J & K CET Medical J & K - CET Medical Solved Paper-2006

  • question_answer
    Vector which is perpendicular to a cos\[\theta \hat{i}\]+ b sin \[\theta \hat{j}\] is:

    A) \[b\sin \theta \hat{i}-a\cos \theta \hat{j}\]        

    B) \[\frac{1}{2}\sin \theta \hat{i}-\frac{1}{b}\cos \theta \hat{j}\]

    C) \[5k\]                                   

    D) all of the above

    Correct Answer: D

    Solution :

    From definition of dot product of vectors, we have \[x.y=xy\cos \theta \] When \[\theta ={{90}^{o}},\cos {{90}^{o}}=0\] \[\therefore \]  \[x.y=0\] Given,   \[x=a\cos \theta \hat{i}+b\sin \theta \hat{j}\]                 \[y=b\sin \theta \hat{i}-a\cos \theta \hat{j}\]                 \[x.y=(a\cos \theta \hat{i}-b\sin \theta \hat{j})\]                 \[(b\sin \theta \hat{i}.a\cos \theta \hat{j})\] \[x.y=ab\sin \theta \cos \theta -ab\sin \theta \cos \theta =0\] Hence, vacuous are perpendicular. Similarly for option and also\[x.y=0\]


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