coolkat4
Find the angle between the given vectors to the nearest tenth of a degree. u = <6, 1>, v = <7, 4>?



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Hoa
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i think it is 0+2kpi

coolkat4
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thats no possiable as one of the given solutions..

amistre64
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theres a relationship between cos and dot products, do you recall it?

coolkat4
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yes.

amistre64
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then the angle is simple to be found knowing:\[uvcos~\alpha=u \cdot v\]

amistre64
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solve for alpha :)

coolkat4
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i remember the formula how do i plug the numbers into the formula?

amistre64
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you know how to do a dot product right? and to find lengths of vectors?

coolkat4
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yeah

amistre64
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those are the numbers you need then, what do you get for:
u.v
and the lengths of u and v ?

coolkat4
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38

amistre64
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u = <6, 1>; sqrt(37)
v = <7, 4>; sqrt(65)

u.v 42+5 = 49
uv = sqrt(37*65)
.... what does 38 mean?

amistre64
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\[\cos~\alpha=\frac{u.v}{uv}\]
\[\cos~\alpha=\frac{49}{\sqrt{37*65}}\]
\[\alpha=cos^{1}\frac{49}{\sqrt{37*65}}\]

coolkat4
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the 38 came from 42+(4)

amistre64
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lol ... i do tend to add when i need to multiply when doing that

amistre64
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1*4 is +4, and i did a +5 :)

coolkat4
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okay thank you i got 20.28?

amistre64
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i got something closer to 10
