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    Don't understand the question and can't make sense of the diagrams given.

    I know OA = (xi,0,0) however don't know how to use info given in bottom part of the question
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    Explanation of diagrams: there's a pole sticking up vertically. There are three "cables" attached to it, spread evenly with angles of 120 degrees. They are distance x from the pole. The vectors \mathbf{i} and \mathbf{j} are shown on the plan view.

    If you still don't get it, the answer to part (a) is in the spoiler below.
    Spoiler:
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    \vec{OA} = x \mathbf{i}
    \vec{OB} = - \frac{1}{2} x \mathbf{i} + \frac{\sqrt 3}{2} x \mathbf{j}
    \vec{OC} = - \frac{1}{2} x \mathbf{i} - \frac{\sqrt 3}{2} x \mathbf{j}
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    Oh, didn't see the last sentence you said. Did you edit that in after I started writing my reply? Sorry, ignore the above post.

    Do you mean part (c)? If so then think of dot products.
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    (Original post by TLDM)
    Explanation of diagrams: there's a pole sticking up vertically. There are three "cables" attached to it, spread evenly with angles of 120 degrees. They are distance x from the pole. The vectors \mathbf{i} and \mathbf{j} are shown on the plan view.

    If you still don't get it, the answer to part (a) is in the spoiler below.
    Spoiler:
    Show
    \vec{OA} = x \mathbf{i}
    \vec{OB} = - \frac{1}{2} x \mathbf{i} + \frac{\sqrt 3}{2} x \mathbf{j}
    \vec{OC} = - \frac{1}{2} x \mathbf{i} - \frac{\sqrt 3}{2} x \mathbf{j}
    How did you get -1/2x and sqrt(3)/2?
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    (Original post by TSRforum)
    How did you get -1/2x and sqrt(3)/2?
    Trigonometry. You can find the angle between \vec{OB} and \mathbf{j}, and you also know that \vec{OB} has length x. You should be able to work it out from there.
 
 
 
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