Angle between two straight lines

  • #1
arpon
236
16

Homework Statement


Find the angle between the straight lines:
##(x^2+y^2)(\cos^2{\theta} \sin^2{\alpha} + \sin^2{\theta})=(x \tan{\theta} - y \sin{\alpha})^2##

Homework Equations


[Not applicable]

The Attempt at a Solution


Dividing by ##x^2##,
## (1+(\frac{y}{x})^2)(\cos^2{\theta} \sin^2{\alpha} + \sin^2{\theta})=(\tan{\theta} - \frac{y}{x} \sin{\alpha})^2 ##
Let, ##\frac{y}{x} =m ##.
## (1+m^2)(\cos^2{\theta} \sin^2{\alpha} + \sin^2{\theta})=(\tan{\theta} - m \sin{\alpha})^2 ##
##(\sin^2{\theta} \cos^2{\alpha}) m^2 + (2 \tan{\theta} \sin{\alpha}) m + (\cos^2{\theta} \sin^2{\alpha} + \sin^2{\theta}-\tan^2{\theta}) = 0 ##
So the solutions of this equation indicate the slopes of the two straight lines. If the solutions are ##m_1## and ##m_2##, then the angle between the two straight lines will be ##\arctan{\frac{m_1-m_2}{1+m_1 m_2}}##;
I came up with a messy equation, as I tried to calculate this. But the answer is very simple, just ##2\theta##. So, I think there is some clever technique to solve this problem.
Any suggestion will be appreciated
 

Answers and Replies

  • #2
36,335
8,294

Homework Statement


Find the angle between the straight lines:
##(x^2+y^2)(\cos^2{\theta} \sin^2{\alpha} + \sin^2{\theta})=(x \tan{\theta} - y \sin{\alpha})^2##

Homework Equations


[Not applicable]

The Attempt at a Solution


Dividing by ##x^2##,
## (1+(\frac{y}{x})^2)(\cos^2{\theta} \sin^2{\alpha} + \sin^2{\theta})=(\tan{\theta} - \frac{y}{x} \sin{\alpha})^2 ##
Let, ##\frac{y}{x} =m ##.
## (1+m^2)(\cos^2{\theta} \sin^2{\alpha} + \sin^2{\theta})=(\tan{\theta} - m \sin{\alpha})^2 ##
##(\sin^2{\theta} \cos^2{\alpha}) m^2 + (2 \tan{\theta} \sin{\alpha}) m + (\cos^2{\theta} \sin^2{\alpha} + \sin^2{\theta}-\tan^2{\theta}) = 0 ##
So the solutions of this equation indicate the slopes of the two straight lines. If the solutions are ##m_1## and ##m_2##, then the angle between the two straight lines will be ##\arctan{\frac{m_1-m_2}{1+m_1 m_2}}##;
I came up with a messy equation, as I tried to calculate this. But the answer is very simple, just ##2\theta##. So, I think there is some clever technique to solve this problem.
Any suggestion will be appreciated
Instead of dividing by x^2, I think you might be better off by replacing x^2 + y^2 by r^2, and replacing x and y by r*cos(θ) and r*sin(θ), respectively. This would get the equation completely into polar form, after which it might be easier to simplify.
 
  • #3
arpon
236
16
Instead of dividing by x^2, I think you might be better off by replacing x^2 + y^2 by r^2, and replacing x and y by r*cos(θ) and r*sin(θ), respectively. This would get the equation completely into polar form, after which it might be easier to simplify.
Thanks for your reply. This also gives me same kind of equation.
 
  • #4
36,335
8,294
Thanks for your reply. This also gives me same kind of equation.
Doing what I suggested, I get a somewhat simpler equation in polar form.

Is there more to this problem than you have posted? Is there an explanation of what ##\theta## and ##\alpha## represent?

Also, is the equation you wrote exactly the same as given in the problem?
 
  • #5
arpon
236
16
Doing what I suggested, I get a somewhat simpler equation in polar form.

Is there more to this problem than you have posted? Is there an explanation of what ##\theta## and ##\alpha## represent?

Also, is the equation you wrote exactly the same as given in the problem?
No, there is no explanation about ##\theta## & ##\alpha##. And the equation I wrote is exactly the same as in the problem.
 
  • #6
36,335
8,294
Can you post an image of the problem? Your solution and my solution both seem to come out very messy, and I'm wondering if there is something we're not seeing.
 

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