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J & K CET Engineering J and K - CET Engineering Solved Paper-2008

  • question_answer
    The line \[\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\] meets the plane \[2x+3y-z=-4\]in the point

    A)  \[(1,2,3)\]

    B)  \[(-1,-1,-1)\]

    C)  \[(2,1,3)\]

    D)  \[(1,1,1)\]

    Correct Answer: B

    Solution :

    Any point on the line \[\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\] (say) ie, \[(2k+1,\,3k+2,\,4k+3)\] Which lies on the plane \[2x+3y-z=-4\] \[\therefore \]   \[2(2k+1)+3(3k+2)-(4k+3)=-4\] \[\Rightarrow \] \[9k=-9\] \[\Rightarrow \] \[k=-1\] \[\therefore \] Required point is \[(-2+1,\,-3+2,\,-4+3)\] ie, \[(-1,\,-1,\,-1)\].


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