# Thread: Denoting vector displacement (solution format enquiry)

1. ## 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?

Thanks as always.

2. ## 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 =

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