Question

Find the angle between the given vectors to the nearest tenth of a degree.

u = <2, -4>, v = <3, -8>

Answer 1

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

Answer 2

look at your dot again

Answer 3

isnt dot multiplying u and v?

Answer 4

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

Answer 5

=?

Answer 6

so 38?

Answer 7

yes

Answer 8

okay and now what do i do?

Answer 9

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

Answer 10

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

Answer 11

find arccos
use calculator!

Answer 12

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

Answer 13

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

Answer 14

.9934640523
arccos=6.55

Answer 15

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

Answer 16

I got .9945054529

Answer 17

yep

that should give a very accurate answer if you do not round yourself into oblivion!

what do you get for \(\theta\) now?

Answer 18

What how do I find theta again?

Answer 19

you say \(cos \theta = .9945054529\)

so ask your calculator for the inverse cosine of \( .9945054529\) like you did last time....

Answer 20

.1023736009

Answer 21

the issue here might be calculator

this is what mine looks like and it is a fairly basic one

Answer 22

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

that guarantees accuracy

Answer 23

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...

Answer 24

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

Answer 25

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

Answer 26

sure

but make a new thread for each quetsion

Answer 27

okie dokie!