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Mathematics 19 Online
OpenStudy (anonymous):

Find the angle between the given vectors to the nearest tenth of a degree. u = <-5, -4>, v = <-4, -3>

OpenStudy (psymon):

So the formula is: \[\cos \theta = \frac{ u*v }{ ||u||*||v|| }\] So this means multiply vectors u and v in the numerator and multiply their magnitudes in the denominator. Do you kno whow to perform either of those operations?

OpenStudy (anonymous):

i did but i can't remember what to do for the denominator

OpenStudy (psymon):

Ah, okay. Well the magnitude is found by doing the pythagorean theorem with i and j. So for vector u, i is -5 and j is -4. So pythagorean theorem with this gives me: \[\sqrt{(-5)^{2}+(-4)^{2}} = \sqrt{41}\] See what i did?

OpenStudy (anonymous):

yeah!

OpenStudy (psymon):

Alright, cool. Now we do that for vector v, which is -4, -3 \[\sqrt{(-4)^{2}+(-3)^{2}} = 5\] So now we have: \[5\sqrt{41}\] in the denominator. You remember how to multiply vectors with dot product?

OpenStudy (anonymous):

i'm following you but i'm not sure what to do next

OpenStudy (psymon):

Well, I just did the denominator part since that was you said you weren't sure what to do with. So we're not quite sure how to do the numeraotr?

OpenStudy (anonymous):

is the top 240?

OpenStudy (psymon):

Way too large.

OpenStudy (psymon):

So if we have two vectors u and v, each ofthe form of i + j. Then uv = the multiplication of the i's + the multiplication of the j's. \[u = u _{1}i + u _{2}j\] \[v = v _{1}i+ v _{2}j\] \[uv = (u _{1})(v _{1}) + (u _{2})(v _{2})\] If that makes any sense.

OpenStudy (anonymous):

so -16?

OpenStudy (psymon):

(-5)(-4) + (-4)(-3) Check again :P

OpenStudy (anonymous):

oh wow i'm losing it haha 32

OpenStudy (anonymous):

so now what do i do

OpenStudy (psymon):

Checking, lol. I'm not getting something pretty xD

OpenStudy (anonymous):

oh boyyyyy hahah

OpenStudy (psymon):

Put it in my calculator wrong xDD

OpenStudy (anonymous):

these are the answer choices :) -9.1° 1.8° 0.9° 11.8°

OpenStudy (psymon):

Yeah, I found the answer, I just put it in the calculator wrong. Okay, so we have the result of all our multiplying. So now we have this: \[\cos \theta = \frac{ 32 }{ 5\sqrt{41} }\] Now just need to solve for theta :P

OpenStudy (anonymous):

0.9?

OpenStudy (psymon):

I got 1.8.

OpenStudy (anonymous):

i got 0.9999 but that would round up :/ i might be plugging it in wrong

OpenStudy (psymon):

Yes, you should get that (don't round). But that is what cos(theta) is equal to. But wedon't want cos(theta), we want the actual theta. Do you know how to get that from here?

OpenStudy (anonymous):

noo :/

OpenStudy (psymon):

Well, right now we have: \[\cos \theta = number\]But we already know the number, we want theta. Well when we want theta, we do this: \[\cos^{-1} (number) = \theta \]

OpenStudy (anonymous):

GOT IT thank you so much

OpenStudy (psymon):

Okay, awesome ^_^

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