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September 5th, 2018, 08:43 AM  #1 
Member Joined: Jul 2018 From: Germany Posts: 34 Thanks: 1  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 mathstudent, 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. 
September 5th, 2018, 06:03 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,316 Thanks: 1230 
letting Mathematica work through the algebra I get $\lambda = \csc (\alpha ) (u \sin (\epsilon )v \cos (\epsilon ))$ 
September 5th, 2018, 06:24 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,301 Thanks: 1971  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. 
September 6th, 2018, 02:34 AM  #4  
Member Joined: Jul 2018 From: Germany Posts: 34 Thanks: 1  Quote:
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...  
September 6th, 2018, 02:47 AM  #5 
Member Joined: Jul 2018 From: Germany Posts: 34 Thanks: 1  
September 6th, 2018, 05:20 AM  #6 
Member Joined: Jul 2018 From: Germany Posts: 34 Thanks: 1 
@romsek How exactly did you solve for lambda? I don't get any solution, when I type in the following: 
September 6th, 2018, 06:52 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,301 Thanks: 1971  
September 6th, 2018, 06:57 AM  #8  
Senior Member Joined: Sep 2015 From: USA Posts: 2,316 Thanks: 1230  Quote:
 
September 6th, 2018, 07:09 AM  #9 
Member Joined: Jul 2018 From: Germany Posts: 34 Thanks: 1  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. 
September 6th, 2018, 07:26 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 20,301 Thanks: 1971 
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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intersection, lines, parameters, point, simplifying 
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