A couple is of moment $\overrightarrow{\mathrm{G}}$ and the force forming the couple is…

A couple is of moment $\overrightarrow{\mathrm{G}}$ and the force forming the couple is $\overrightarrow{\mathrm{P}}$. If $\overrightarrow{\mathrm{P}}$ is turned through a right angle the moment of the couple thus formed is $\overrightarrow{\mathrm{H}}$. If instead, the force $\overrightarrow{\mathrm{P}}$ are turned through an angle $\alpha$, then the moment of couple becomes
  1. $\overrightarrow{\mathrm{H}} \sin \alpha-\overrightarrow{\mathrm{G}} \cos \alpha$
  2. $\overrightarrow{\mathrm{G}} \sin \alpha-\overrightarrow{\mathrm{H}} \cos \alpha$
  3. $\overrightarrow{\mathrm{H}} \sin \alpha+\overrightarrow{\mathrm{G}} \cos \alpha$
  4. $\overrightarrow{\mathrm{G}} \sin \alpha+\overrightarrow{\mathrm{H}} \cos \alpha$

Solution

$\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}} ;|\overrightarrow{\mathrm{a}}|=\mathrm{rp} \sin \theta$ $|\overrightarrow{\mathrm{H}}|=\operatorname{rpcos} \theta \quad\left[\because \sin \left(90^{\circ}+\theta\right)=\cos \theta\right\rfloor$ $\mathrm{G}=\mathrm{rp} \sin \theta$
$\mathrm{H}=\mathrm{rp} \cos \theta$
$\mathrm{x}=\mathrm{rp} \sin (\theta+\alpha)$
From (1), (2) \& (3), $\mathrm{x}=\overrightarrow{\mathrm{a}} \cos \alpha+\overrightarrow{\mathrm{H}} \sin \alpha$

Asked in: JEE Main 2003

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