find the angle between vector u: and vector v:

Question

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

jdoe0001 · Answer

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

anonymous · Answer

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

jdoe0001 · Answer

hehe, yes it can get ambiguous

jdoe0001 · Answer

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

jdoe0001 · Answer

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

anonymous · Answer

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

anonymous · Answer

@ses11 they are the norms of the vectors

anonymous · Answer

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

anonymous · Answer

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