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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Afternoon)

  • question_answer
                A perpendicular is drawn from a point on the line \[\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{1}\]to the plane \[x+y+z=3\] such that the foot of the perpendicular Q also lies on the plane \[xy+z=3.\]Then the co-ordinates of Q are : [JEE Main 10-4-2019 Afternoon]

    A) \[\left( 2,0,1 \right)\]

    B) \[\left( 4,0,1 \right)\]

    C) \[\left( 1,0,4 \right)\]                 

    D) \[\left( 1,0,2 \right)\]

    Correct Answer: A

    Solution :

                Let point P on the line is \[(2\lambda +1,\lambda 1,\lambda )\] foot of perpendicular Q is given by \[\frac{x-2\lambda -1}{1}=\frac{y+\lambda +1}{1}=\frac{z-\lambda }{1}=\frac{-(2\lambda -3)}{3}\] \[\therefore \] Q lies on \[x+y+z=3\And xy+z=3\] \[\Rightarrow x+z=3\And y=0\] \[y=0\Rightarrow \lambda +1=\frac{-2\lambda +3}{3}\Rightarrow \lambda =0\]\[\Rightarrow Q\]is (2,0,1)           


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