Okay, this problem may be ridiculously easy, but I have been at it for a while, and I keep getting something different. It goes like this:
Two vectors $a$ and $b$ in the plane. It is known that
- The angle between $a$ and $b$ is 60 degrees.
- $|a|=3$
- $|a+b|=4$
What is $|b|$?
I have tried drawing parallelograms, used the sine and cosine relations, but I keep getting something different. I hope someone can help me with this.
My attempt at solving the problem:
$|a+b|^2 = (a+b) \cdot (a+b)=|a| + 2\cdot |a| \cdot |b| \cdot cos(v) +|b|$
$4^2=3+2\cdot 3\cdot |b| \cdot cos(60)+ |b|$
$13=3\cdot|b|+|b|$
$13/4=|b|=3,25$
Hint: You know that $\lVert a+b\rVert^2=16$ and you also know that$$\lVert a+b\rVert^2=\lVert a\rVert^2+\lVert b\rVert^2+2\cos(60^\circ)\lVert a\rVert.\lVert b\rVert.$$