Question

Determine whether u and v are orthogonal, parallel, or neither.u = <15,45>
v= <-5,12>

Answer 1

Is it neither?

Answer 2

Do you know any tests for orthogonality?

Answer 3

I know it's not orthogonal

Answer 4

It doesn't equal zero when you multiply them out

Answer 5

alright, what about parallel? how can you tell when two vectors are pointing in the same direction?

Answer 6

?

Answer 7

To test if they're parallel, I have to see if Cos in the equation equals to 1 ot -1

Answer 8

*or

Answer 9

That works too. If cos(theta) = 1 or -1, that means the angle between the two vectors is either 0 or pi (180), and the vectors are pointing in the same direction, or in opposite directions (negative direction)

Answer 10

When I used the equaltion: u ▪ v = ||u|| ||v||cos I didn't get 1 or -1. I need to know if I'm right

Answer 11

If you didnt get 1 or -1, then you already know they arent parallel, so it must be neither. They are skew vectors.

Answer 12

I need someone to check my work xD just in case

Answer 13

its right. those vectors are pointing in totally different directions. There's no way they are parallel.

Answer 14

can you show your work to prove it?

Answer 15

lets see....
\[||u||=13,||v||=15\sqrt{10}\]\[u\cdot v=-75+540=465\]\[\frac{u\cdot v}{||u||\cdot ||v||}\ne \pm 1\]

Answer 16

yes

Answer 17

And then you would have 465/ 517.45

Answer 18

but you can draw the two vectors and see they arent parallel.

Answer 19

then you set it up like:
Cos theta = 465/ 517.45
=26 degress?

Answer 20

*Degrees.
Anyway, that's why I'm confused?

Answer 21

Shouldn't it = 1 or -1 when you take the cos of it?

Answer 22

as soon as you show that cos(theta) isnt 1 or -1, you are done. The question didnt ask for the angle between the two.
You showed they arent perpendicular since their dot product isnt 0.
You showed they arent parallel since cos(theta) doesnt equal 1 or -1.
So it must be neither. There is nothing more you need to show.

Answer 23

SO...I WAS RIGHT :D

Answer 24

Thank you for reassuring me