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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    The projection of the vector\[2i+a\text{ }j-k\]on the vector\[i-2j+k\] is\[\frac{-5}{\sqrt{6}}\]. Then, the value of a is equal to

    A)  1                            

    B)         2

    C)  \[-2\]                   

    D)         3

    E)  \[-3\]

    Correct Answer: D

    Solution :

    Let\[a=2i+aj-k\]and\[b=i-2j+k\] the projection of a on b is \[=\frac{a.b}{|b|}\] \[\because \]     \[\frac{-5}{\sqrt{6}}=\frac{(2i+aj-k).(i-2j+k)}{\sqrt{1+4+1}}\] \[\Rightarrow \]               \[\frac{-5}{\sqrt{6}}=\frac{2-2a-1}{\sqrt{6}}\] \[\Rightarrow \]               \[-2a+1=-5\]       \[\Rightarrow \]               \[2a=6\] \[\Rightarrow \]               \[a=3\]


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