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Math Help - Parametric Equations

  1. #1
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    Parametric Equations

    Find parametric equations of the line of intersection of the two planes
    -8 x - 2 y + 3 z = 9
    and
    5 x + 4 y + 2 z = -8

    Assign your direction vector, as a Maple list, to the name DirectionVector. Use t as the parameter along the line and assign the equations of the line to the list EquationsOfLine in the form [x=.... , y = ..... , z = ....] .




    DirectionVector :=

    EquationsOfLine :=

    How do i start this? Should i take the cross product and then use the results as the answer, or should i solve for one equation and substitute into the other?
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let z=t and solve the system formed by the two equations in respect to x and y.
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  3. #3
    MHF Contributor Calculus26's Avatar
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    you could in fact set z = 0 to find a pt on the line and as you suggested

    take the cross product of the normals -8i -2j +3k and 5i + 4j +2k to find the vector parallel to the line of intersection
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  4. #4
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    Quote Originally Posted by purplerain View Post
    Find parametric equations of the line of intersection of the two planes
    -8 x - 2 y + 3 z = 9
    and
    5 x + 4 y + 2 z = -8

    Assign your direction vector, as a Maple list, to the name DirectionVector. Use t as the parameter along the line and assign the equations of the line to the list EquationsOfLine in the form [x=.... , y = ..... , z = ....] .




    DirectionVector :=

    EquationsOfLine :=

    How do i start this? Should i take the cross product and then use the results as the answer, or should i solve for one equation and substitute into the other?
    I hate these things were you are told to just plug values into "Maple" or some other machine! What are you expected to learn from that? The equations of the planes are -8x- 2y+ 3z= 9 and 5x+ 4y+ 2z= -8. Multiply the first equation by 2 to get -16x- 4y+ 6z= 18 and add to the second to eliminate y: -11x+ 8z= 10. Then 8z= 11x+ 10 and z= (11/8)x+ 5/4. Putting that back into the first equation, -8x- 2y+ 3((11/8)x+ 9/4)= -8x- 2y+ (33/8)x+ 27/4= 9 or (-31/8)x- 2y= 9/4 so -2y= (31/8)x+ 9/4 and y= (-31/16)x- 9/8. In particular, if we take x= 16t, we have y= -31t- 9/8 and z= 22x+ 5/4.
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  5. #5
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    Ok, when you mean vector parallel to the line of intersection, do you mean unit vector? I solved for x, y, and z and got x =-10/11, y =-19/22, and z = 3. Would the equation of the line be 5 (-10/11)t+4(-19/22)t+2(3t)?
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  6. #6
    MHF Contributor Calculus26's Avatar
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    When you take the cross product of -8i -2j +3k and 5i + 4j +2k

    you obtain -16 i + 31 j - 22k

    this is the direction vector of the line since it is perpindicular to both planes it is parallel to the line of intersection --not a unit vector

    In letting z = 0 you obtain x = -10/11 and y = -19/22

    So (-10/11,-19/22,0) is a point on the line

    We have then x = -10/11 -16t

    y= -19/22 +31t

    z = -22t
    Obviously thre is no unique answer
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  7. #7
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    Thanks guys for the help, it cleared up a lot of confusion that i had.
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