \cos \theta \hat{j}) \cdot(Q \cos \theta \hat{i}+Qsin\theta \hat{j})$
$=P Q \sin \theta \cos \theta-P \cos \theta- Q\sin \theta$
$=0$
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