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Math Help - Distance between line and plane

  1. #1
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    Distance between line and plane

    Find distance between line x-y-z =2, 2x+y+z=4 and the plane x-y=5

    My problem is that the book is saying the minimum distance between the plane and line is zero but l am getting 6/\sqrt{2}. Can someone tell me where did l get this one wrong

    Attempt to solution :


    If l intersect the first two planes given l will get the equation of the line :


    x-y-z=2................(1)
    2x+y+z=4...............(2)


    Add (1) and (2) ===> 3x+2z=6


    z=3-3x/2.............(3)

    Substitute (3) into (1)

    y=1-x/2


    and x=x

    Equations of the line :

    x=t

    y=1-t/2

    z=3-3t/2

    Get point on the line and the normal vector for plane :

    (0,1,3)

    Formula for distance between point and plane x-y=5 :

    normal vector for the plane : (1,-1,0)

    distance=\frac{ax+by+cz+d}{\sqrt{(a^2+b^2+c^2)}}

    when l solve for distance l get -6/sqrt(2).
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  2. #2
    MHF Contributor
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    Quote Originally Posted by nyasha View Post
    x-y-z=2................(1)
    2x+y+z=4...............(2)


    Add (1) and (2) ===> 3x+2z=6
    Please try that addition again. Check the 'z', in particular.

    when l solve for distance l get -6/sqrt(2).
    Please never write that again. Distance isn't usually negative. Sometimes, I suppose, but you should be very, very clear that is what you mean.
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    Please try that addition again. Check the 'z', in particular.



    Please never write that again. Distance isn't usually negative. Sometimes, I suppose, but you should be very, very clear that is what you mean.

    Okay thanks for pointing out the mistake. For the distance l forgot to put absolute value symbol
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