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Thread: Angle at Ox

  1. #1
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    Angle at Ox

    I'm really confused on which angle here is "the angle at Ox", and how you find out.



    That's a quick image I made of it. I have that as the vector $\displaystyle |-3i + 2j|$ but I'm not sure whether the angle is $\displaystyle a$ or $\displaystyle a + b$ or $\displaystyle b + c$ or $\displaystyle a + b + c$.
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  2. #2
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    Quote Originally Posted by Viral View Post
    That's a quick image I made of it. I have that as the vector $\displaystyle |-3i + 2j|$ but I'm not sure whether the angle is $\displaystyle a$ or $\displaystyle a + b$ or $\displaystyle b + c$ or $\displaystyle a + b + c$.
    From your graphic, the usual understanding of $\displaystyle O_x=b+c$.
    The that angle is $\displaystyle \pi + \arctan \left( {\frac{2}{{ - 3}}} \right)$.

    However, you need to check your text material for the wanted definition.
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  3. #3
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    That's the problem =\ , the only information given is:

    Given that $\displaystyle a = i + j$ and $\displaystyle b = 2i - j$ find the magnitude of each of the following vectors and the angle each makes with Ox.
    So, for the above example, it's the same as $\displaystyle 180 - arctan(\frac{2}{3})$ ?

    What about with this example: ?



    I've been answering those ones as just $\displaystyle \theta = arctan(\frac{j}{i})$

    EDIT: The a and b referred to in the information given is not referring to the angles, I should have given them different variables...
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  4. #4
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    It appears to me as if your text material is not standard.
    So you really need to read it closely to find these definitions.

    I can give a standard approach. Consider the vector $\displaystyle p\vec{i}+q\vec{j}$.
    The angle associated with that vector is: $\displaystyle \left\{ {\begin{array}{rl} {\arctan \left( {\frac{q}{p}} \right),} & {p > 0} \\ {\pi + \arctan \left( {\frac{q}{p}} \right),} & {p < 0\;\& \;q > 0} \\ { - \pi + \arctan \left( {\frac{q}{p}} \right),} & {p < 0\;\& \;q < 0} \\ \end{array} } \right.$
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