Question

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

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

Answer 1

Equation to solve this would be:

u ▪ v = ||u|| ||v||cos theta

Answer 2

Yeah. What are you stuck at?

Answer 3

I'm attempting to solve it through the given equation and through the given examples (I can make a print screen of them as well), but I don't seem to be getting any answers that are provided.

Answer 4

Hmm. Did you note that \(u = \dfrac{2}{3}v\)?

Answer 5

So that mean these vectors are in same direction, right?

Answer 6

Yes, I noticed that they are. But where did you get 2/3v from?

Answer 7

Well, I used this formula and I got \(\dfrac{u\cdot v}{||u||~||v||} = 1 =\cos\theta\), so I knew these vectors are in same direction.

Answer 8

Noted.

Answer 9

Hmm actually I am wrong... Sorry, though it is very close to 1...

Answer 10

let's just do calculations

Answer 11

What is ||u|| and ||v|| in the equation, if you don't mind me asking?

Answer 12

\(||u||\) is magnitude of \(u\)

Mathematically if \(u = \langle u_x,u_y\rangle\), then \[\large ||u|| = \sqrt{u_x^2+u_y^2} \]

Answer 13

That makes a lot more sense now.

Answer 14

Does that clear things up for you?

Answer 15

Great, now you know, so try and recalculate.

Answer 16

Now when you say \[u \frac{ 2 }{ x }\], what would the u equal?

Answer 17

Now just realizing that's not a fraction.

Answer 18

About \(u = \dfrac{2}{3}v\)... It's wrong. Because \(\langle2,-4\rangle\neq\dfrac{2}{3}\langle3,-8\rangle = \langle2,-\dfrac{8}{3}\rangle\)

I just made mistake that's all...

Answer 19

Right, but I'm making reference to ||u||.

Answer 20

Exactly what do you mean?

Answer 21

When you gave the equation \[||u|| =\sqrt{u^2_x...}\] I'm unsure of what the values in the parenthesis would mean.

Answer 22

*Not parenthesis, square root.I'm sorry.

Answer 23

These are just x and y components of vector