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JEE Main & Advanced JEE Main Solved Paper-2015

  • question_answer
    The distance of the point \[\left( 1,0,2 \right)\]from the point of intersection of the line \[\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}\]and the plane \[x-y+z=16,\] is : [JEE Main Solved Paper-2015 ]

    A) \[3\sqrt{21}\]                                   

    B) 13

    C) \[2\sqrt{14}\]                                   

    D) 8

    Correct Answer: B

    Solution :

    Parametric co-ordinate of any point on the line\[\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}\]is\[x=3\lambda +2,\]\[y=4\lambda -1,z=12\lambda +2\] If this point lies on the plane \[x-y+x=15\] then\[(3\lambda +2)-(4\lambda -1)+(12\lambda +2)=16\] \[\Rightarrow \]\[11\lambda =11\Rightarrow \lambda =1\] \[\Rightarrow \] The point at intersection = ( 5, 3, 14) \[\Rightarrow \]distance = 13.                


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