2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Evening)

  • question_answer
    Let \[\sqrt{3}\hat{i}+\hat{j},\hat{i}+\sqrt{3}\hat{j}\]and \[\beta \hat{i}+(1-\beta )\hat{j}\]respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is \[\frac{3}{\sqrt{2}},\]then the sum of all possible values of \[\beta \] is [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

    A) 1                                 

    B) 4     

    C) 3                                 

    D)   2

    Correct Answer: A

    Solution :

    Angle bisector of OA and OB where O is the origin is given by \[x-y=0\]                          ...(i) Since, the distance of C from (i) is \[\frac{3}{\sqrt{2}}\] \[\therefore \]      \[\frac{|\beta -(1-\beta )|}{\sqrt{2}}=\frac{3}{\sqrt{2}}\] \[\Rightarrow \]\[|2\beta -1|=3\]\[\Rightarrow \]\[\beta =2\]or \[-1\]


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