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Finding equation of plane containing triangle

  1. Jun 16, 2012 #1
    The problem statement, all variables and given/known data

    B(-6,-7,-3) C(2,7,3) D(-4,-1,-11)
    Find the equations of the plane containing the triangle BCD.

    The attempt at a solution
    I've gotten to the end and got an equation, but the normal vector it has is just making me really paranoid.

    Vector BC = [2+6, 7+7, 3+3] = [8,14,6]
    Vector BD = [-4+6,-1+7,-11+3] = [2,6,-8]

    BC x BD = (-112 -36)i - (-64 -12)j + (-64 -28)k
    Normal vector = [-148, 76, -92]


    Equation:
    -148(x+6) +76(y+7) -92(z+3) =0
    -148x +76y +92z -632 = 0


    The normal vector is what's bothering me. Some later problems require it, and it's making all of them look kinda gross as well, so I just wanted to make sure I didn't make any mistakes.
     
  2. jcsd
  3. Jun 16, 2012 #2

    Mark44

    Staff: Mentor

    For a quick check, take the dot product of your normal vector with BC and then with BD. If you get 0 for both, your normal is perpendicular to both.
     
  4. Jun 16, 2012 #3
    Alright, so I did that and neither of them were 0.
    Now I have no idea where I went wrong.
     
  5. Jun 16, 2012 #4

    Mark44

    Staff: Mentor

    I think you have a mistake in your cross product. When I dot BC with the normal I don't get 0.

    BC = <8, 14, 6> = 2<4, 7, 3>
    N = <-148, 76, -92> = 4<-37, 19, -23>

    BC ## \cdot ## N = 2<4, 7, 3> ## \cdot ## 4<-37, 19, -23>
    = 8 * <4, 7, 3> ## \cdot ## <-37, 19, -23>
    = 8 * 4(-37) + 7(19) + 3(-23)
    = 8 * (-148 + 133 - 69) ## \neq ## 0
     
  6. Jun 16, 2012 #5
    I think I found the problem. When I set up the cross product, I accidentally put int a -8 instead of a 6.

    N should be <-148, 76, 20>
    And N ⋅ BC does = 0, as well as N ⋅ BD

    Thank you for your help Mark44.
     
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