1
$\begingroup$

Show that two lines are parallel if and only if for two distinct points $v_1$ and $v_2$ on the first line, and two distinct points $w_1$ and $w_2$ on the second line, the difference $w_1 - w_2$ is a multiple of $v_1 - v_2$.

I am wondering if my answer is rigorous enough and my thinking didn't skip any steps. I think it feels solid, but I also feel that there might be some circular reasoning going on too.


Let $u_1$ and $u_2$ be vectors defined as $u_1 = v_1 - v_2$ and $u_w = w_1 - w_2$. This makes $u_1$ and $u_2$ be vectors from $v_1$ to $v_2$ and $w_1$ to $w_2$ respectively. We can rewrite $u_1$ and $u_2$ so that these vectors are represented parametrically $$\begin{align*} u_1 &= tu_1 - v_2 \\ u_2 &= tu_2 - w_2 \end{align*}$$

To show that these two lines are parallel, then $u_1$ and $u_w$ must be linear combinations such that there are solutions for $$\begin{align*} u_1 &= tu_2 - v_2 \\ u_2 &= tu_1 - w_2 \end{align*}$$

$\endgroup$
1
  • $\begingroup$ What did you really mean by $u_1=tu_1-v_2$? The only way this can be true is if $v_2$ is a multiple of $u_1$, which is clearly not true in general. $\endgroup$
    – amd
    Sep 1, 2019 at 23:07

2 Answers 2

2
$\begingroup$

I am sure that you will be allowed to assume that two non-zero vectors are parallel if and only if one is a multiple of the other.

You have correctly obtained a vector $v_1-v_2$ which is parallel to one line and $w_1-w_2$ which is parallel to the other line. You can therefore conclude that the lines are parallel if and only if these vectors are parallel i.e. if and only if one is a multiple of the other.

$\endgroup$
0
$\begingroup$

I would do this as follows:

Let $$y=a+bx ~~~~(1)$$$$w=c+dv~~~~(2)$$ be linear mappings that represent the two lines, where $a, b, c, d \in \mathbb{R}$. Let $v_1={v_{11} \choose v_{12}}$ and $v_2={v_{21} \choose v_{22}}$ which both fulfil $y=a+bx$. As $v_1$ and $v_2$ are distinct vectors it follows that their coordinates must be pairwise different (otherwise you get a contradiction). The same holds for $w_1={w_{11} \choose w_{12}}$ and $w_2={w_{21} \choose w_{22}}$ regarding $w=c+dv$.

"$\Rightarrow$": We assume both lines to be parallel.

We can conclude from (1) and (2) that $$v_1 -v_2={ v_{11}-v_{21} \choose b(v_{11}-v_{21})}=(v_{11}-v_{21}){ 1 \choose b}~~~~~~~~~~~~~~~(3)$$ $$w_1 -w_2={ w_{11}-w_{21} \choose b(w_{11}-w_{21})}=(w_{11}-w_{21}){ 1 \choose d}.~~~~~~~~(4)$$ As both lines are parallel it holds $b=d$. Combining $(3)$ and $(4)$ and rearranging yields: $(v_1 - v_2) = \frac{v_{11}-v_{21}}{w_{11}-w_{21}}(w_1 - w_2)$. Hence, $(v_1 - v_2)$ and $(w_1 - w_2)$ are linearly dependent.

"$\Leftarrow$": We assume $(v_1 - v_2) = z\cdot(w_1 - w_2)$, with $z \neq 0$.

Setting $z=\frac{v_{11}-v_{21}}{w_{11}-w_{21}}$ and using $(3)$ and $(4)$ yields: ${ 1 \choose b}= { 1 \choose d}$ which shows that $b=d$. Hence, both lines must be parallel. q.e.d.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.