# Vector Problem

• October 9th 2008, 08:01 AM
daragh8008
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
• October 11th 2008, 05:04 PM
Tony2710
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?
• October 13th 2008, 05:29 AM
daragh8008
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.
• October 13th 2008, 07:31 AM
Plato
It follows at once from this.
$S = A + B \Rightarrow \left( {S \cdot S} \right) = \left( {A + B} \right) \cdot \left( {A + B} \right) = A \cdot A + 2A \cdot B + B \cdot B$
• October 13th 2008, 07:46 AM
daragh8008
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