Find the angle between the vectors u and v if u = (1, 2) and v = (- 4, - 2)

6 years agoUse the formula \[\Large \theta = \arccos\left(\frac{u\cdot v}{|u||v|}\right)\] where u and v are vectors.

6 years agowhich number in my points do i use for each... for example which number in (1,2) would I use for which u

6 years agoIn the numerator of that fraction, you're computing the dot product between u and v

6 years agoSo what is the dot product between u = (1, 2) and v = (- 4, - 2)?

6 years ago-8

6 years agogood

6 years agonow you need to find |u| and |v|

6 years agohow do I do that?

6 years agouse the formula |x| = sqrt(a^2+b^2) where the vector x is x = (a,b)

6 years agoso it is sqrt(1^2+2^2) for u's and then sqrt(-4^2+-2^2) for v's?

6 years agoyes, so what do you get?

6 years agoi think u's is sqrt(5) and v's is 2sqrt(5)

6 years agoYou got it, nice work So what is |u||v|?

6 years ago10... so then its arccos(-8/10)?

6 years agoyes

6 years agowhich turns out to be 143.13degrees

6 years agoyou nailed it

6 years agoYou keep helping me sooo much. Thanks, I am taking a Pre-Calculus course by myself and sometimes it can be difficult to figure out all of the questions by myself :).

6 years agoyou're very welcome, that must be pretty crazy lol

6 years agoyep... I'm almost done though (thankfully haha)

6 years agowell that's good

6 years agois there a difference between a u surrounded by two of the straight lines as opposed to the just one in the question you just helped me with?

6 years agoWell technically it should have been \[\Large \|u\|\] since the two vertical bars denote the norm or magnitude of a vector. But the absolute value bars say the same thing (they're just used in a more general sense). So either work in my opinion.

6 years agooh, ok... thanks for clarifying. I am working on another type of question like that and wasn's sure if there was a difference.

6 years agothere's not much of a difference really

6 years ago