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The sum of two forces  \vec{P} and \vec{Q} is \vec{R} such that \left | \vec{R} \right |=\left | \vec{P} \right |. The angle \theta (in degrees) that the resultant of 2\vec{P} and \vec{Q} will make with  \vec{Q} is ,
Option: 1 90
Option: 2 120
Option: 3 45
Option: 4 30
 

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As

 \\ |\vec{P}+\vec{Q}|=|\vec{P}| \\ \because |\vec{R}|=|\vec{P}| \\\Rightarrow P^2+Q^2+2PQcos\theta =P^2\\ \Rightarrow Q+2Pcos\theta =0...(1)

Now Vector P is doubled but vector Q is the same so the angle between 2\vec{P} \ and \ \vec{Q} will be also \theta

And If resultant of 2\vec{P} \ and \ \vec{Q} is \vec{R'}

Then from the figure angle between \vec{R'} \ and \ \vec{Q} is \alpha

and tan \alpha =\frac{2Psin\theta }{Q+2Pcos\theta }\\ From \ equation \ (1) \\ \text{we get tan} \alpha =\infty\Rightarrow \alpha =90^0

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Ritika Jonwal

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