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Thread: Denoting vector displacement (solution format enquiry)

  1. #1
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    Denoting vector displacement (solution format enquiry)

    Hey dudes (and dudettes).

    I actually know how to solve this problem, however the authors provide their answers in a different format (groups of square roots) whereas I just solve to the nearest hundredth using trig functions. I don't understand where the authors get their square root values from in the solution, and it's relevant that I know as it's likely the same solution format will be tested in the exam. Could someone please clarify this for me?
    Denoting vector displacement (solution format enquiry)-q8.png
    Denoting vector displacement (solution format enquiry)-a8.png

    Thanks as always.
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  2. #2
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    Re: Denoting vector displacement (solution format enquiry)




    the solution doesn't require use of a calculator ... exact trig values come from the special angles on a unit circle.

    $300 \, km$ Southeast =

    $300\cos(-45)i + 300\sin(-45)j = 300 \cdot \dfrac{\sqrt{2}}{2} i + 300 \cdot \left(-\dfrac{\sqrt{2}}{2}\right) j = 150\sqrt{2}i - 150\sqrt{2}j = 150\sqrt{2}(i-j)$

    $150 \, km$ 30 deg W of N =

    $150\cos(120)i + 150\sin(120)j = 150\left(-\dfrac{1}{2}\right)i + 150 \cdot \dfrac{\sqrt{3}}{2} j = -75i + 75\sqrt{3}j= 75(-i + \sqrt{3} j)$

    Last edited by skeeter; Jun 16th 2017 at 03:35 AM.
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