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How can we easily see that these two lines are not parallel? I normally compare the direction numbers to see if they are not a multiple of each other, but here I can not directly see the direction numbers of line F.

$D: \frac{x+1}{4}=3-y=\frac{1-z}{3}$

$F:\left\{\begin{array}{l}x-2 y+2 z=3 \\ 3 y-z=7\end{array}\right.$

Ricardi
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    Line $F$ is the intersection of two planes, so one way to get a vector parallel to its direction is to take the cross product of the normal vectors of the two planes. – Jakob Streipel Sep 12 '22 at 14:51
  • You may eliminate the $z$ in $x-2y+2z=3$, and get one equation linking $x$ and $y$, and another linking $y$ and $z$. Then $F$ may be manipulated to have a form like $D$. – peterwhy Sep 12 '22 at 15:04

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$M=(x,y,z)\in D \iff \begin{cases}x+1=4(3-y) \\3(x+1)=4(1-z) \end{cases} \iff \begin{cases}x=11-4y \\z=-8+3y \end{cases} \iff M\in (11,0,-8)+\mathbb R(-4,1,3)$

$M=(x,y,z)\in F \iff \begin{cases} x-2y+2z=3\\3y-z=7 \end{cases} \iff \begin{cases} x=17-4y\\z=-7+3y\end{cases} \iff M\in (17,0,-7)+\mathbb R(-4,1,3)$

$D$ and $F$ are parallel : they have the same direction, $\mathbb R(-4,1,3)$.

Stéphane Jaouen
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