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Homework Help: Equation of a plane

  1. Aug 7, 2008 #1
    1. The problem statement, all variables and given/known data

    Ex: Find the equation of the plane passing through A(1, 0, 1), B(0, 1, 2) and C(1, 3, 2).

    Solution: In this case, we first need to establish a normal vector for the plane. Note that
    AB and AC are both vectors on (or parallel to) the plane. Since their cross product is
    perpendicular to both AB and AC, it must be a normal vector to the plane as well. Thus,
    we have n = AB × AC = [−1, 1, 1] × [0, 3, 1] = [−2, 1,−3]. Hence, the general equation
    takes the form −2x + y − 3z = d, where d can be found by substituting the point A, i.e.
    d = −2(1) + (0) − 3(1) = −5. Hence, the equation of the plane is −2x + y − 3z = −5.

    2. Relevant equations
    3. The attempt at a solution

    I'm having a hard time understanding the answer - I was hoping someone could clarify a few questions that I had. It feels like im missing gigantic gaps in my vector mathematics knowledge.

    Note that AB and AC are both vectors on (or parallel to) the plane.
    How do I know this?

    I'd normally find out if vectors are parallel if I use the dot product and end up with a 0 or 180 degree angle.

    Since their cross product is perpendicular to both AB and AC, it must be a normal vector to the plane as well.

    Using the cross product method, I'm left with the vector (-2, 1, 3). Using the dot product method on bot AB and AC, i find that they are equal to zero, therefore perpendicular. But, why is this normal to the plane?

    I understand the rest of the solution though, but I'm majorly stuck with the above parts.

    Is there an easier way to solve this? I've had to do around 8 calculations to solve this question without making major assumptions.
  2. jcsd
  3. Aug 7, 2008 #2


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    Science Advisor

    Yes, that is exactly correct. As a check note that with x=0, y= 1, z= 2, -2(0)+ 1- 3(-2)= 1- 6= -5 so this is also correct for point B and with x= 1, y= 3, c= 2, -2(1)+ 3- 3(2)= -2+ 3- 6= -5 so this is also correct for point C. Points A, B, and C are all satisfy the equation so this is the equation of the plane they are in.

    Because the endpoints A, B, and C are given as being in the plane and this is a plane

    Just as 3 points determine a plane, two independent vectors determine a plane. Any vector in the plane can be written as a[−1, 1, 1] + b[0, 3, 1]. A vector having 0 dot product with those two vectors will have 0 dot product with any vector in the plane: [itex]\vec{v}\cdot (a[-1, 1, 1]+ b[0, 3, 1])= a (\vec{v}\cdot [-1, 1, 1])+ b(\vec{v}\cdot [0, 3, 1])= a(0)+ b(0)= 0[/itex]. Since the vector is perpendicular to any vector in the plane, it is perpendicular to any line in the plane: that's the definition of "perpendicular to the plane".

    What kind of "calculations" do you mean? The hardest was finding the cross product and that should take very little time.
    Last edited by a moderator: Aug 7, 2008
  4. Aug 7, 2008 #3
    So.. Basically the question tells me that a plane passes through these points, so automatically, I know that these points reside on the plane.

    Realizing that I need a vector which is normal to the plane in order to find the plane equation, I go about getting a normal vector by using the cross product on two vectors on the plane (in this case, AB and AC), or parallel to it. A third vector results being in the direction that is normal to the plane..

    Is that right? Awesome explanation by the way. You've helped me out a lot.
  5. Aug 7, 2008 #4


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    Homework Helper

    Yes that's right.
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