Confused on how to do this: Find the angle between u = and v =

Question

Confused on how to do this: Find the angle between u = <-3, 2> and v = <2, 3>

sleepyjess · Answer

@Loser66

loser66 · Answer

you have formula for this, it is cos (u and v) =.....

sleepyjess · Answer

I know $\sf cos\theta = \dfrac{u*v}{||u||~||v||}$, but I'm not really sure how to use that.

loser66 · Answer

so, u dot v, not u*v, what do you get for u dot v(the numerator)

sleepyjess · Answer

-3 * 2 + 2*3
-6 + 6
-

sleepyjess · Answer

*0

anonymous · Answer

Well, take arccos of both sides. $$
\theta = \arccos\left(\frac{\mathbf u\cdot \mathbf v}{\|\mathbf u\|\|\mathbf v\|}\right)
$$

loser66 · Answer

so, the whole thing is 0, cos (what?) = 0?

sleepyjess · Answer

cos(90) = 0

sleepyjess · Answer

but what about the denominator of the fraction? That is what is confusing me

loser66 · Answer

DAT SIT.

anonymous · Answer

$$
\|\mathbf u\| = \sqrt{\mathbf u \cdot \mathbf u}
$$

kainui · Answer

Maybe this helps, just imagine the vector as a hypotenuse of a right triangle, so we can use the pythagorean theorem to get its length. $$\Large \Large v =  \langle x,y,z \rangle \ \Large ||v|| = \sqrt{x^2+y^2+z^2}$$

sleepyjess · Answer

Ok, so the denominator would be $\sqrt{-3^2 + 2^2}\sqrt{2^2 + 3^2}$?

anonymous · Answer

$$||u|| = \sqrt{(-3)^2+(2)^2}$$

anonymous · Answer

But yes, you got the right idea

sleepyjess · Answer

ok ^_^ thank you all

anonymous · Answer

That's one messed up triangle

sleepyjess · Answer

How does this relate to a triangle? *confused*

kainui · Answer

All vectors are just the hypotenuse of triangles. |dw:1422744906792:dw|