Why do the Lagrange points lie only on the orbital plane of the two co-rotating bodies? I could not find a detailed explanation of the fact that Lagrange points lie only on the orbital plane of the 2 co-rotating bodies. As per the wikipedia page calculations are here.
There is no indication in the equations, as to the planar nature of the solution. Can someone shed some light on this topic? Or suggest a detailed pdf perhaps?
 A: Call the orbital plane of Earth and the moon the $xy$ plane, where $z=0$. Suppose you put a peanut at a hypothetical Lagrange point above the orbital plane, where $z>0$. Then the gravitational force from both Earth and the moon on the object has a negative $z$-component. IE, both gravitational forces pull the peanut back towards the orbital plane. The peanut will accelerate in the $-z$-direction. That means it couldn't have been in a Lagrange point to begin with, because being at a Lagrange point means the peanut needs to keep the same relative position.
If it isn't obvious that the force of gravity from both Earth and the moon has a negative $z$ component, consider the equation
$$\mathbf{F}_{grav} = -\frac{GMm}{r^2} \hat{r}$$
In this case $\hat{r}$ has a positive $z$ component because it points in the direction from Earth / moon (where $z=0$) to the peanut (where $z>0$). So with the minus sign in there, the force has a negative $z$ component.
A: A Lagrange point is a point where a small object maintains a fixed position with respect to two large objects. The combined gravitational forces from the large objects cancels the centrifugal force felt by the small object. 
Suppose a Lagrange point was to one side of the plane. Then it would stay that distance away. It would orbit in a plane parallel to the orbital plane. Centrifugal force would be parallel to the plane. 
But both of the other masses would pull it toward the plane. The gravitational force is not parallel to the plane, and cannot balance that centrifugal force. 
