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    Let the Line L1 be defined as -

    x = 1+t
    y = 2-t
    z = 1+2t

    Let the Line L2 be defined as -

    x = 3+t
    y = 4-t
    z = 1-2t


    Find the equations of two parallel planes A and B such that
    L1 ⊆ A and L2 ⊆ B


    How do I go about this? Help Please
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    1. Do you see how a point on L1 is (1,2,1) and a point on L2 is (3,4,1)?
    2. Do you see that if a plane is parallel to two lines, then it has to be able to extend towards BOTH lines' directional vectors?
    3. Do you know that the vector equation of a plane is a+d1+d2 (position + direction1 + direction2)?
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    (Original post by aznkid66)
    1. Do you see how a point on L1 is (1,2,1) and a point on L2 is (3,4,1)?
    2. Do you see that if a plane is parallel to two lines, then it has to be able to extend towards BOTH lines' directional vectors?
    3. Do you know that the vector equation of a plane is a+d1+d2 (position + direction1 + direction2)?
    Taking the cross product would give me a normal to the required plane I suppose.

    After that, I plug in the points (1,2,1) and (3,4,1) to get two different constants c in the plane equation x+y+z= c

    Does that sound about right?
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    Yup, that would also work. I'm guessing you need to find the equation of the plane in cartesian form? I was suggesting a different form in my post ^^;
 
 
 
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