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Thread: [SOLVED] Angle between vectors

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Angle between vectors

    The vectors a, b and c are all unit vectors (length=1). The angle between all the pairs is $\displaystyle \pi/3$.

    What is the angle between $\displaystyle u = a+b+c$ and $\displaystyle v=3c-a-b$
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  2. #2
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    Quote Originally Posted by Spec View Post
    The vectors a, b and c are all unit vectors (length=1). The angle between all the pairs is $\displaystyle \pi/3$.

    What is the angle between $\displaystyle u = a+b+c$ and $\displaystyle v=3c-a-b$
    Hi

    Compute the dot product of u and v
    Compute the length of u and the length of v

    $\displaystyle \cos(u,v) = \frac{u.v}{||u||.||v||}$
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  3. #3
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    Hello, Spec!

    I have a start on the problem . . . but that's all.


    The vectors $\displaystyle \vec a, \vec b, \vec c$ are all unit vectors (length=1).
    The angle between all the pairs is $\displaystyle \tfrac{\pi}{3}$

    Find the angle between $\displaystyle \vec u \:=\:\vec a+\vec b+\vec c$ and $\displaystyle \vec v\:=\:3\vec c-\vec a-\vec b$
    The angle $\displaystyle \theta$ between vectors $\displaystyle \vec u\text{ and }\vec v$ is given by: .$\displaystyle \cos\theta \:=\:\frac{|\vec u\cdot\vec v|}{|\vec u||\vec v|} $

    Let $\displaystyle \alpha$ be the angle between $\displaystyle \vec b\text{ and }\vec c\!:\;\;\cos\alpha \:=\:\frac{\vec b\cdot\vec c}{|\vec b||\vec c|}$

    Since .$\displaystyle \alpha = \tfrac{\pi}{3},\;|\vec b| = |\vec c| = 1$, we have: .$\displaystyle {\color{blue}\vec b\cdot\vec c \:=\:\tfrac{1}{2}}$

    . . Similarly, we have: .$\displaystyle {\color{blue}\vec a\cdot\vec b \:=\:\tfrac{1}{2}}\:\text{ and }\;{\color{blue}\vec a\cdot\vec c \:=\:\tfrac{1}{2}}$


    And I'm certain those will be used in subsequent calculations.

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  4. #4
    Senior Member Spec's Avatar
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    I solved it, but it wasn't pretty.

    $\displaystyle (a+b+c)\cdot (a+b+c) = a \cdot (a+b+c) + b \cdot (a+b+c) + c \cdot (a+b+c) = $
    $\displaystyle =a \cdot a + a \cdot b + a\cdot c +b \cdot a + b \cdot b + b\cdot c + c \cdot a + c \cdot b + c\cdot c=6 \implies |a+b+c| = \sqrt{6}$

    I'm not going to post all calculations, but that's basically how I solved it (and using the second dot product formula above).
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  5. #5
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    That's it !
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