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Math Help - Equation of a plane given point and norm.

  1. #1
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    Unhappy Equation of a plane given point and norm.

    Hi guys I am stuck on something,

    Find the equation of the plane p containing the point P(0;2;3) and normal to the line with parametric
    form x = 1 + 2t; y = 0; z = t.

    So norm is (1,0,0) + t(2,0,1)?

    Next do I have to define another point say, R on plane so-

    vector PR will be perpendicular to norm direction vector and-

    PR x n = 0 (cross product of 2 perp. vectors)..??

    If this is the right logic how do I use the norm in the form I have? I think I'd be able to do this if I had the norm as a 3 number direction vector, not (1,0,0) + t(2,0,1)..

    Am I right in saying that as a plane is Ax + By + Cz = D, norm (x,y,z) provides the coefficients in this formula and point P(x,y,z) gives the rest of the information to algebraically solve?
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  2. #2
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    Re: Equation of a plane given point and norm.

    Quote Originally Posted by camjerlams View Post
    Find the equation of the plane p containing the point P(0;2;3) and normal to the line with parametric
    form x = 1 + 2t; y = 0; z = t.
    So norm is (1,0,0) + t(2,0,1)?
    Next do I have to define another point say, R on plane so-
    vector PR will be perpendicular to norm direction vector and-
    Am I right in saying that as a plane is Ax + By + Cz = D, norm (x,y,z) provides the coefficients in this formula and point P(x,y,z
    The norm is simply a vector. In this case, N=<2,0,1>

    A point on the plane is R=(x,y,z). Then \overrightarrow {PR}  = \left\langle {x - 0,y-2,z - 3} \right\rangle is a vector "in" the plane.
    As you say that \overrightarrow {PR} must be perpendicular to the normal N.

    Thus N\cdot\overrightarrow {PR}=0 or 2(x)+1(z-3)=0
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