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JEE Main & Advanced JEE Main Online Paper (Held On 23 April 2013)

  • question_answer
                    If \[\overset{\to }{\mathop{a}}\,\] and \[\overset{\to }{\mathop{b}}\,\] are non-collinear vectors, then the value of \[\alpha \]for which the vectors \[\overset{\to }{\mathop{\operatorname{u}}}\,=(\alpha -2)\overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,\]and \[\overset{\to }{\mathop{\operatorname{v}}}\,=(2+3\alpha )\overset{\to }{\mathop{a}}\,-\overset{\to }{\mathop{3b}}\,\] are collinear is:     JEE Main Online Paper ( Held On 23  April 2013 )

    A)                   \[\frac{3}{2}\]                                 

    B)                                          \[\frac{2}{3}\]

    C)                                          \[-\frac{3}{2}\]                                               

    D)                                          \[-\frac{2}{3}\]

    Correct Answer: B

    Solution :

                      Since, \[\vec{u}\] and \[\vec{v}\] are collinear, therefore \[k\vec{u}+\vec{v}=0\] \[\Rightarrow \]\[[k(\alpha -2)+2+3\alpha ]\vec{a}+(k-3)\vec{b}=0\]      ?(i) Since \[\vec{a}\] and \[\vec{b}\]are non-collinear, then for some constant m and n, \[m\vec{a}+n\vec{b}=0\Rightarrow m=0,n=0\] Hence from equation (i) \[k-3=0\Rightarrow k=3\] And\[k(\alpha -2)+2+3\alpha =0\] \[\Rightarrow \]\[3(\alpha -2)+2+3\alpha =0\Rightarrow \alpha =\frac{2}{3}\]


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