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Finding a plane given 3 points

  1. Apr 10, 2008 #1
    1. The problem statement, all variables and given/known data

    Find a plane through the points (6,-7,2), (-4,7,-4), (4,8,-2)

    3. The attempt at a solution

    A = <-10, 14, -6>

    B = <-2, 15, -4>

    i j k
    -10 14 -6
    -2 15 -4

    i(-146) - j(28) + k(-178)

    -146(x - 6) - 28(y + 7) - 178(z + 2)

    -146x - 28y - 178z = 324

    Actual answer: `34*x -28*y -122*z = 156`

    ugh
     
  2. jcsd
  3. Apr 10, 2008 #2
    u evaluated the determinant of the cross product matrix incorrectly, it should be 34i-28y-122z, did u use cramer's rule?
     
  4. Apr 10, 2008 #3
    Okay, here's my work for a similar problem with the same concept. Does this look right?

    P = (-8, -3, 7)

    Q = (-8, 7, 4)

    R = (-2, -4, -8)

    PQ = (0, 10, -3)

    PR = (6, -1, -15)

    i j k
    0 10 -3
    6 -1 -15

    i(-150 + 3) - j(-18) + k(60) = 0

    147(x + 8) + 18(y - 3) + 60(z - 7) = 0

    147x + 18y + 60z = -1176 + 54 + 420 = -702
     
  5. Apr 10, 2008 #4
    Ugh!!! It's wrong...what am I doing wrong?!
     
  6. Apr 10, 2008 #5
    You have the right idea, but perhaps you are calculating the determinant too fast, it should go like this:
    i((10*-15)-(-3*-1))-j((0*-15)-(-3*6))+k((0*-1)-(10*6)=
    i(-150-3)-j(0-(-18))+k(0-60)=
    -153i-18j-60k
    Now, as you are having the right idea, take the dot product of our newfound normal vector with P and set it equal to zero since, by definition, it is normal:
    (-153i-18j-60k).(i(x-(-8))+j(y-(-3))+k(z-7))=
    153(x+8)-18(y+3)-60(z-7)=0
    You should be able to solve it from there. Let me know if this answer does not coincide with the book. There are lots of double negatives that you seemed to have misplaced in your calculations...be careful!
     
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