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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.enter image description here

I want to know x and y, it rotates taking the reference point of 0,0

Thanks

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    It´s a problem that my sister could not resolve and i did not remember how to it :/2012-07-26

4 Answers 4

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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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    @Rudy_TM as along as your sister can already deal with them...2012-07-26
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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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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).