Question

Finding angle between vector F and S.
F= (4,-4) and S=(4,3).
I tried doing Cosθ= f⋅s/ lFl x lSl
I ended up getting θ=90?
but thats because absolute value of F ended as 0 ( sqrt[(4^2)+(-4^2)]= 0 but i don't think thats right.. lol help please?

Answer 1

|F| = sqrt( 4^2 + 4^2 )
|F| = sqrt( 16 + 16 )
|F| = sqrt( 2*16 )
|F| = sqrt( 16*2 )
|F| = sqrt(16)*sqrt(2)
|F| = 4*sqrt(2)

Not sure how you got |F| = 0 which implies that vector F is 0 units long

Answer 2

|S| = sqrt(4^2 + 3^2)
|S| = sqrt(16+9)
|S| = sqrt(25)
|S| = 5

Answer 3

F dot S = <4, 4> dot <4,3>
F dot S = 4*4 + 4*3
F dot S = 16+12
F dot S = 28

Answer 4

do you see how to finish up?

Answer 5

Sorry ! I had a typo, my Fy=-4 not 4 sorry i just corrected it

Answer 6

Not much will change

|F| = sqrt( 4^2 + (-4)^2 )
|F| = sqrt( 16 + 16 )
|F| = sqrt( 2*16 )
|F| = sqrt( 16*2 )
|F| = sqrt(16)*sqrt(2)
|F| = 4*sqrt(2)

Answer 7

F dot S = <4, -4> dot <4,3>
F dot S = 4*4 + (-4)*3
F dot S = 16-12
F dot S = 4

Answer 8

Ah. ok I see what I did, I input my equation into the calculator wrong :( thank you very much!

Answer 9

np