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JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Line and Plane

  • question_answer
    The point of intersection of the line \[\frac{x-1}{3}=\frac{y+2}{4}=\frac{z-3}{-2}\] and plane \[2x-y+3z-1=0\] is [Orissa JEE 2005]

    A)            \[(10,\,\,-10,\,3)\]

    B)            \[(10,\,\,10,\,-3)\]

    C)            \[(-10,\,\,10,\,3)\]

    D)            None of these

    Correct Answer: B

    Solution :

                        Given line is, \[\frac{x-1}{3}=\frac{y+2}{4}=\frac{z-3}{-2}=k\], (say)                    \  Point  on the line is                     \[x=3k+1,y=4k-2,\]\[z=-2k+3\]                 .....(i)                    This point must satisfies the equation of plane                    \ \[2(3k+1)-(4k-2)+3(-2k+3)-1=0\Rightarrow k=3\]                    From (i),\[(x,y,z)=(10,\,10,\,-3)\].


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