JEE Main & Advanced JEE Main Solved Paper-2014

  • question_answer
    The image of the line \[\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}\]in the plane \[2x-y+z+3=0\]is the line:   JEE Main  Solved  Paper-2014

    A) \[\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}\]

    B) \[\frac{x+3}{-3}=\frac{y-5}{-1}=\frac{z+2}{5}\]

    C) \[\frac{x-3}{3}=\frac{y+5}{1}=\frac{z-2}{-5}\]

    D) \[\frac{x-3}{-3}=\frac{y+5}{-1}=\frac{z-2}{5}\]

    Correct Answer: A

    Solution :

    Equation of AB is\[\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=\lambda \] Co-ordinate of point B is \[\Rightarrow \]\[x=1+2\lambda \]point satisfy the equation of plane \[y=3-\lambda \] \[y=4+\lambda \] \[2(1+2\lambda )-(3-\lambda )+(4+\lambda )+3=0\] \[\Rightarrow \]\[\lambda =-1\] \[\Rightarrow \]Co-ordinate of point B(−1, 4, 3) \[\Rightarrow \]Co-ordinate of point C(−3, 5, 2) \[\Rightarrow \]equation of line passing through ?C? is \[\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}\]

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