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September 5th, 2018, 08:43 AM   #1
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Simplifying intersection point of lines with parameters

I'm kind of lost on simplifying this formula:

$\displaystyle
\frac{u+\lambda*\cos(\epsilon + \alpha)}{\cos(\epsilon)}-

\frac{v+\lambda*\sin(\epsilon + \alpha)}{\sin(\epsilon)}=

0
$

I'm no math-student, just trying to play around with interference patterns, anyways...

I try to figure out what $\displaystyle \lambda$ is and feel like I lack a certain trick/knowledge to simplify this formula, I don't want to end up with a horribly overcomplicated thing. I tried to squish everything into one fraction having $\displaystyle \sin(\epsilon)*\cos(\epsilon)$ as my denominator, and tried to factor out $\displaystyle \lambda$, I think it doesn't make sense to write all of my tryouts down, as they just seem to make the problem more complicated.

The equation above is the result of me trying to intersect these two lines:

$\displaystyle
l_0=\lambda*\binom{\cos(\epsilon)}{\sin(\epsilon)}
$

$\displaystyle
t_0=\binom{u}{v}+\lambda*\binom{\cos(\epsilon + \alpha)}{\sin(\epsilon+ \alpha)}
$

I'm also new to $\LaTeX$, if there is anything wrong with my notation please let me know as well.

I need to simplify four of these kinds of formulas, so if someone could give me a hint or explain how to do so, that would be greatly appreciated.

Last edited by greg1313; September 5th, 2018 at 07:17 PM.
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September 5th, 2018, 06:03 PM   #2
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letting Mathematica work through the algebra I get

$\lambda = \csc (\alpha ) (u \sin (\epsilon )-v \cos (\epsilon ))$
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September 5th, 2018, 06:24 PM   #3
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Quote:
Originally Posted by theHeavyLobster View Post
. . . these two lines
It would seem that $\lambda$ is just a parameter in the equations for the lines, so its value for the first line needn't equal its value for the second line, even for their point of intersection.

When using LaTeX, you should put a "\" immediately before a standard function name such as cos or sin. Also, insert a space occasionally, else a minor bug may prevent correct interpretation.
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September 6th, 2018, 02:34 AM   #4
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Quote:
Originally Posted by skipjack View Post
It would seem that $\lambda$ is just a parameter in the equations for the lines, so its value for the first line needn't equal its value for the second line, even for their point of intersection.
Both lines do have different multiplicators, I messed that up in my original post, for one line I use lambda and for the other one I use µ. In my understanding they are somewhat special parameters, as they are the multiplicators of the directional vectors on those lines. If I set lambda to a number, the formula will give me a point at the directional vector times lambda.

Same goes for the second line with µ, so yeah, then lambda and µ can definitely be different.

Here's the correct way of describing these lines, because I messed up before (sorry):

$\displaystyle

l_0 = \binom{0}{0} + \lambda * \binom{ \cos(\epsilon) }{ \sin(\epsilon) }

$

$\displaystyle

t_0 = \binom{u}{v} + \mu * \binom{ \cos(\epsilon + \alpha) }{ \sin(\epsilon + \alpha) }

$

then for the intersection:

$\displaystyle
Top: \lambda( \cos(\epsilon)) = u + \mu(\cos(\epsilon + \alpha))
$
$\displaystyle
Btm: \lambda( \sin(\epsilon)) = v + \mu(\sin(\epsilon + \alpha))
$

$\displaystyle
\lambda = \frac{u + \mu( \cos( \epsilon + \alpha ))}{\cos(\epsilon)}
$

$\displaystyle
\lambda = \frac{v + \mu( \sin( \epsilon + \alpha ))}{\sin(\epsilon)}
$

Here I need to find $\displaystyle \mu$! as I subtract one from the other.

In my original post I switched mu for lambda, for unknown reasons...
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September 6th, 2018, 02:47 AM   #5
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Quote:
Originally Posted by romsek View Post
letting Mathematica work through the algebra I get

$\lambda = \csc (\alpha ) (u \sin (\epsilon )-v \cos (\epsilon ))$
...and that worked like a charm!



Is it Mathematica from Wolfram? I'll definitely look into it.
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September 6th, 2018, 05:20 AM   #6
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@romsek

How exactly did you solve for lambda?

I don't get any solution, when I type in the following:

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September 6th, 2018, 06:52 AM   #7
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Quote:
Originally Posted by theHeavyLobster View Post
Here I need to find $\displaystyle \mu$ as I subtract one from the other.
You get a linear equation for $\mu$ that's easy to solve.
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September 6th, 2018, 06:57 AM   #8
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Quote:
Originally Posted by theHeavyLobster View Post
@romsek

How exactly did you solve for lambda?

I don't get any solution, when I type in the following:

I used the actual Mathematica, not the WolframAlpha web site
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September 6th, 2018, 07:09 AM   #9
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Quote:
Originally Posted by skipjack View Post
You get a linear equation for $\mu$ that's easy to solve.
Maybe. I do have problems with it though. I can get that equation for $\mu$, but my formula sadly was way too bloated or in other words, not simplified. As I need to work with it further, I didn't want to have a monstrous equation for a simple intersection between two lines.

For me it was not easy to reduce it. Well, I try to learn though, I'll give it another try knowing what it should look like, to understand the intersection a bit better.
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September 6th, 2018, 07:26 AM   #10
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One gets $\displaystyle \mu = \frac{u \sin(\epsilon) - v \cos(\epsilon)}{\cos(\epsilon)\sin(\epsilon + \alpha) - \cos(\epsilon + \alpha)\sin(\epsilon)} = \frac{u \sin(\epsilon) - v \cos(\epsilon)}{\sin((\epsilon + \alpha) - \epsilon)} = \frac{u \sin(\epsilon) - v \cos(\epsilon)}{\sin(\alpha)}$.
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