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

1. anonymous

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

2. IrishBoy123

3. anonymous

isnt dot multiplying u and v?

4. IrishBoy123

2x3 + (-4)x(-8)

5. IrishBoy123

=?

6. anonymous

so 38?

7. IrishBoy123

yes

8. anonymous

okay and now what do i do?

9. Michele_Laino

better is if we write the lengths of our vectors, like this: $\Large \begin{gathered} \left\| u \right\| = \sqrt {4 + 16} = \sqrt {20} \hfill \\ \left\| v \right\| = \sqrt {9 + 64} = \sqrt {73} \hfill \\ \end{gathered}$

10. IrishBoy123

$$\large cos \theta = \frac{\vec u \bullet \vec v}{|\vec u||\vec v|}$$

11. IrishBoy123

find arccos use calculator!

12. anonymous

Im so confused lol what do I put for cos theta= like what goes in the u and v?

13. IrishBoy123

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

14. anonymous

.9934640523 arccos=6.55

15. IrishBoy123

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

16. anonymous

I got .9945054529

17. IrishBoy123

yep that should give a very accurate answer if you do not round yourself into oblivion! what do you get for $$\theta$$ now?

18. anonymous

What how do I find theta again?

19. IrishBoy123

you say $$cos \theta = .9945054529$$ so ask your calculator for the inverse cosine of $$.9945054529$$ like you did last time....

20. anonymous

.1023736009

21. IrishBoy123

the issue here might be calculator this is what mine looks like and it is a fairly basic one

22. IrishBoy123

see how all information is in the calculator, so i am not rounding anything. that guarantees accuracy

23. IrishBoy123

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...

24. anonymous

Yeah whoa accuracy haha yeah my calculator doesn't read what yours reads I wonder why

25. anonymous

thank you so much can you help me in a last question though?

26. IrishBoy123

sure but make a new thread for each quetsion

27. anonymous

okie dokie!