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Math Help - parametric equation

  1. #1
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    Question parametric equation

    Question
    If Lsub1 has parametric equation x=1+3t, y=1+t, z=4-t, and Lsub2 has parametric equation x=7-6t, y=-2t, z=3+2t, then Lsub1 and Lsub2 are parallel.

    The answer is true.

    Please show me why it is true. Thank you very much.
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  2. #2
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    Quote Originally Posted by Jenny20 View Post
    Question
    If Lsub1 has parametric equation x=1+3t, y=1+t, z=4-t, and Lsub2 has parametric equation x=7-6t, y=-2t, z=3+2t, then Lsub1 and Lsub2 are parallel.

    The answer is true.
    Please show me why it is true. Thank you very much.
    Hello Jenny,

    your equations describe two straight lines in \mathbb{R}^3

    L_1: [x,y,z]=\underbrace{[1,1,4]}_{\text{fixed point}}+t\cdot \underbrace{[3,1,-1]}_{\text{direction}}

    L_2: [x,y,z]=\underbrace{[7,0,3]}_{\text{fixed point}}+t\cdot \underbrace{[-6, -2, 2]}_{\text{direction}}

    By comparison you can see, that

    [-6, -2, 2]=(-2) \cdot [3,1,-1]. That means the direction vectors are collinear: They have the same direction but different length.
    Therefore L_1 and L_2 are at least parallel. To proof if they are actually the same you have to show that the fixed point of L_1 belongs to L_2 too.

    EB
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  3. #3
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    Thank you very much , earboth!
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  4. #4
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    Hello, Jenny!

    If L_1 has parametric equations: . \begin{Bmatrix}x\:= & 1+3t\\  y\:= & 1+t \\ z\:= & 4-t\end{Bmatrix}

    and L_2 has parametric equations: . \begin{Bmatrix} x\:= & 7-6t \\ y\:= & -2t \\ z\:= & 3+2t\end{Bmatrix}

    then L_1 and L_2 are parallel.

    Two lines are parallel if their direction vectors are parallel.
    . . (They do not have to be collinear.)

    L_1 has direction vector: \vec{u}\:=\:\langle 3,1,\text{-}1\rangle
    L_2 has direction vector: \vec{v}\:=\:\langle\text{-}6,\text{-}2,2\rangle \:=\:\text{-}2\langle3,1,\text{-}1\rangle

    Since \vec{v} = -2\vec{u}\!:\;\;\vec{u} \parallel \vec{v} . . . . Q.E.D.

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  5. #5
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    Two lines are parallel if their direction vectors are parallel.
    . . (They do not have to be collinear.)

    L_1 has direction vector: \vec{u}\:=\:\langle 3,1,\text{-}1\rangle
    L_2 has direction vector: \vec{v}\:=\:\langle\text{-}6,\text{-}2,2\rangle \:=\:\text{-}2\langle3,1,\text{-}1\rangle

    Since \vec{v} = -2\vec{u}\!:\;\;\vec{u} \parallel \vec{v} . . . . Q.E.D.


    Hi Soroban,
    Thank you very much!
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