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Math Help - Planar Intersections-Please help

  1. #1
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    Planar Intersections-Please help

    Find the equation of the plane that passes through the line of intersection of the planes x - 3y - 2z - 1 = 0 and 2x + 4y + z - 5 = 0 and parallel to the x-axis.


    Find the line of intersection for the two planes:
    1: 0t 3𝑦 − 2𝑧 − 1 = 0
    2: 2𝑥 + 4𝑦 + 𝑧 − 5 = 0
    let 𝑥=𝑡 (a scalar)
    0𝑡+3𝑦 − 2𝑧 − 1 = 0(1)
    2𝑡 + 4𝑦 + 𝑧 − 5 = 0..(2)
    use substitution to find the value of y from equation (1):
    0𝑡+3𝑦 − 2𝑧 − 1 = 0..(1)
    3𝑦= 2𝑧+ 1
    𝑦=2𝑧+13(3)
    eliminate the variable y from the equations:
    First Multiply equation (1) by 4, and multiply equation (2) by -3:
    0𝑡+12𝑦 −8𝑧 − 4 = 0
    −6𝑡−12𝑦−3𝑧+15 = 0
    Use the elimination method:
    0𝑡+12𝑦 −8𝑧 − 4 = 0
    −6𝑡−12𝑦−3𝑧+15 = 0
    −6𝑡−11𝑧+11=0
    −6𝑡−11𝑧=−11(4)
    −11𝑧=−11+6𝑡
    𝑧=−6𝑡/11+1
    y:
    Substitute the value of z into equation (3):
    𝑦=2𝑧+13
    𝑦=2(−6𝑡/11+1)
    𝑦=(−4𝑡−11)/11

    I am very confused by the decimal values...Does my working make sense? I would appreciate any help
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  2. #2
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    Nov 2009
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    Quote Originally Posted by spoc21 View Post
    Find the equation of the plane that passes through the line of intersection of the planes x - 3y - 2z - 1 = 0 and 2x + 4y + z - 5 = 0 and parallel to the x-axis.


    Find the line of intersection for the two planes:
    1: 0t 3�� − 2�� − 1 = 0
    2: 2�� + 4�� + �� − 5 = 0
    let ��=�� (a scalar)
    0��+3�� − 2�� − 1 = 0(1)
    2�� + 4�� + �� − 5 = 0..(2)
    use substitution to find the value of y from equation (1):
    0��+3�� − 2�� − 1 = 0..(1)
    3��= 2��+ 1
    ��=2��+13(3)
    eliminate the variable y from the equations:
    First Multiply equation (1) by 4, and multiply equation (2) by -3:
    0��+12�� −8�� − 4 = 0
    −6��−12��−3��+15 = 0
    Use the elimination method:
    0��+12�� −8�� − 4 = 0
    −6��−12��−3��+15 = 0
    −6��−11��+11=0
    −6��−11��=−11(4)
    −11��=−11+6��
    ��=−6��/11+1
    y:
    Substitute the value of z into equation (3):
    ��=2��+13
    ��=2(−6��/11+1)
    ��=(−4��−11)/11

    I am very confused by the decimal values...Does my working make sense? I would appreciate any help
    Any equation of a plane passing through the intersection of 2 planes is given by (x-3y-2z-1)+(lambda)(2x+4y+z-5)=0 where lambda=l be any constant.Simplifying we get x(1+2l)+y(4l-3)+z(l-2)-6l=0-------(1).But it is given that the resulting plane is parallel to the x-axis. therefore the coefficient of y in (1) must be 0. thus 4l-3=0 =>l=3/4. Putting this value in (1) we should the value of the required plane which according to my calculations is:10x-5z-18=0. Hope this was helpful!!!
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