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KVPY Sample Paper KVPY Stream-SX Model Paper-7

  • question_answer
    Let \[\vec{a}=2\hat{i}+{{\lambda }_{1}}\hat{j}+3\hat{k},\] \[\hat{b}=4\hat{i}+(3-{{\lambda }_{2}})\hat{j}+6\hat{k}\]  and \[\vec{c}=3\hat{i}+6\hat{j}+({{\lambda }_{3}}-1)\hat{K}\] be three vectors such that \[\vec{b}=2\vec{a}\] and \[\vec{a}\] is perpendicular to \[\vec{c}.\] Then a possible value of \[({{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}})\] is:

    A) (1, 3, 1)            

    B) \[\left( -\frac{1}{2},4,0 \right)\]

    C) \[\left( \frac{1}{2},4-2 \right)\]                 

    D) (1, 5, 1)

    Correct Answer: B

    Solution :

    Because\[b=2\vec{a},\]
    So        \[3-{{\lambda }_{2}}=2{{\lambda }_{2}}...(i)\]
    Because a is perpendicular to c so
    \[6+6{{\lambda }_{1}}+3({{\lambda }_{3}}-1)=0....(ii)\]
    \[\Rightarrow \]   \[({{\lambda }_{1}},{{\lambda }_{2}},{{\lambda }_{3}})=({{\lambda }_{1}},3-2{{\lambda }_{1}},-1-2{{\lambda }_{1}})\]
    where\[{{\lambda }_{1}}\in R\]
    \[\Rightarrow \]\[\left( -\frac{1}{2},4,0 \right)\] satisfies above triplet.


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