  Question

The resultant of two vectors $$\overrightarrow P$$ and $$\overrightarrow Q$$ is $$\overrightarrow R$$ . If $$\overrightarrow Q$$ is doubled then the new resultant vector is perpendicular to $$\overrightarrow P$$ . The magnitude of $$\overrightarrow R$$ is :-

A
P2Q22PQ  B
Q  C
PQ  D
P+QPQ  Solution

The correct option is A $$\displaystyle Q$$After the vector $$Q$$ is doubled ,The resultant is $$90^{o}$$ with $$P$$ hence angle between $$P$$ and $$2Q$$ is obtuse Let the angle be $$90+\theta$$ and $$P$$ be along X-axis then resultant would be along Y-axisResolving $$2Q$$ along X-axis and Y-axis$$Qcos\theta =$$resultant$$2Qsin \theta = P$$Hence, $$sin\theta =\dfrac{P}{2Q}$$Initially the magnitude of resultant $$R$$ is $$=\sqrt{P^{2}+Q^{2}+2PQcos(90+\theta)}$$$$=\sqrt{P^{2}+Q^{2}-2PQsin\theta}$$                                                                      $$=\sqrt{P^{2}+Q^{2}-2PQ\times \dfrac{P}{2Q}}$$                                                                     $$=\sqrt{Q^{2}}$$$$=Q$$Physics

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