question_answer

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)                  \[\frac{16}{\sqrt{5}}\]

Correct Answer: B

Solution :

                 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 Eq. (i), we get                 \[{{(8)}^{2}}={{(2p)}^{2}}+{{p}^{2}}=4{{p}^{2}}+{{P}^{2}}=5{{p}^{2}}\]                 or            \[5{{p}^{2}}=64\]                 \[{{P}^{2}}=\frac{64}{5}\]                 \[\therefore \]  \[p=\frac{8}{\sqrt{5}}\]