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Math Help - angle between lines

  1. #1
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    angle between lines

    given the two lines calculate the angle that is formed by the intersection of the pair of lines

    L1:x=-1+3t,y=1+4t,z=-2t
    L2:x=-1+2s,y=3s,z=-7+s

    i have tried everything even converting these to symmetric equation and figuring out the Cartesian coordiantes for both lines which are for L1(-1,1,0) and for L2(-1,0,7)

    then of course taking the dot product and dividing by the magnitude of both and the negative cosine to get the angle
    Last edited by william; March 10th 2010 at 12:46 PM.
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  2. #2
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    Quote Originally Posted by william View Post
    given the two lines
    L1:x=-1+3t,y=1+4t,z=-2t
    L2:x=-1+2s,y=3s,z=-7+s
    Find the angle between <3,4,-2>~\&~<2,3,1>
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  3. #3
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    Quote Originally Posted by Plato View Post
    These two are skew lines.
    So what do you mean by "an angle between them"?
    what are skew lines?

    the question states calculate the angle that is formed by the intersection of each pair of lines

    pardon me for not clearly stating the question, my mistake
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  4. #4
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    Quote Originally Posted by william View Post
    what are skew lines?

    the question states calculate the angle that is formed by the intersection of each pair of lines

    pardon me for not clearly stating the question, my mistake
    Sorry I made a mistake. They intersect.
    I corrected the reply.
    BTW: I made the mistake because your equations are hard to read.
    Why don’t you bother to learn basic LaTeX??
    http://www.mathhelpforum.com/math-he...-tutorial.html
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  5. #5
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    Quote Originally Posted by Plato View Post
    Find the angle between <3,4,-2>~\&~<2,3,1>
    i get \frac{16}{\sqrt 29 \sqrt 14}=\frac{16}{20.14}=0.79=37.4^o after taking the dot product and dividing by both lines magnitudes

    also i have learned basic LaTex, i apologize for my laziness

    also, thanks for the help

    also, can you verify my work?
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  6. #6
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    \arccos \left( {\frac{{16}}<br />
{{\sqrt {29} \sqrt {14} }}} \right) = 0.653 = 37.433^ \circ  <br />
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  7. #7
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    Quote Originally Posted by Plato View Post
    \arccos \left( {\frac{{16}}<br />
{{\sqrt {29} \sqrt {14} }}} \right) = 0.653 = 37.433^ \circ  <br />
    thanks a lot for your help, perhaps you can have a look at my very similar problem that would be must appreciated

    http://www.mathhelpforum.com/math-he...een-lines.html
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  8. #8
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    Hello, william!

    Given the two lines, calculate the angle formed by the intersection of the lines.

    . . L_1\!:\;\;\begin{Bmatrix}x&=&-1+3t \\ y&=&1+4t \\ z&=& -2t\end{Bmatrix}\qquad L_2\!:\;\begin{Bmatrix}x&=&-1+2s \\ y&=&3s \\z&=&-7+s \end{Bmatrix}
    \text{Formula: }\;\cos\theta \;=\;\frac{\vec u \cdot \vec v}{|\vec u|\,|\vec v|}


    We have: . \begin{array}{ccc}\overrightarrow{L_1} &=& \langle 3,4,\text{-}2\rangle \\ \\[-4mm] \overrightarrow{L_2} &=& \langle 2,3,1\rangle \end{array}

    Hence: . \cos\theta \;=\;\frac{\langle3,4,-2\rangle \cdot\langle 2,3,1\rangle}{\sqrt{9+16+4}\,\sqrt{4+9+1}} \;=\;\frac{6+12-2}{\sqrt{29}\,\sqrt{14}} \;=\;\frac{16}{\sqrt{406}} \;=\;0.794066667


    Therefore: . \theta \;=\;37.43281131^o \;\approx\;37.4^o

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