Question

A vector $ Q\to $ which has a magnitude of 8 is added to the vector $ P\to $ , which lies along the X-axis. The resultant of these two vectors is a third vector $ R\to $ , which lies along the Y-axis and has a magnitude twice that of $ P\to $ .The magnitude of $ P\to $ is :

• A. $ \frac{6}{\sqrt{5}} $
• B. $ \frac{8}{\sqrt{5}} $
• C. $ \frac{12}{\sqrt{5}} $
• D. $ \frac{16}{\sqrt{5}} $

Answer 1

The Correct Option is B

Given: $ Q=8\,units $ $ \vec{R}=2\vec{P} $ Since, $ \vec{R} $ is along Y-axis and $ \vec{P} $ is along $ x- $ axis. Therefore, $ \vec{P} $ and $ \vec{R} $ is perpendicular vectors. Hence, $ {{Q}^{2}}={{R}^{2}}+{{P}^{2}} $ Putting the given values in E (i), we get $ {{(8)}^{2}}={{(2p)}^{2}}+{{p}^{2}}=4{{p}^{2}}+{{P}^{2}}=5{{p}^{2}} $ or $ 5{{p}^{2}}=64 $ $ {{P}^{2}}=\frac{64}{5} $ $ \therefore $ $ p=\frac{8}{\sqrt{5}} $