Let $u,v$ be vectors such that $||u|| = 2, ||v|| = \sqrt3$ , and $u \cdot v = 1$. Find ||u + v||.
So far I calculated $\dfrac{u\cdot v}{||u||*||v||}=\cos(\theta)$ and then with one angle and two sides in the triangle known, I can calculate the length of $(u-v)$ as the third side of the triangle.
On
I assumed it was a 2 component vector for each vector. $u=<a,b>$ and $v=<c,d>$
$||u||=2$ means $\sqrt{a^2+b^2}=2$
$||v||=\sqrt{3}$ means $\sqrt{c^2+d^2}=\sqrt{3}$
$u \cdot v=1$ means $a c+b d=1$
Finding $||u+v||$ now
$||u+v||=\sqrt{(a+c)^2+(b+d)^2}$
Expand inside the square root a bit
$||u+v||=\sqrt{a^2+b^2+c^2+d^2+2(ac+bd)}$
$$ || u + v ||^2 = (u + v) \cdot (u + v) = \ldots $$