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Math Help - Vectors in 3 space

  1. #1
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    Vectors in 3 space

    Find the equation of the plane through point (-1, -2, 3) and perpendicular to both the planes x - 3y + 2z = 7 and 2x - 2y - z = -3. Please help, final tomorrow and I'm confused on this. I know how to find the equation given 3 points but not with 2 planes. Thanks!
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  2. #2
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    Quote Originally Posted by pakman View Post
    Find the equation of the plane through point (-1, -2, 3) and perpendicular to both the planes x - 3y + 2z = 7 and 2x - 2y - z = -3. Please help, final tomorrow and I'm confused on this. I know how to find the equation given 3 points but not with 2 planes. Thanks!
    If two planes are parrallel then their normals are parrallel.
    These two plane normals are not parallel. *)
    Thus, the planes are not parallel.
    Assume you can draw a third plane (this problem) parallel to both.
    Then the first one is parallel with third one.
    And the second one is parallel with third one.
    Thus, the first one is parallel with second one.
    But that is a contradiction!
    Thus there is no answer to your question.

    *)The vectors are not scalar multiples of each other.
    ----------
    Do you mean that I should find two seperate planes, one parallel with first one and another parallel with second one. So they pass through the point (-1,-2,3).
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  3. #3
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    I copied the question directly from the book, it was a suggested problem to prepare for the final. The answer to the problem is 7x + 5y + 4z = -5. It says perpendicular though?
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  4. #4
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    Actually I just found the problem in my notes.

    So we have the point is (-1, -2, 3) we need the normal vector

    a = <1, -3, 2> and b = <2,-2,-1> derived from the plane equations provided

    then use cross products...
    v = a x b

    I don't know how to show it using ASCII but it'll come out to (3-(-4))i - (-1-4)j + (-2-(-6))k = <7,5,4> = v = normal vector

    plane equation is 7(x+1) + 5(y+2) + 4(z-3) = 0
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