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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- Dec 6th 2006, 09:20 PMJenny20parametric 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. - Dec 6th 2006, 10:19 PMearboth
Hello Jenny,

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

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

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

By comparison you can see, that

$\displaystyle [-6, -2, 2]=(-2) \cdot [3,1,-1]$. That means the direction vectors are collinear: They have the same direction but different length.

Therefore $\displaystyle L_1$ and $\displaystyle L_2$ are at least parallel. To proof if they are actually the same you have to show that the fixed point of $\displaystyle L_1$ belongs to $\displaystyle L_2$ too.

EB - Dec 6th 2006, 11:43 PMJenny20
Thank you very much , earboth! :)

- Dec 7th 2006, 04:54 AMSoroban
Hello, Jenny!

Quote:

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

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

then $\displaystyle L_1$ and $\displaystyle L_2$ are parallel.

Two lines are parallel if their direction vectors are parallel.

. . (They do not have to be collinear.)

$\displaystyle L_1$ has direction vector: $\displaystyle \vec{u}\:=\:\langle 3,1,\text{-}1\rangle$

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

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

- Dec 7th 2006, 08:51 AMJenny20
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! :)