Question

find the angle between vector u: <-2, 5> and vector v:<-1, 3>

Answer 1

well, first off get the cosine of the angle
then get the arcCos of it
\(\bf cos(\theta)=\cfrac{u \cdot v}{||u||\times ||v||}\)

Answer 2

@jdoe0001 careful using \(\times\) there -- people may confuse it for a cross product

Answer 3

hehe, yes it can get ambiguous

Answer 4

I meant to leave it out, but then again some folks may wonder if it's a product or what

Answer 5

\(\bf cos(\theta)=\cfrac{u \cdot v}{||u||\times ||v||} \implies \cfrac{\text{dot product}}{\text{product of magnitudes}}\)

Answer 6

wait sorry... how do you find the magnitudes?

Answer 7

@ses11 they are the norms of the vectors

Answer 8

\[||u||=\sqrt{\sum_{k=1}^n (u_k)^2}\]
for \(u=\langle u_1,...,u_n\rangle\).

Answer 9

@oldrin.bataku so IIuII=\[\sqrt{-2^{2}+5^{2}}\]