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Math Help - Normal to Plane

  1. #1
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    Normal to Plane

    Find the equation of the plane in R^3 that is perpendicular to (1,2,3) and contains the point (3,2,1).

    So the normal to the plane is (1,2,3)

    So: x+2y+3z=d

    Sub the point in and solve for d.

    (3) + 2(2) + 3(1) + d = 0
    10 + d = 0
    d = -10

    So the equation is
    x + 2y + 3z -10 = 0

    Is this the correct method?
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  2. #2
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    Quote Originally Posted by millerst View Post
    Find the equation of the plane in R^3 that is perpendicular to (1,2,3) and contains the point (3,2,1).

    So the normal to the plane is (1,2,3)

    So: x+2y+3z=d

    Sub the point in and solve for d.

    (3) + 2(2) + 3(1) + d = 0
    10 + d = 0
    d = -10

    So the equation is
    x + 2y + 3z -10 = 0

    Is this the correct method?
    The equation of the plane \pi which prep. to the vector <a,b,c> and contains <x_o,y_0,z_o> is:
    \pi: a(x-x_o)+b(y-y_o)+c(z-z_o)=0

    Your plane:
    (x-3)+2(y-2)+3(z-1)=0
    or:
    x+2y+3z=10.
    As you said.
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