
Vector Problem
So I have this small problem. Two vectors A and B added together give the vector S. Show that S (which I'm assuming is the scalar of S) is equal to (A^2 + B^2 + 2AB Cos (theta) ) / 2 remembering that S . S = S^2 and S= A+B. I'm really not sure how to make a start on it. I'm thinking that it has something to do with the rule a.b= [a][b]cos theta, but I can't seem to make the first leap. Any suggestions

Hi,
I'm not sure the question quite makes sense  vectors can't usually be "squared"  you can multiply them with the dot or cross product, but the notation in your question suggests otherwise.
Does the answer refer to the magnitude of the two vectors?

I think your right about the squaring of the vector, not being possible. However I don't theink the squraed parts are refering to the vectors. They weren't in bold, which I believe is the usual convention for vectors. So I'm assuming that the are the values or the scalar. ie Ai + Aj or possible the scalar product. Sorry I'm really at a loss about this stuff.

It follows at once from this.
$\displaystyle S = A + B \Rightarrow \left( {S \cdot S} \right) = \left( {A + B} \right) \cdot \left( {A + B} \right) = A \cdot A + 2A \cdot B + B \cdot B$

So the complete derivation would be something like this:
S.S= (A+B).(A+B)
=> A.A + B.B + 2(A.B)
=> A.A = [A][A] cos theta = [A]^2 because cos 0 = 1
=> B.B = [B][B] cos theta = [B]^2 as above
=> 2(A.B) = 2[A][B] cos theta
putting all the parts together
A^2 + B^2 + 2AB cos theta
then taking the square root bcause S.S = S^2
giving S = (A^2 + B^2 + 2AB cos theta)^1/2
Is that how it goes? Thanks for the help