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JEE Main & Advanced Mathematics Vector Algebra Question Bank Modulus of vector Algebra of vectors

  • question_answer
    In a trapezium, the vector \[\overrightarrow{BC}=\lambda \overrightarrow{AD}.\] We will then find that \[\mathbf{p}=\overrightarrow{AC}+\overrightarrow{BD}\] is collinear with \[\overrightarrow{AD},\] If \[\mathbf{p}=\mu \overrightarrow{AD},\] then

    A) \[\mu =\lambda +1\]

    B) \[\lambda =\mu +1\]

    C) \[\lambda +\mu =1\]           

    D) \[\mu =2+\lambda \]

    Correct Answer: A

    Solution :

    We have, \[\mathbf{p}=\overrightarrow{AC}+\overrightarrow{BD}=\overrightarrow{AC}+\overrightarrow{BC}+\overrightarrow{CD}=\overrightarrow{AC}+\lambda \overrightarrow{AD}+\overrightarrow{CD}\]           \[=\lambda \,\overrightarrow{AD}+(\overrightarrow{AC}+\overrightarrow{CD})=\lambda \,\overrightarrow{AD}+\overrightarrow{AD}=(\lambda +1)\overrightarrow{AD}.\] Therefore \[\mathbf{p}=\mu \overrightarrow{AD}\Rightarrow \mu =\lambda +1.\]       


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