Question

if v1=(2,-5) and v2=(4,-3) then the angle between the two vectors is_____ degrees. (Round your answer to two decimal places)

Answer 1

@Hero can you help me?

Answer 2

dot product them

Answer 3

would it be -23

Answer 4

that's part of the process but where did you get minus sign?

Answer 5

well when you add-8 and -15, you get -23, or did I do it wrong?

Answer 6

-8 and - 15?? same again, where are the signs coming from ?!

Answer 7

2*4 + (-5)*(-3)

Answer 8

what I did was (-2)(4)+(5)(-3) and that is how I -23

Answer 9

\(\vec a \bullet \vec b = <a_{x}, a_{y}>\bullet <b_{x},b_{y}> = a_{x}*b_{x}+a_{y}*b_{y}\)

Answer 10

you have

\(\vec v_1=(2,-5)\) and

\( \vec v_2=(4,-3)\)

Answer 11

actually they are (-2,5) and v2=(4,-3),

Answer 12

can you just check that again because you posted differently above...

Answer 13

oh never mind, I am going stupid, sorry about that, I got mixed up!

Answer 14

so is the opening post right, then?

Answer 15

or is it (-2,5) and v2=(4,-3) ?

Answer 16

yeah the opening post is that right one, soory again!

Answer 17

so the dot prod is 23, right?!?!

next, the dot product is also this:

\(\vec a \bullet \vec b = |\vec a | \ | \vec b| \ cos \theta\), right?

so you need \( |\vec a |\) and \(| \vec b|\)....

Answer 18

\(|\vec a| = \sqrt {a_x^2 + a_y ^2}\)

Answer 19

do that for v1, and v2

Answer 20

and then

\(\large \theta = cos^{-1}( \frac {23}{|\vec v_1| \ \vec v_2|})\)

gotta dash...

Answer 21

This is hard i am getting really confused!

Answer 22

Wait for v1 and v2 in the bottom of the equation, what do i sub them in for?

Answer 23

i showed you above how to calc \(|\vec a|\), which is the magnitude of \(\vec a\)

you do that for v1 and v2

then put them into the equation

i'll do v1

\( |\vec v_1| = \sqrt { 2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt {29}\)

can you do the same for v2?