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3-1.Vectors
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The sum of two forces $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ is $\overrightarrow{\mathrm{R}}$ such that $|\overrightarrow{\mathrm{R}}|=|\overrightarrow{\mathrm{P}}| .$ The angle $\theta$ (in degrees) that the resultant of $2 \overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ will make with $\overrightarrow{\mathrm{Q}}$ is
A
$80$
B
$90$
C
$85$
D
$95$
(JEE MAIN-2020)
Solution
$|\overrightarrow{ P }+\overrightarrow{ Q }|=|\overrightarrow{ P }|$
$P ^2+ Q ^2+2 PQ \cos \theta= P ^2$
$= Q +2 P \cos \theta=0$
$\cos \theta=\frac{- Q }{2 P }$
$\tan \alpha=\frac{2 P \sin \theta}{2 P \cos \theta+Q}=\infty$ $[2 P \cos \theta+Q=0]$
$\alpha=90^{\circ}$
Standard 11
Physics
