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    If \vec{OA} is the unit vector l_1i+m_1j+n_1k and \vec{OB} is the unit vector l_2i+m_2j+n_2k, by using the cosine formula in triangle OAB find the angle between \vec{OA} &\vec{OB}..

    I have tried expressing them as direction cosines , but none of that is working..can anyone point me in the right direction..
    Thanks
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    Do you know the formula {\bf a . b} = |{\bf a}||{\bf b}| \cos \theta, where \theta is the angle between a and b?
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    yes , but the book has not introduced scalar products yet..all I know is unit vector.|V|=V and direction cosines, and eq of straight lines..
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    dot/scalar product?
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    (Original post by rbnphlp)
    yes , but the book has not introduced scalar products yet..all I know is unit vector.|V|=V and direction cosines, and eq of straight lines..
    OK, then I can only assume you're supposed to be proving that formula.

    Which is rather hard for an exercise I would have thought. See

    http://en.wikipedia.org/wiki/Dot_pro...interpretation
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    (Original post by DFranklin)
    OK, then I can only assume you're supposed to be proving that formula.

    Which is rather hard for an exercise I would have thought. See

    http://en.wikipedia.org/wiki/Dot_pro...interpretation
    thanks ..this the last question in the excercise before the introduction of scalar product ..
    also the answer is cos\theta =l_1l_2+m_1m_2+n_1n_2 ..
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    am i missing something here? using cosine rule in OAB with angle between OA and OB =A
    a=|AB|,b=|OB|,c=|OA|

    a^2=b^2+c^20-2bc cosA

    but b=c=1 as unit vectors.
    also (l1)^2+(m1)^2+(n1)^2=1 and (l2)^2+(m2)^2+(n2)^2

    result follows
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    (Original post by mathz)
    am i missing something here? using cosine rule in OAB with angle between OA and OB =A
    a=AB,b=OB,c=OA

    a^2=b^2+c^20-2bc cosA

    but b=c=1 as unit vectors.
    also (l1)^2+(m1)^2+(n1)^2=1 and (l2)^2+(m2)^2+(n2)^2

    result follows
    Thanks much easier and probably the preferred method ..will rep tommorow
 
 
 
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