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

1. anonymous

@Hero can you help me?

2. IrishBoy123

dot product them

3. anonymous

would it be -23

4. IrishBoy123

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

5. anonymous

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

6. IrishBoy123

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

7. IrishBoy123

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

8. anonymous

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

9. IrishBoy123

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

10. IrishBoy123

you have $$\vec v_1=(2,-5)$$ and $$\vec v_2=(4,-3)$$

11. anonymous

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

12. IrishBoy123

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

13. anonymous

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

14. IrishBoy123

so is the opening post right, then?

15. IrishBoy123

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

16. anonymous

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

17. IrishBoy123

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 $$| \vec b|$$....

18. IrishBoy123

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

19. IrishBoy123

do that for v1, and v2

20. IrishBoy123

and then $$\large \theta = cos^{-1}( \frac {23}{|\vec v_1| \ \vec v_2|})$$ gotta dash...

21. anonymous

This is hard i am getting really confused!

22. anonymous

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

23. IrishBoy123

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?