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Math Help - Intersection of a Normal and a plane

  1. #1
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    Intersection of a Normal and a plane

    A normal to 3x-y+4z=12 passes through A(5,1,6). Find the point of intersection of the normal and the plane.

    So from what I know the normal would be:
    n = (3,-1,4)

    But I don't know how to go about answering this....Any help is appreciated...
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  2. #2
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    Quote Originally Posted by lil_cookie View Post
    A normal to 3x-y+4z=12 passes through A(5,1,6). Find the point of intersection of the normal and the plane.
    The normal line through A is \left\{ \begin{gathered}  x = 5 + 3t \hfill \\  y = 1 - t \hfill \\  z = 6 + 4t \hfill \\ \end{gathered}  \right.
    Replace the x,~y,~z~ in the equation of the plane and solve for t.
    Last edited by Plato; June 23rd 2009 at 07:53 AM.
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by lil_cookie View Post
    A normal to 3x-y+4z=12 passes through A(5,1,6). Find the point of intersection of the normal and the plane.

    So from what I know the normal would be:
    n = (3,-1,4)

    But I don't know how to go about answering this....Any help is appreciated...
    Hello : the nomal is definid by : x-5=3t
    y-1=-t
    z-6=4t
    x=5+3t and y=1-t and z=6+4t
    Remplace in plan equation : 3(5+3t) -(1-t)+4(6+4t)=12
    solve : I'have t= -1
    remplace in system : x=5+3( -1 ) and y =1-(-1 ) and z= 6+4( -1)
    coclusion: the point M( 2 , 2 , 2 )
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  4. #4
    Super Member dhiab's Avatar
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    Quote Originally Posted by Plato View Post
    The normal line through A is \left\{ \begin{gathered} x = 5 + 3t \hfill \\ y = 1 - t \hfill \\ z = 6 + 5t \hfill \\ \end{gathered} \right.
    Replace the x,~y,~z~ in the equation of the plane and solve for t.
    Hello / you are z= 6+4t .... Thank you
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  5. #5
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    I understand now, thank you so much
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