Question

If the original coordinate axes are rotated 30° to obtain the x' and y' axes, what is the value of x in terms of x' and y'?

Answer 1

\[x=\sqrt{\frac{ 3 }{ 2}}x'-\frac{ 1 }{ 2}y' \]

Answer 2

\[x=\sqrt{\frac{ 3 }{ 2}}x'+\frac{ 1 }{ 2}y'\]

Answer 3

\[x=\frac{ 1 }{ 2}x'-\sqrt{\frac{ 3 }{ 2}}y'\]

Answer 4

\[x=\frac{ 1}{ 2}x'+\sqrt \frac{ 3 }{ 2}y'\]

Answer 5

That is a b c and D

Answer 6

@Michele_Laino

Answer 7

I have found a solution :)

Answer 8

http://www.wolframalpha.com/input/?i=rotation+matrix

Answer 9

as an alternative :p



with x,y and x', y' as UNIT vectors [can't seem to get vector notation working in latex]
\[x' = x\cos \theta + y\sin \theta \]
\[y' = -x \sin \theta i + y \cos \theta \]
\[\left(\begin{matrix}x' \\ y'\end{matrix}\right) = \left[\begin{matrix}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{matrix}\right]\left(\begin{matrix}x \\ y\end{matrix}\right)\]
\[\left(\begin{matrix}x \\ y\end{matrix}\right) = \left[\begin{matrix}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{matrix}\right]^{-1}\left(\begin{matrix}x' \\ y'\end{matrix}\right)\]
\[\left(\begin{matrix}x \\ y\end{matrix}\right) = \left[\begin{matrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{matrix}\right]\left(\begin{matrix}x' \\ y'\end{matrix}\right)\]
\[x = \frac{\sqrt{3}}{2}x' - \frac{1}{2}y'\]