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Math Help - identity

  1. #1
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    identity

    How do you prove the following: Given the direction cosines of two lines  \cos \alpha, \ \cos \beta, \ \cos \gamma and  \cos \alpha ', \cos \beta ', \cos \gamma ', , then the cosine of the angle  \theta between the two lines is  \cos \theta = \cos \alpha \cos \alpha ' + \cos \beta \cos \beta ' + \cos \gamma \cos \gamma ' ?
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  2. #2
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    Hello, heathrowjohnny!

    Given the direction cosines of two lines: . [\cos\alpha,\cos\beta,\cos\gamma] and [\cos\alpha',\cos\beta',\cos\gamma'] ,
    then the cosine of the angle  \theta between the two lines is:
    . . \cos\theta \;= \;\cos\alpha\cos\alpha' + \cos\beta\cos\beta' + \cos\gamma\cos\gamma'
    Line L_1 has vector: . \vec{u} \:=\:\langle\cos\alpha,\,\cos\beta,\,\cos\gamma\ra  ngle . . . and: . |\vec{u}| \:=\:1

    Line L_2 has vector: . \vec{v} \:=\:\langle\cos\alpha',\,\cos\beta',\,\cos\gamma'  \rangle . . . and: . |\vec{v}| \:=\:1

    . . That is, a set of direction cosines always forms a unit vector.


    The angle \theta between two vectors \vec{u}\text{ and }\vec{v} is given by: . \cos\theta \;=\;\frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|  }
    So we have: . \cos\theta \;=\;\frac{\langle\cos\alpha,\cos\beta,\cos\gamma\  rangle\cdot\langle\cos\alpha',\beta',\gamma'\rangl  e}{(1)(1)}

    Therefore: . \cos\theta \;=\;\cos\alpha\cos\alpha' + \cos\beta\cos\beta' + \cos\gamma\cos\gamma'

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