Calculating an Angle from $2$ points in space Given two points $p_1$ , $p_2$ around the origin $(0,0)$ in $2D$ space, how would you calculate the angle from $p_1$ to $p_2$?
How would this change in $3D$ space?
 A: I will assume that you mean the angle of the line from $p_1$ to $p_2$ with respect to the $x$-axis
This is the best I can do given the information you have provided.
In any case, the official mathsy way would be to find the dot product between the two, and divide by the magnitude of $p_1-p_2$ and take the arccossine.
$$
\begin{aligned}
v &= (\text{normalized vector from } p_1 \text{ to } p_2) \\
\theta &= \arccos( v \cdot \langle1,0\rangle) \qquad\qquad\qquad\qquad (\text{dot product})
\end{aligned}
$$
You can normalize a vector by dividing every term by the magnitude (length) of the entire vector.
For 3D, the same thing applies:
$$ \theta = \arccos( v \cdot \langle1,0,0\rangle ) \qquad\qquad (\text{dot product}) $$

You could also possibly mean the angle between the line from the origin to $p_1$ and the line from the origin to $p_2$.
You can do this with dot products, as well; but both vectors must be normalized.
$$ \theta = \arccos( a \cdot b ) \qquad\qquad (\text{dot product}) $$
where $a$ is the normalized vector from the origin to $p_1$ and $b$ is the normalized vector from the origin to $p_2$.
A: Assuming this is relative to the origin (as John pointed out): Given two position vectors $\vec p_1$ and $\vec p_2$, their dot product is:
$$\vec p_1\cdot \vec p_2 = |\vec p_1| \cdot |\vec p_2| \cdot \cos \theta$$
Solving for $\theta$, we get:
$$\theta = \arccos\left(\frac{\vec p_1 \cdot \vec p_2}{|\vec p_1| \cdot |\vec p_2|}\right)$$
In a 2D space this equals:
$$v = \arccos\left(\frac{x_1x_2 + y_1y_2}{\sqrt{(x_1^2+y_1^2) \cdot (x_2^2+y_2^2)}}\right)$$
And extended for 3D space:
$$v = \arccos\left(\frac{x_1x_2 + y_1y_2 + z_1z_2}{\sqrt{(x_1^2+y_1^2+z_1^2) \cdot (x_2^2+y_2^2+z_2^2)}}\right)$$
A: In case you want to implement it on some programming language.
Usually there is a function called like $\operatorname{atan2}(y, x)$ which returns oriented angle between points $(x, y)$ and $(1, 0)$. In that case use could use
$$\operatorname{atan2}(\text{vector product}, \text{scalar product}). $$
This is usually more stable than using just $\arccos$ or $\arcsin$.
A: Let the two points be $(x_1,y_1)$ and $(x_2,y_2)$ then angle $\theta$ between them $=$
$$\theta=\tan^{-1}{\frac{y_2}{x_2}}-\tan^{-1}{\frac{y_1}{x_1}}$$
