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Math Help - Vestors - finding an equaiton of a plane

  1. #1
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    Vestors - finding an equaiton of a plane

    HI,

    I have the plane equation that is

    2x-y+4z =2

    and have to find the equation of the plane that is parallel to this equation and passes through the points (2, 1, -1)

    To work this out do i just cross multiply the (2,-1,4)x(2,1,-1) and this is the equation of parallel plane?
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  2. #2
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    Quote Originally Posted by kayne View Post
    I have the plane equation that is 2x-y+4z =2
    and have to find the equation of the plane that is parallel to this equation and passes through the points (2, 1, -1)
    For any k the plane 2x-y+4z=k will be parallel to the given plane.
    To find k subsititute the point.
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  3. #3
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    plane equation 2x-y+4z =2\ normal\ vector\ \begin{bmatrix} 2\\-1\\4 \end{bmatrix}
    Any plane with this normal vector will parallel with it.
    So new plane equation : 2(x-2)-(y-1)+4(z+1) =0 \rightarrow 2x-y+4z=-1
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