How to get a new point of a vector when rotated. I want to obtain the new point of a vector that I rotate like this.
When I rotate them, I have the angle of rotation.
I want to know x and y, it rotates taking the reference point of 0,0 
Thanks
 A: To give a general answer, you take your position vector $\vec{v}\in\mathbb{R}^{n}$, and you multiply it by the appropriate rotation matrix ${\bf M}\in\mathbb{R}^{n\times n}$. So we have:
$$\vec{v}'={\bf M}\vec{v}$$
This will give you the position vector under the rotation described by ${\bf M}$.
So let's take your example, of the vector $\vec{v}\in\mathbb{R}^{2}$, where $\vec{v}=\left[55,0\right]$, and multiply it by the matrix ${\bf M}\in\mathbb{R}^{2\times2}$, where ${\bf M}=\left[\begin{smallmatrix}\cos{30^{\circ}} & -\sin{30^{\circ}} \\ \sin{30^{\circ}} & \cos{30^{\circ}} \end{smallmatrix}\right]$. So we have:
$$\vec{v}'=\underbrace{\begin{bmatrix}\frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{bmatrix}}_{\bf M}\underbrace{\left[55\atop 0\right]}_{\vec{v}}\approx\left[47.63 \atop 27.5\right]$$
Hope this helps!
A: Easy. Let $v=55+0i\in\mathbb{C}$ be your original vector embedded in the complex plane, $w\in\mathbb{C}$ the rotated vector and $\theta=30^\circ$.
Surely, $w=ve^{i\theta\,\pi/180}=\frac{55\sqrt 3+55i}{2}$.
A: If your starting vector is along the $x$ axis, it is $(L,0)$, the rotated one is $(L \cos \theta, L \sin \theta)$.  If you don't start along the $x$ axis, you could see Wikipedia
A: HINTS: 


*

*Assume that the lines in the lower picture have length $1$.

*Complete the lower figure to a rectangular triangle. 

*Draw a coordinate system with the lower side lying on the $x$ axis.


Now: What are the coordinates of the upper corner? Use trigonometric functions!
Finally, mulitply by $55$ (in your special case).
