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

Answer: (B)

$\frac{8}{\sqrt{5}}$

Answer 2

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 $5{{p}^{2}}=64$ ${{P}^{2}}=\frac{64}{5}$ $\therefore$ $p=\frac{8}{\sqrt{5}}$