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 $\displaystyle \mathbb{R}^3$
$\displaystyle L_1: [x,y,z]=\underbrace{[1,1,4]}_{\text{fixed point}}+t\cdot \underbrace{[3,1,-1]}_{\text{direction}}$
$\displaystyle L_2: [x,y,z]=\underbrace{[7,0,3]}_{\text{fixed point}}+t\cdot \underbrace{[-6, -2, 2]}_{\text{direction}}$
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
Hello, Jenny!
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.
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!