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Math Help - Help with Lines/Planes/Vectors

  1. #1
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    Help with Lines/Planes/Vectors

    I'm having trouble with these following problems. Appreciate all the help,

    1.) Write an equation of the indicated plane through the origin and parallel to the plane with equation 3x + 4y = z + 10


    2.) Write an equation of the plane that contains both the point P and the line L: P (2,4,6); L: x = 7 - 3t , y = 3 + 4t, z = 5 + 2t

    3.) Determine whether the line L and the plane P intersect or are parallel. If they interest, find the point of intersection.

    L: x = 7 - 4t, y = 3 + 6t, z = 9 + 5t
    P: 4x + y + 2z = 17
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  2. #2
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    Quote Originally Posted by messianic View Post
    I'm having trouble with these following problems. Appreciate all the help,

    1.) Write an equation of the indicated plane through the origin and parallel to the plane with equation 3x + 4y = z + 10
    3x + 4y -z = 10 is the equation for a plane with normal vecor n = 3,4,-1.

    A parallel plane P(x,y,z) must have the same normal vector (any scalar multiple but 0). If it is supposed to go through the origin, then P(0,0,0)=0.
    If the equation for a plane is Ax +By +Cz + D = 0, then it is easy to find D so it passes through the origin.

    Quote Originally Posted by messianic View Post
    2.) Write an equation of the plane that contains both the point P and the line L: P (2,4,6); L: x = 7 - 3t , y = 3 + 4t, z = 5 + 2t
    A vector parallel to the line L is v1=(dx/dt, dy/dt, dz,dt). You have a another point P, which is not located on the line L,
    so you can find a vector between any point on the line L and P, lets call it v2.

    Now you have 2 vectors, v1 and v2, they lie in the plane containing P and L, so if you can find a vector perpendicular to both of them,
    then you have a normal vector for the plane. Then it is easy to find D by substituting e.g. P in the resulting equation.

    Quote Originally Posted by messianic View Post
    3.) Determine whether the line L and the plane P intersect or are parallel. If they interest, find the point of intersection.

    L: x = 7 - 4t, y = 3 + 6t, z = 9 + 5t
    P: 4x + y + 2z = 17
    If they intersect then there is a point K =(7-4b, 3+6b, 9+5b), that satisfies the equation for the plane P(x,y,z) = 17 = P(K).

    Hope that helps
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  3. #3
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    for the 2nd question, can you explain how you found the 2nd vector?
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  4. #4
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    Sorry for not replying sooner, I missed the notification.

    for the 2nd question, can you explain how you found the 2nd vector?
    Like I said:
    A vector parallel to the line L is v1=(dx/dt, dy/dt, dz,dt). You have a another point P, which is not located on the line L,
    so you can find a vector between any point on the line L and P, lets call it v2.
    You could take the vector from L, when t=0, that is: P2 = ( 7, 3, 5 ), and find the vector:
    v2 = P2 - P = (7-2 , 3-4, 5-6) = (5, -1, -1).

    But that doesn't matter, you can take v2 to be vector from P to any point P2 on L. They all lie in the same plane, so it does not matter which vector you choose.
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  5. #5
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    Quote Originally Posted by messianic View Post
    3.) Determine whether the line L and the plane P intersect or are parallel. If they interest, find the point of intersection.
    L: x = 7 - 4t, y = 3 + 6t, z = 9 + 5t
    P: 4x + y + 2z = 17
    Note that \left\langle {4,1,2} \right\rangle  \cdot \left\langle { - 4,6,5} \right\rangle  = 0. What does that tell you?
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