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$\vec A$ has the magnitude $13.5$ m and is angled $68.0°$ counterclockwise from the positive direction of the x axis of an xy coordinate system. Also, $\vec B$ $= (11.5 m)i + (7.15 m)j$ on that same coordinate system. We now rotate the system counterclockwise about the origin by $20.0°$ to form an $x'y'$ system.

(a) On this new system, what are $\vec A$ and $\vec B$ in unit-vector notation?

I don't understand what needs to be done.

Am I supposed to:

$13.5$ times $68$ times cos minus $20$ ?

$\sqrt 11.5^2 + 7.15 ^2 - 20$?

Darío A. Gutiérrez
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user366783
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  • Have you drawn a picture? Show the original and new axes and $\vec A$ and $\vec B$ relative to the original axes. What is the angle of $\vec A$ from the new $x$ axis? The magnitude stays the same, but you need to assess the components along the new axes. For $\vec B$ you need to compute its angle from the old axes and then its angle from the new axes. – Ross Millikan Sep 22 '16 at 03:59
  • I don't understand. How could I draw a picture if I don't have any x,y coordinates? How do I determine angles and components and other angles? – user366783 Sep 22 '16 at 04:05
  • You can draw the axes, and you can draw $\vec A$ by following the directions. Find an angle $68^\circ$ counterclockwise from $+x$ and draw a vector of length $13.5$. You can imagine swinging a compass. You can also calculate the components with $x=r \cos \theta, y=r \sin \theta. You have the components for $\vec B$. You will need to find its angle from the arctangent. – Ross Millikan Sep 22 '16 at 04:30

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