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Thread: Quick linear algebra question

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    Quick linear algebra question

    Hey guys,

    I didn't know who to ask and I need an answer quickly :S. It should be simple, I just cant seem to remember how I did this problem.

    What is the equation of the plane parallel to the vector (1, 1, -2) which passes through the points (1, 5, 18) and (4, 2, -6).

    The answer is 5x - 3y + z = 8

    How do I go about solving this?

    Thanks for yout time!

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    Junior Member hairball's Avatar
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    Re: Quick linear algebra question

    Go to math forum then...

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    vbuggy krtxmrtz's Avatar
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    Re: Quick linear algebra question

    Quote Originally Posted by INF3RN0666
    Hey guys,

    I didn't know who to ask and I need an answer quickly :S. It should be simple, I just cant seem to remember how I did this problem.

    What is the equation of the plane parallel to the vector (1, 1, -2) which passes through the points (1, 5, 18) and (4, 2, -6).

    The answer is 5x - 3y + z = 8

    How do I go about solving this?

    Thanks for yout time!
    The equation in vector form is:

    R = r + av1 + bv2

    where bold means vector and a,b are scalars (numbers).

    r is a vector from the coordinate origin to a point in the plane, say to (1,5,18)
    v1 and v2 are any two (linear independent) vectors contained on the plane. For example, v1 can be the given vector (1,1,-2) and v2 can be the vector joining the 2 given points:

    v2 = (4,2,-6) - (1,5,18) = (3,-3,-24)

    So the equation becomes:

    R = (x,y,z) = (1,5,18) + a(1,1,-2) + b(3,-3,-24)

    Expressing each component separately:

    x = 1 + a +3b
    y = 5 + a -3b
    z = 18 -2a -24b

    Now you make substitutions to find z in terms of x and y:

    a = x - 1 - 3b
    y = 5 + x -1 -3b -3b = 4 + x -6b
    b = (4 + x - y) / 6

    z = 18 -2(x - 1 -3b) - 24b = 18 - 2x + 2 + 6b - 24b = 20 - 2x - 16(4 + x - y) / 6 = 20 - 2x -3(4 + x - y) = 20 - 2x - 12 - 3x + 3y = 8 - 5x + 3y

    Or, rearranging,

    5x - 3y + z = 8
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