Find the angle between the given vectors to the nearest tenth of a degree. u = , v = ?

Question

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

anonymous · Answer

i think it is 0+2kpi

anonymous · Answer

thats no possiable as one of the given solutions..

amistre64 · Answer

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

anonymous · Answer

yes.

amistre64 · Answer

then the angle is simple to be found knowing:$$|u||v|cos~\alpha=u \cdot v$$

amistre64 · Answer

solve for alpha :)

anonymous · Answer

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

amistre64 · Answer

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

anonymous · Answer

yeah

amistre64 · Answer

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

anonymous · Answer

38

amistre64 · Answer

u = <6, -1>; sqrt(37) v = <7, -4>; sqrt(65) --------------- u.v 42+5 = 49 |u||v| = sqrt(37*65) .... what does 38 mean?

amistre64 · Answer

$$\cos~\alpha=\frac{u.v}{|u||v|}$$
$$\cos~\alpha=\frac{49}{\sqrt{37*65}}$$
$$\alpha=cos^{-1}\frac{49}{\sqrt{37*65}}$$

anonymous · Answer

the 38 came from 42+(-4)

amistre64 · Answer

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

amistre64 · Answer

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

anonymous · Answer

okay thank you i got 20.28?

amistre64 · Answer

i got something closer to 10

amistre64 · Answer