Equations of lines in 3d space - validity of a shortcut for determining if they describe the same line or are parallel

Question

My math teacher said that I could not use a method I tried to get the right answer because it was invalid and I got the answer purely by chance. I used this method on four different equations and got the right answer every time. I'm not sure if this was a fluke or not so I'm here to ask.

Essentially, we were given two lines, in this example they were 2d, but this works with 3d vectors as well as far as I know:

$(x,y) = (4,2) + t(-3,1)$

$x=4-3t$

$y=2+t$

and

$(x,y) = (10,-3) + s(6,-2)$

$x = 10 + 6s$

$y= -3-2s$

since (-3,1) & (6,-2) are multiples of each other, I knew they were either parallel or coincident lines.

I equated the equations, and made s=t:

$10 + 6t = 4-3t$

$-3-2t = 2+t$

I found that the values of t did not match, and concluded that as a result, they were not coincident but parallel. I used this method on other equations to determine they were coincident as well. Is this a legitimate way to solve the problem or similar problems?

Answer 1

Before I comment on your method, the example lines in your question are

$(x,y) = (4,2) + t(-3,1)$
$(x,y) = (10,-3) + s(6,-2)$

Checking direction vector, you know they are either parallel or the same line. If they are the same line, $(10, - 3)$ will also be on first line, right?

So why not simply do, $10 = 4 - 3 t \ $ and $-3 = 2 + t$. As they are not consistent, you know lines are not same, just parallel.

Now coming to your method, what you are saying is correct but the mistake that you are making is to replace $s$ with $t$. At the least, it is not a standard approach. Take an example,

$l1: (1,1) + r (1,2)$
$l2: (3,5) + s (1,2)$
$l3: (3,4) + t (1,2)$

Say you need to find which line of $l2$ and $l3$ is parallel to $l1$ and which is the same line as $l1$.

Now for checking $l1$ and $l2$, try changing $s$ to $r$ or vice versa. You cannot find value of $r$ and $s$, can you?

So if you are going this path, the right way will be to equate $x$ coordinate of points on both lines. In that case, $y$ coordinates should also be equal if they are the same line. Let's do it for $l1$ and $l2$.

Equating $x$, $1 + r = 3 + s \implies r - s = 2$

Equating $y$, $1 + 2r = 5 + 2s \implies r - s = 2$

Can you see why they are the same line? Or we could have taken value of $r$ from the first which is $(2 + s)$ and plugged into second and that gives us $(5 + 2s)$ which matches with $y$ component of $l2$.

So $l1$ and $l2$ are the same line.

You can now check $l1$ and $l3$ yourself. You can simply use what I showed in the beginning.

Answer 2

Your trick does not work.

If the direction vectors are proportional, the directions are indeed parallel. But equating the parameters does not make sense. Because you can re-parameterize any line (for instance replace $t$ by $at+b$) without changing that line.

What you can do is to pick an arbitrary point on a line (say $t=0$) and see if if can match the equation of the second line (you have a system of $2$ - or $3$ - equations in $1$ unknown).