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Thread: 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!
    Attached Thumbnails Attached Thumbnails normal line point on surface-finalreview2.png  
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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 = \displaystyle\frac{\partial{z}}{\partial{x}}$$\displaystyle \space$i$\displaystyle + \displaystyle\frac{\partial{z}}{\partial{y}}$$\displaystyle \space$j$\displaystyle - $$\displaystyle \space$k

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

    N$\displaystyle = y$$\displaystyle \space$i$\displaystyle + x$$\displaystyle \space$j$\displaystyle - $$\displaystyle \space$k

    from the given line we have the symmetric equations:

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

    which gives us:

    $\displaystyle \space$ $\displaystyle -2$i$\displaystyle \space$$\displaystyle +$$\displaystyle \space$$\displaystyle 5$j$\displaystyle \space$$\displaystyle +$$\displaystyle \space$$\displaystyle 3$k
    Last edited by jpritch422; Apr 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:

    $\displaystyle -2$$\displaystyle \space$i$\displaystyle + 5$$\displaystyle \space$j$\displaystyle + 3$$\displaystyle \space$k

    the normal line to the surface coincides with the vector:

    $\displaystyle y$$\displaystyle \space$i$\displaystyle + x$$\displaystyle \space$j$\displaystyle - $k

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

    $\displaystyle y$$\displaystyle \space$i$\displaystyle + x$$\displaystyle \space$j$\displaystyle - $k$\displaystyle = \displaystyle\frac{2}{3}$$\displaystyle \space$i$\displaystyle - \displaystyle\frac{5}{3}$$\displaystyle \space$j$\displaystyle - $$\displaystyle \space$k
    Last edited by jpritch422; Apr 29th 2013 at 04:33 AM.
    Thanks from topsaa
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