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Math Help - angle between straight lines

  1. #1
    Member
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    Wink angle between straight lines

    Hi folks,

    I am trying to show that the angle between two straight lines is given by:

     \tan\theta = \frac{m_{1} - m_{2}}{1 +  m_{1}m_{2}}

    I am using vector methods but my problem is the algebra. If the vector equation of a line is given by \overline{r_{1}} = \overline{a_{1}} + \lambda \overline{b_{1}}. And the second line by subscript 2, the angle between the lines is given by:

    \overline{b_{1}}.\overline{b_{2}} = |b_{1}||b_{2}|  \cos\theta

    for a straight line of the form y = mx + c we have:

    r_{1}  = \left( \begin{array}{c} x \\ y \end{array} \right) = <br />
\left(  \begin{array}{c} 0 \\ c_{1}  \end{array} \right) + \lambda <br />
\left(  \begin{array}{c} 1 \\ m_{1}  \end{array} \right)

    or,
    r_{1}  = 0i + c_{1}j + \lambda (i + m_{1}j)
    r_{2} = 0i + c_{2}j  + \lambda (i + m_{2}j)

    So, the angle between the two lines is given by

    (1i + m_{1}j).(1i + m_{2}j) = \sqrt{1 +  m^{2}_{1}}\sqrt{1 + m^{2}_{2}}. \cos\theta

    1 +  m_{1}m_{2} = \sqrt{(1 + m^{2}_{1})(1 + m^{2}_{2})}. \cos\theta

    At this stage I decided to square both sides and look for a chance to combine 1 and \cos^2\theta to give \sin^2  \theta then simplify and take a square root to get back to  \tan\theta

    instead things got very messy indeed. Am I on the right track? I know it's murder trying to type lots of algebra but I suspect there would be less algebra if it was done right! A few lines in the right direction would be greatly appreciated!

    best regards
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  2. #2
    MHF Contributor

    Joined
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    Quote Originally Posted by s_ingram View Post
    Hi folks,

    I am trying to show that the angle between two straight lines is given by:

     \tan\theta = \frac{m_{1} - m_{2}}{1 +  m_{1}m_{2}}

    I am using vector methods but my problem is the algebra. If the vector equation of a line is given by \overline{r_{1}} = \overline{a_{1}} + \lambda \overline{b_{1}}. And the second line by subscript 2, the angle between the lines is given by:

    \overline{b_{1}}.\overline{b_{2}} = |b_{1}||b_{2}|  \cos\theta

    for a straight line of the form y = mx + c we have:

    r_{1}  = \left( \begin{array}{c} x \\ y \end{array} \right) = <br />
\left(  \begin{array}{c} 0 \\ c_{1}  \end{array} \right) + \lambda <br />
\left(  \begin{array}{c} 1 \\ m_{1}  \end{array} \right)

    or,
    r_{1}  = 0i + c_{1}j + \lambda (i + m_{1}j)
    r_{2} = 0i + c_{2}j  + \lambda (i + m_{2}j)

    So, the angle between the two lines is given by

    (1i + m_{1}j).(1i + m_{2}j) = \sqrt{1 +  m^{2}_{1}}\sqrt{1 + m^{2}_{2}}. \cos\theta

    1 +  m_{1}m_{2} = \sqrt{(1 + m^{2}_{1})(1 + m^{2}_{2})}. \cos\theta

    At this stage I decided to square both sides and look for a chance to combine 1 and \cos^2\theta to give \sin^2  \theta then simplify and take a square root to get back to  \tan\theta

    instead things got very messy indeed. Am I on the right track? I know it's murder trying to type lots of algebra but I suspect there would be less algebra if it was done right! A few lines in the right direction would be greatly appreciated!

    best regards
    What you say you are doing is correct. Since

    cos(\theta)= \frac{1+ m_1m_2}{\sqrt{1+ m_1^2}\sqrt{1+ m_2^2}}

    cos^2(\theta)= \frac{1+ 2m_1m_2+ m_1^2+ m_2^2}{1+ m_1^2+ m_2^2+ m_1m_2^2}

    so
    sin^2(\theta)= 1- cos^2(\theta)= \frac{1+ m_1^2+ m_1^2+ m_1^2m_2^2- (1+ 2m_1m_2+ m_1^2m_2^2)}{(1+ m_1^2)(1+ m_2^2)}

    and the numerator reduces to m_1^2- 2m_1m_2+ m_2^2, a perfect square!

    By the way, if you are nor required to use vectors, you might find it easier to use the trig identity tan(\theta+ \phi)= \frac{tan(\theta)+ tan(\phi)}{1+ tan(\theta)tan(\phi)}.
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  3. #3
    Member
    Joined
    May 2009
    Posts
    91
    Thanks again, HallsofIvy!

    I tried it your way and it works fine. The problem I am having is that I don't always go the right way, particularly with algebra!

    This is what I did and as you can see it rapidly gets out of hand even though I am not making any mistakes!

    (1 + m_{1}m_{2})^2 = (1 + m^2_{1} + m^2_{2} + m^2_{1}m^2_{2})\cos^2\theta

    1 + 2m_{1}m_{2} + m^2_{1}m^2_{2} = \cos^2\theta + (m^2_{1} + m^2_{2}) \cos^2\theta + m^2_{1}m^2_{2}cos^2\theta

    1 - \cos^2\theta = -m^2_{1}m^2_{2}(1 - \cos^2\theta) -2m_{1}m_{2} + (m^2_{1} + m^2_{2})\cos^2\theta

    sin^2\theta = -m^2_{1}m^2_{2}\sin^2\theta -2m_{1}m_{2} + (m^2_{1} + m^2_{2})\cos^2\theta

    \sin^2\theta(1 + m^2_{1}m^2_{2}) = -2m_{1}m_{2} + \cos^2\theta(m^2_{1} + m^2_{2})

    and as you can see I have this hanging 2m_{1}m_{2} term and no clear way to get to the required solution. Have I made a mistake and if not can you see how to get my final line into the answer we know and love?
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