2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 9 April 2014)

  • question_answer
    If\[\left| \overset{\to }{\mathop{a}}\, \right|=2,\left| \overset{\to }{\mathop{b}}\, \right|=3\]and\[\left| 2\overset{\to }{\mathop{a}}\,-\overset{\to }{\mathop{b}}\, \right|=5,\]then\[\left| 2\overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\, \right|\] equals:   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) 17                                          

    B) 7

    C) 5                                             

    D) 1

    Correct Answer: C

    Solution :

                    Given \[|2\vec{a}-\vec{b}|=5\] \[\sqrt{{{(2|\vec{a}|)}^{2}}+|\vec{b}{{|}^{2}}-2\times |2\vec{a}||\vec{b}|\cos \theta }=5\] Putting values of \[|\vec{a}|\]and \[|\vec{b}|,\]we get \[{{(2\times 2)}^{2}}+{{(3)}^{2}}-24\cos \theta =25\] \[\Rightarrow \]\[\cos \theta =0\]\[\Rightarrow \]\[\theta =\frac{\pi }{2}\] \[|2\vec{a}+\vec{b}|=\sqrt{16+9+24\cos \theta }=\sqrt{25}=5\]


You need to login to perform this action.
You will be redirected in 3 sec spinner