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Given two points around an origin $(0,0)$ in $2D$ space, how would you calculate an angle from $p_1$ to $p_2$?

How would this change in $3D$ space?

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There is no angle between two points... Do you mean the angle between the points at the origin? Maybe? – John Gietzen Jul 20 '10 at 19:44
Regarding the 3d space problem, you will basically need to measure the angle in spherical co-ordinates, so two angles. This is after adding in the twin requirements of an origin point and origin vector to measure angle from (as alluded to in the previous comment) – workmad3 Jul 20 '10 at 19:50
Yes, it's safe to assume a given origin (0,0). – David McGraw Jul 20 '10 at 19:57
up vote 6 down vote accepted

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:

$$v = \arccos\left(\frac{x_1x_2 + y_1y_2 + z_1z_2}{\sqrt{(x_1^2+y_1^2+z_1^2) \cdot (x_2^2+y_2^2+z_2^2)}}\right)$$

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I will assume that you mean the angle of the line from p1 to p2 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 p1-p2 and take the arccossine.

v = (normalized vector from p1 to p2)
theta = arccos( v * <1,0> )    (dot product)

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 * <1,0,0> )   (dot product)

You could also possibly mean the angle between the line from the origin to p1 and the line from the origin to p2.

You can do this with dot products, as well; but both vectors must be normalized.

theta = arccos( a * b ) (dot product)

where a is the normalized vector from the origin to p1 and b is the normalized vector from the origin to p2.

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In case you want to implement it on some programming language.

Usually there is a function called like atan2(y, x) which returns oriented angle between points (x, y) and (1, 0). In that case use could use

atan2(vector product, scalar product).

This is usually more stable than using just arccos or arcsin.

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Im not quite sure what your answer is. And speaking in terms of mechanical programming languages does not really express the mathematical concept and understanding that us mathematicians are going for. – CogitoErgoCogitoSum Feb 12 '13 at 2:07

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