Question

Find the angle between the vectors u and v if u = 3i +2j and v = 4i - 2j. Round answer to two decimal places.

Answer 1

The dot product is\[\overrightarrow{u}*\overrightarrow{v}=|u||v|\cos\theta\]where |u| and |v| are the magnitudes of u and v, respectively.
So we can solve this for theta...

Answer 2

Can you do a dot product?
Can you find the magnitude of the vectors u and v?

Answer 3

yes i can, i just don't know how i would do it with the i and j. do i just leave them out?

Answer 4

\[i*i=1\]\[j*j=1\]\[i*j=0\]so that is what will happen in the dot product. That way you wind up multiplying the individual components only.

Answer 5

ohh i see! see i use the formula cos\[\cos \Theta= (u)(v)/ \left| u \right|\left| v \right|\]

Answer 6

right, where \[(u)(v)\]is the dot product

so inverse cosine both sides solves for theta

Answer 7

i know how to do the dot product thing, just to get the denominator part how?

Answer 8

The magnitude of a vector is essentially the length of the vector.

So considering that each vector is made up of components at right angles to each other, we can use the Pythagorean theorem and treat the length of the vector as the hypotenuse.

Answer 9

therefor\[|u|=\sqrt{(u_i)^2+(u_j)^2}\]where u_i and u_j are the components.
We can do similarly for v.

Answer 10

so what is\[|u|\]in your problem?

Answer 11

u=3i+2j

Answer 12

That is the vector itself.
Use the formula above for the magnitude.
Treat each component as legs of a right triangle and find the hypotenuse.

Answer 13

look at the components:\[u_i=3\]\[u_j=2\]now use the formula above to find the magnitude.

Answer 14

\[\sqrt{13}\]

Answer 15

and v would be\[\sqrt{20}\]

Answer 16

good :)
so what is\[|v|\]???

Answer 17

good

Answer 18

now you can use\[\overrightarrow{u}*\overrightarrow{v}=|u||v|\cos\theta\]and solve for the angle.

Answer 19

what do you have for\[\overrightarrow{u}*\overrightarrow{v}\]???

Answer 20

i got 12-4=8

Answer 21

right
So we are ready to go:\[\overrightarrow{u}*\overrightarrow{v}=|u||v|\cos\theta\]solve for theta...

Answer 22

so 8/ (\[\sqrt{13}\]\[\sqrt{20}\]

Answer 23

inverse cosine

Answer 24

theta equals 60.3

Answer 25

yup!
in degrees sounds about right...
Nice job

Answer 26

Thanks soo much!! :) that is one of the options! Thanks for all your help!

Answer 27

anytime :)