1. ## 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.

Please show me why it is true. Thank you very much.

2. Originally Posted by Jenny20
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.

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

3. Thank you very much , earboth!

4. 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.

5. 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!