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Equation of a plane that passes thorugh the line of intersection of [answer check]

  1. Sep 1, 2008 #1
    Find the equation of the plane that passes through the line of intersection of the planes 4x - 3y - z - 1 = 0 and 2x + 4y + z - 5 = 0 and passes through A(1, -3, 2).

    I have answered the question but I am not sure it is correct; if someone can review it their comments would be appreciated!
    Thank you!


    The plane that passes through the line of intersection of the two planes has the equation

    4x – 3y – z – 1 + k(2x +4y + z – 5) = 0

    Substitute the point A(1, -3, 2) into the equation to find k

    4(1) – 3(-3) – 2 – 1 + k (2(1) + 4(-3) + 2 – 5) = 0
    10 + k(-13) = 0
    k = 10 / 13

    substitute the value of k into the equation of the plane

    4x – 3y – z – 1 + (10/13)(2x +4y + z – 5) = 0
    4x – 3y – z – 1 + (20/13)x + (40/13)y + (10/13)z – (50/13) = 0

    Therefore the equation of the plane that passes through the line of intersection of the two planes is

    (72/13)x + (1/13)y – (3/13)z – (63/13) = 0
     
  2. jcsd
  3. Sep 1, 2008 #2

    Defennder

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    Re: Equation of a plane that passes thorugh the line of intersection of...[answer che

    Yup it looks ok.
     
  4. Sep 2, 2008 #3
    Re: Equation of a plane that passes thorugh the line of intersection of...[answer che

    Great! :D Thank you!
     
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