Finding third charge coordinates in an equilibrium position

[AdrianMachin]

Homework Statement

Here are the problem statement and the solution. I'm stuck at where the book suggests the formulas for the x and y coordinations (highlighted in yellow) of the third charge. Any explanations or proof on how they came to the conclusion for the third charge coordinations would be much appreciated. Is there an easier solution for this problem?

The given solution:

Homework Equations

x3=x2-rcosθ
y3=y2-rsinθ

The Attempt at a Solution

I think it must have something to do with the "point-slope" form of the equation of a straight line, but stuck in finding a reasonable proof.

 

[BvU]

Lazy, lazy, eh ? Perhaps you can at least make a drawing of the situation and post it, together with a non-empty attempt at solution. You know the guidelines at PF I hope !

 

[haruspex]

You highlighted collinear, so I assume you do not understand why they must be. Draw yourself a diagram of three charges not collinear and consider the forces on one of them. Can those forces be in balance?

 

[AdrianMachin]

OK, I've progressed a bit. Now, I don't know why there is also a negative sign before rsinθ in y3=y2-rsinθ.

 

[haruspex]

y3=y2-rsinθ

What does the book solution state in parentheses immediately after that? Does that accord with the way you have drawn θ?

 

[AdrianMachin]

What does the book solution state in parentheses immediately after that? Does that accord with the way you have drawn θ?

So it means that the negative sign is because of the fact that sin(-θ)=-sin(θ)?

 

[BvU]

Theta is the angle wrt the negative x-direction. That's all.

 

[haruspex]

So it means that the negative sign is because of the fact that sin(-θ)=-sin(θ)?

Yes.
They define θ as the angle the vector from q₃ towards q₁ makes to +ve x axis. By convention, that would mean anticlockwise from that axis. Your diagram shows θ measured clockwise from the axis, so your θ is minus their θ.