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Math Help - normal line point on surface

  1. #1
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    normal line point on surface

    Would anyone know how to begin a problem where a function z(x,y) is given and x(t),y(t),z(t).

    Thanks!
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  2. #2
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    Re: normal line point on surface

    the normal line to the surface is expressed as:

    N  = \displaystyle\frac{\partial{z}}{\partial{x}} \spacei  + \displaystyle\frac{\partial{z}}{\partial{y}} \spacej  - \spacek

    \displaystyle\frac{\partial{z}}{\partial{x}} = y; \space \displaystyle\frac{\partial{z}}{\partial{y}} = x

    N  = y \spacei  + x \spacej  - \spacek

    from the given line we have the symmetric equations:

    \displaystyle\frac{x - 3}{-2} = \displaystyle\frac{y - 4}{5} = \displaystyle\frac{z - 3}{3}

    which gives us:

    \space -2i \space + \space 5j \space + \space 3k
    Last edited by jpritch422; April 28th 2013 at 07:53 PM.
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  3. #3
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    Re: normal line point on surface

    so we would solve for where <y,x,-1> is equal to <-2,5,3>, wouldn't that be never because of k, -1 cannot equal 3
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  4. #4
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    Re: normal line point on surface

    Quote Originally Posted by topsaa View Post
    so we would solve for where <y,x,-1> is equal to <-2,5,3>, wouldn't that be never because of k, -1 cannot equal 3
    what this is saying is that the given line coincides with vector:

    -2 \spacei  + 5 \spacej  + 3 \spacek

    the normal line to the surface coincides with the vector:

    y \spacei  + x \spacej  - k

    since the two vectors are parallel, that would make the two lines parallel. So then,

    y \spacei  + x \spacej  - k  = \displaystyle\frac{2}{3} \spacei  - \displaystyle\frac{5}{3} \spacej  - \spacek
    Last edited by jpritch422; April 29th 2013 at 04:33 AM.
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