What is the definition of the vector if you only now the angle between another vector, and its length?

Question

Assume that you have four scenarios.

What we know:

1. We know the length of the blue vector $|B|$, red vector $|A|$ green vector $|R|$.
2. We know $x_1, y_1, x_2, y_2$.
3. We know the angle $V$ between red vector and blue vector
4. We know the angle $W$ between the red vector and the green vector.

Assumptions:

1. If the sum of the angle between red vector and green vector + the angle between blue vector and green vector is larger than angle $V$, then the green vector is not between the red vector and the blue vector.
2. If the angle between the red vector and green vector is larger than the angle between the blue vector and green vector, then the rotation is clockwise, else anti clock wise.
3. The centrum of the circle have the coordinates $(0, 0)$
4. $R1, R2, R3$ have different lengths.

What we don't know:

1. We don't know the vector coordinates for the green vector $x_3, y_3$

Question:

How can find a definition to find the coordinates $x_3, y_3$ for the green vector?

Answer 1

To calculate the green vector all you need is the angle between some other vector and the green vector and the length of the green vector. So I'll choose the blue vector $\vec{R}$, just for convinience. The angle between $\vec{R}$ and $\vec{G}$ is $W$ (wierd way to name an angle, but hey, I'll go with it).
First you should normalise the vector $\vec{R}=(x_2,y_2)$ (you've changed its name to $R_1$ later in the question). To normalise it you just divide all the coordinates by its norm, $|\vec{R}|=\sqrt{x_1^2+y_1^2}$. Now you have a normalized vector $\displaystyle\vec{R_0}=\left(\frac{x_1}{\sqrt{x_1^2+y^2_1}},\frac{y_1}{\sqrt{x_1^2+y^2_1}}\right)$. Now you want to rotate this vector $R_0$, by the angle $W$. You can use rotation matrix for this: $$\begin{bmatrix} \cos W & -\sin W\\ \sin W & \cos W \end{bmatrix}$$ Now multiplying this matrix with $\vec{R_0}$ you'll get a vector $\displaystyle \vec{G_0}=\left(\frac{x_1 \cos W-y_1\sin W }{\sqrt{x_1^2+y^2_1}},\frac{x_1 \sin W+y_1\cos W }{\sqrt{x_1^2+y^2_1}}\right)$. This vector $\vec{G_0}$ is the normalized vector of $G$. Which means its length is equal to $1$ and it has the same direction as vector $G$ you're looking for. To get vector $G$ you just multiply $\vec{G_0}$ by the length of vector $\vec G$ which you already know and you'll get its coordinates.

Note
Be careful when using rotation matrix, you have to make sure that the angle $W$ is always on the right of the red vector. For example in the second and third pictures you would need to use the angle $360^{\circ}-W$.