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Mathematics 22 Online
OpenStudy (anonymous):

Find a ⋅ b. a = <2, 4>, b = <2, 5>

OpenStudy (anonymous):

these are the answer choices <4, 9> 24 <4, 20> 16

OpenStudy (amistre64):

and your choice is?

OpenStudy (anonymous):

well I thought it was <4,20> but that clearly isn't an answer choice

OpenStudy (amistre64):

hmm, a dotproduct produces a number, not a vector

OpenStudy (amistre64):

stack, multiply, add the resutls

OpenStudy (amistre64):

a = <2, 4> b = <2, 5> 4 + 20

OpenStudy (anonymous):

so would it be 24?

OpenStudy (amistre64):

i organize them like this to avoid misplacing the information all i have to do then is multiply down the columns and add the resutls

OpenStudy (amistre64):

yes, 24

OpenStudy (anonymous):

ah ok I see how you do this...

OpenStudy (amistre64):

:) practice makes perfect ... and then dementia sets in lol

OpenStudy (anonymous):

haha :) would you mind helping me with a few more?

OpenStudy (amistre64):

one more, need to start heading to bed

OpenStudy (anonymous):

Find the angle between the given vectors to the nearest tenth of a degree. u = <2, -4>, v = <3, -8>

OpenStudy (amistre64):

this involves the definition of a dotproduct \[|u||v|cos(a)=u\cdot v\]

OpenStudy (amistre64):

so we can start with dotting u to v, what do we get?

OpenStudy (anonymous):

what is a?

OpenStudy (amistre64):

a is alpha, or angle .... its what we will end up solving for also notice that when we dot a vector to itself: (x , y) (x , y) ---------- x^2 + y^2 the results are almost the length .... sqrt(x^2 + y^2) defines the length of a vector

OpenStudy (anonymous):

I don't really understand...

OpenStudy (amistre64):

do the dotproducts: u.v u.u v.v we will use these in the "formula" in order to solve for the required angle

OpenStudy (anonymous):

what am I multiplying? Idk at all

OpenStudy (amistre64):

the first think you posted asked you to find the dot product of 2 vectors, i demonstarted what to do .... you said you "ah ok I see how you do this..." now given: u = <2, -4> and v = <3, -8> find u dot v and u dot u and v dot v

OpenStudy (anonymous):

ao you basically do them separately and then add them together at the end?

OpenStudy (amistre64):

do them seperately, and then we will use them as needed. at the moment we simply need to know what they are.

OpenStudy (anonymous):

ok so u.v is 38 uu is 20 and vv is 73?

OpenStudy (amistre64):

let me chk :) u = <2, -4> v = <3, -8> 6 + 42 = 40 u = <2, -4> and v = <3, -8> u = <2, -4> and v = <3, -8> 4+16 = 20 9 + 64 = 73 correct

OpenStudy (anonymous):

how do I solve for the angle then?

OpenStudy (amistre64):

now we can work the problem\[|u||v|cos(a)=u\cdot v\] \[|a|=\sqrt{a\cdot a}\] sooo \[\sqrt{20}~\sqrt{73}~cos(a)=40\] divide off the sqrts and inverse the cosine to detemrine a \[cos(a)=\frac{40}{\sqrt{20}~\sqrt{73}}\] \[a=cos^{-1}\left(\frac{40}{\sqrt{20}~\sqrt{73}}\right)\]

OpenStudy (anonymous):

ok well I tried plugging it into my calculator but it keeps saying error

OpenStudy (amistre64):

thats cause the result is bigger than 1 ....

OpenStudy (anonymous):

what would that mean?

OpenStudy (anonymous):

3.0° 6.0° -7.0° 16.0° these are the answer choices

OpenStudy (amistre64):

u = <2, -4> v = <3, -8> 6 + 32 = 38 you had the right number, i messed it up ... said it was 40

OpenStudy (amistre64):

\[a=cos^{-1}\left(\frac{38}{\sqrt{20}~\sqrt{73}}\right)\] much better

OpenStudy (anonymous):

so 6 degrees?

OpenStudy (amistre64):

yep

OpenStudy (amistre64):

all that work defining the dots, and i used 40 instead of 38 :) i mentioned the dementia part right?

OpenStudy (anonymous):

thank you so much for your help:) and yes I remember you mentioning dementia:p lol well have a good night you should probably go to bed:)

OpenStudy (amistre64):

thnx, and good luck ;)

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