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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    The distance of the point of intersection of the line\[\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}\]and me plane \[x-y+z=5\]from the point\[(-1,-5,-10)\]is

    A)  13                                         

    B)  12

    C)  11                         

    D)         8

    E)  7

    Correct Answer: A

    Solution :

    Given line is\[\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=k(say)\] Any point on the line is\[(3k+2,4k-1,12k+2)\] This point lies on the plane \[x-y+z=5\] \[\therefore \]\[3k+2-(4k-1)+12k+2=5\] \[\Rightarrow \]\[11k=0\]\[\Rightarrow \]           \[k=0\] \[\therefore \]Intersection point is\[(2,-1,2)\]. \[\therefore \]Distance, between points\[(2,-1,2)\]and \[(-1,-5,-10)\] \[=\sqrt{{{(-1-2)}^{2}}+{{(-5+1)}^{2}}+{{(-10-2)}^{2}}}\] \[=\sqrt{9+16+144}=13\]


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