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Math Help - Finding an equation for the plane....

  1. #1
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    Finding an equation for the plane....

    I am unsure of how to solve this linear algebra problem...

    Find an equation for the plane that passes through (2,4,-1) and contains the line of intersection of the planes x-y-4z=2 and -2x+y+2z=3

    Help would be greatly appreciated!
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  2. #2
    MHF Contributor
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    The normal vectors to the planes x - y - 4z = 2 and -2x + y + 2z = 3 are <1, -1, -4> and <-2, 1, 2>. If you take the cross-product of these two vectors, you will get a vector that points in the same direction of the line. Then, all you have to do is find two points A and B on that line, and let C = (2, 4, -1). Then the cross-product of the vectors AC and BC will be a normal vector <p, q, r> to the desired plane, whose equation will be px + qy + rz = s, and s can be determined by putting point C into the equation: 2p + 4q - r = s.
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  3. #3
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    how can you find the equation of the plane in ax+by+cz=d format through the point (1,2,3) for example with normal vector (2,-1,4)
    thanks for any help
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  4. #4
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    sorry just dont understand wher yo use the normal vector after gettin x+y+z= s
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