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WB JEE Medical WB JEE Medical Solved Paper-2010

  • question_answer
    The value of \[\lambda \] for which the two vectors \[\vec{a}=5\hat{i}+\lambda \hat{j}+\hat{k}\]and \[\vec{b}=\hat{i}-2\hat{j}+\hat{k}\] are perpendicular to each other is

    A)  2                                            

    B)  -2

    C)  3                            

    D)  -3

    Correct Answer: C

    Solution :

                     For two vectors\[\vec{a}\] and \[\vec{b}\]to be perpendicular, \[\vec{a}.\vec{b}=0.\] Thus,     \[(5\hat{i}+\lambda \hat{j}+\hat{k}).(\hat{i}-2\hat{j}+\hat{k})\]                 \[=5(\hat{i}.\hat{i})-2\lambda (\hat{j}.\hat{j})+1(\hat{k}.\hat{k})\] \[0=5-2\lambda +1\]                 \[\Rightarrow \]               \[0=6-2\lambda \Rightarrow \lambda =3\]


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