Question

Please help me finish this pre-calculus question!!

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

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

This is how far I got:

cos(theta)= u * v/ IIuII IIvII
cos(theta)= <2, -4> * <3, -8>/ II<2, -4>II II<3, -8>II
cos(theta)= (2)(-4) + (3)(-8)/ sqrt((2)^2 + (-4)^2)) sqrt((3)^2 + (-8)^2))
cos(theta)= -8+24/sqrt(20) sqrt(73)
cos(theta)= 16/ sqrt(20) sqrt(73)
cos(theta)= 16/2*sqrt(365)
cos(theta)= 8/sqrt(365)

I dont know how to get a degree value from this? Please help!

Answer 1

\[\tan^{-1} (\frac{ y_2-y_1 }{ x_2-x_1 }\]

Answer 2

I use tan^-1= y2-y1/x2-x1 to find a degree value for cos(theta)= 8/sqrt(365)? Isnt that the slope formula? @saseal

Answer 3

imagine it this way

Answer 4

Okay. @saseal

Answer 5

y2-y1 / x2-x1 gives you the Opposite/Adjacent

Answer 6

-8-(-4)/(3-2)
= -4/1
=-4
Okay @saseal

Answer 7

you made a mistake here

`cos(theta)= (2)(-4) + (3)(-8)/ sqrt((2)^2 + (-4)^2)) sqrt((3)^2 + (-8)^2))`

the numerator should be (2)*(3) + (-4)*(-8)

Answer 8

tan^-1 (-4) for the angle

Answer 9

Thank you for catching that @jim_thompson5910 . Is my denominator correct though?

Answer 10

yes it is

Answer 11

Okay, so if I simplify it further, my end result should be cos(theta)=19/sqrt(365). But I am stuck here because the question is asking for the angle between the given vectors. Im not sure what to do next to get that angle. @jim_thompson5910

Answer 12

you apply the arccos function to both sides

Answer 13

So would it be theta= arccos(19/(sqrt(365))? Would I just have to solve for that? @jim_thompson5910

Answer 14

now evaluate arccos(19/sqrt(365)) with a calculator

Answer 15

some calculators use \(\Large \cos^{-1}\) in place of arccos

Answer 16

I got 6.009 degrees, or 6 degrees if I round it to the nearest tenth. Thats one of my answer choices @jim_thompson5910

Answer 17

correct

Answer 18

Thank you so much for taking the time to help me with this question, its been nagging at me for quite a while now! Thanks a lot @jim_thompson5910

Answer 19

you're welcome

Answer 20

\[u=2 i-4j\]
\[v=3i-8j\]
\[u.v=\left( 2i-4j \right)\left( 3i-8j \right)=\left( 2 \right)\left( 3 \right)+\left( -4 \right)\left( -8 \right)\]
=6+32=38
\[\left| u \right|=\sqrt{\left( 2 \right)^2+\left( -4 \right)^2}=\sqrt{20}=2\sqrt{5}\]
\[\left| v \right|=\sqrt{\left( 3 \right)^2+\left( -8 \right)^2}=\sqrt{73}\]
\[u.v=\left| u \right|\left| v \right|\cos \theta,~where ~\theta~is~the~\angle~\between~u~and~v\]
\[38=2\sqrt{5}\sqrt{73}\cos \theta \]
\[\cos \theta=\frac{ 19 }{ \sqrt{5}\sqrt{73} }\]
\[\theta=\cos^{-1} \left( \frac{ 19 }{ \sqrt{365} } \right)\]