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Math Help - Vector Geometry

  1. #1
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    Question Vector Geometry

    Determine the equation of the plane which distance to the origin is equal to 3 and passing through the intersection of the planes x+y+z-11 = 0 and
    x-4y+5z-10=0

    Help please!
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  2. #2
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    Re: Vector Geometry

    Hey bastiante.

    Can you show us what you have tried? (Hint: If it goes through both planes then the area of intersection will be where P1 = P2 or Plane1 = Plane2).
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  3. #3
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Vector Geometry

    Equation of system of planes containing the line of intersection of the two planes:

    x+y+z-11+k(x-4y+5z-10)=0

    Expanding and collecting terms planes are:

    -11-10 k+(1+k) x-(-1+4 k) y+(1+5 k) z=0

    Distance of point (0,0,0) from these planes:

    d=\left|\frac{-10 k+0 (k+1)+0 (5 k+1)+0 (1-4 k)-11}{\sqrt{(k+1)^2+(5 k+1)^2+(1-4 k)^2}}\right| = 3

    Solve to find k={1, -47/139}. Hence equation of two planes with distance 3 and containing line of intersection of the two given planes are:

    -21+2 x-3 y+6 z=0

    and:

    -1059+92 x+327 y-96 z = 0

    NB.: Check the 2 planes if both are solutions
    Last edited by MaxJasper; October 19th 2012 at 10:12 PM.
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  4. #4
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    Re: Vector Geometry

    Thank you very much! I did not know that equation for intersection of two planes
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