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3-1.Vectors
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A vector $\vec Q$ which has a magnitude of $8$ is added to the vector $\vec P$ which lies along $x-$ axis. The resultant of two vectors lies along $y-$ axis and has magnitude twice that of $\vec P$. The magnitude of is $\vec P$
A
$\frac {6}{\sqrt 5}$
B
$\frac {8}{\sqrt 5}$
C
$\frac {12}{\sqrt 5}$
D
$\frac {16}{\sqrt 5}$
Solution
Now $Q \sin \theta=P$
$\&\,\,\, \quad Q \cos \theta=2 P$
$\tan \theta=1 / 2 \quad \text { so, } \sin \theta=1 / \sqrt{5}$
$P=8(1 / \sqrt{5})=\frac{8}{\sqrt{5}}$
Standard 11
Physics
