Vector Proofs using vector components

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The discussion revolves around proving the formula r = √(a² + b² + 2abcosθ) for the magnitude of the resultant vector R formed by two vectors a and b at an angle θ. The user expresses confusion about incorporating the term 2abcosθ into their calculations, recognizing its relation to the dot product of the vectors. They successfully derive part of the proof using vector components, noting that R² = A² + B² + 2A·B. The conversation emphasizes the necessity of using components as specified in the problem, despite the user's initial uncertainty. Understanding the relationship between vector addition and the dot product is crucial for completing the proof.
lizette
Hi! I'm new to the forums. I'm taking an introduction to physics class this semester and I've been having some difficulty with it. Oh, I also wanted to let you know that it's been a while since I've taken calculus or any other math class for that matter. But I need physics to graduate. Anywho ... the question that I have deals with vector components.

Two vectors of magnitudes a and b make an angle theta (which I'll represent as @) with each other when placed tail to tail. Prove, by taking components along two perpendicular anes, that

r = the square root of (a^2 + b^2 + 2abcos@)

gives the magnitude of the sum vector R (vector R = r with that arrow above it) of the two vectors.

Well this is what I have so far:

vector A = Axi + Ayi
vector B = Bxi + Byi
vector R = vector A + vector B

A^2 = Ax^2 + Ay^2
B^2 = Bx^2 + By^2
R^2 = A^2 + B^2

A dot B = A*B = ABcos@

I can see how r = square root of (A^2 + B^2) but where does the 2ABcos@ come in. I have a feeling that it deals with the A*B product, but I don't know how to fit it in.
 
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A dot A=a^2 (a is magn A) A- vector
B dot B=b^2 -the same-
=>
(A+B) dot (A+B)=a^2+b^2+2*A dot B=r^2 (evidently)
...so r^2 =a^2+b^2+2*a*b*cos@...
...do you really have to use components ?
 
well, the problem said to use components .. so i believe so.

thanks for your help!
 
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