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Math Help - Vector Angle

  1. #1
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    Vector Angle

    As below
    Attached Thumbnails Attached Thumbnails Vector Angle-vector.png  
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  2. #2
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    Re: Vector Angle

    Hello, BabyMilo!

    \text{The vector }\overline{OP}\text{ makes angle of }60^o\text{ with the positive }x\text{-axis}
    . . \text{and }45^o\text{ with the positive }y\text{-axis.}
    \text{Find the possibles angles the vector can make with the positve }z\text{-axis.}
    . . . . . . . . . . \text{Theorem}

    \text{If }\alpha,\,\beta,\,\gamma\text{ are the direction angles of a vector,}
    . . . . then: . \cos^2\!\alpha + \cos^2\!\beta + \cos^2\!\gamma \;=\;1


    We have: . \left(\cos 60^o)^2 + (\cos45^o)^2 + \cos^2\gamma \;=\;1

    . w . . . . . . . . . . . \left(\tfrac{1}{2}\right)^2 + \left(\tfrac{1}{\sqrt{2}}\right)^2 + \cos^2\gamma \;=\;1

    . a . . . . . . . . . . . . . . . . . \tfrac{1}{4} + \tfrac{1}{2} + \cos^2\gamma \;=\;1

    . m . . . . . . . . . . . . . . . . . . . . . . \cos^2\gamma \;=\;\tfrac{1}{4}

    . . n . . . . . . . . . . . . . . . . . . . . . . \cos\gamma \;=\; \pm\tfrac{1}{2}

    . . . . . . . . . . . . . . . . . .Therefore: n . \gamma \;=\;60^o,\,120^o

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