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

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

Mathematics
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I got this much so far, is it right? u=sqrt. 2^2+-4^2 v=sqrt.3^2+-8^2 u=4.5 v=8.5 dot=38.25
look at your dot again
isnt dot multiplying u and v?

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Other answers:

2x3 + (-4)x(-8)
=?
so 38?
yes
okay and now what do i do?
better is if we write the lengths of our vectors, like this: \[\Large \begin{gathered} \left\| u \right\| = \sqrt {4 + 16} = \sqrt {20} \hfill \\ \left\| v \right\| = \sqrt {9 + 64} = \sqrt {73} \hfill \\ \end{gathered} \]
\(\large cos \theta = \frac{\vec u \bullet \vec v}{|\vec u||\vec v|} \)
find arccos use calculator!
Im so confused lol what do I put for cos theta= like what goes in the u and v?
calculate the RHS first then, ie using your numbers: \(\large \frac{38}{4.5 \times 8.5}\) what do you get we NEED TO optimise later but do this for now
.9934640523 arccos=6.55
yeah! that's a correct answer for "your" numbers but as @Michele_Laino pointed out, your rounding is going to let you down use these instead \(\huge \frac{38}{\sqrt{20}\times \sqrt{73}}\) your calculator will now compute these square roots for you to a very high dgeree of accuracy, and you will get a much more accurate answer
I got .9945054529
yep that should give a very accurate answer if you do not round yourself into oblivion! what do you get for \(\theta\) now?
What how do I find theta again?
you say \(cos \theta = .9945054529\) so ask your calculator for the inverse cosine of \( .9945054529\) like you did last time....
.1023736009
the issue here might be calculator this is what mine looks like and it is a fairly basic one
see how all information is in the calculator, so i am not rounding anything. that guarantees accuracy
i've also inadvertently given you the answer (!!) .... but anyways compare that to the 6.55 you originally calculated. the question wants you to be accurate to within 1/10 th of a degree, so 6.55 would have failed...
Yeah whoa accuracy haha yeah my calculator doesn't read what yours reads I wonder why
thank you so much can you help me in a last question though?
sure but make a new thread for each quetsion
okie dokie!

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