# 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

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Multiply by a rotation matrix, see, eg, en.wikipedia.org/wiki/Rotation_matrix – copper.hat Jul 26 '12 at 19:19
It´s a problem that my sister could not resolve and i did not remember how to it :/ – Rudy_TM Jul 26 '12 at 19:26

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!

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Thanks, i did´nt remember the rotation matrices :) Thank You – Rudy_TM Jul 26 '12 at 19:30
@Rudy_TM as along as your sister can already deal with them... – draks ... Jul 26 '12 at 20:03
Thanks :) yes XD – Rudy_TM Jul 26 '12 at 20:11

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, $w=ve^{i\theta\,\pi/180}=\frac{55\sqrt 3+55i}{2}$.

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+1 for the most complex solution ;-) – draks ... Jul 26 '12 at 20:04

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

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HINTS:

1. Assume that the lines in the lower picture have length $1$.
2. Complete the lower figure to a rectangular triangle.
3. 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).

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