Question

Find the angle between vectors u=<-1,5> and v=<3,-2>. Help!!! Please!! Will give MEDAL!!

Answer 1

angle between two vectors is the dot product, isnt' it?

http://www.mathsisfun.com/algebra/vectors-dot-product.html

Answer 2

So 4.9??????

Answer 3

I got 135.1 deg

what did you get for your dot product of the two vectors?

and what did you get for |a| * |b|?

Answer 4

I got a=4 an B=2.2 I know that wrong so how did you get A and B?

Answer 5

so its like this:
u (dot) v = |u| * |v| cos(angle between)



u (dot) v = x_u * x_v + y_u * y_v

if vector u=<-1,5>
if vector v=<3,-2>

then
u (dot) v =
= -1*3 + 5*2
=?

Answer 6

7

Answer 7

ya

now do the cross product of u and v

Answer 8

|u| x |v|

Answer 9

like <-1,5>*<3,-2>???<3,-10>

Answer 10

to do a cross product, you find the magnitutde of both vectors and multiply them together

i'll find the magnitude of vetor u,
sqr(-1^1 + 5^2)
= sqr(1+25)
= 5.099

whats the magnitude of vector v?

Answer 11

2.236???

Answer 12

sqrt(3^2 + -2^2)
=sqrt(9+4)
=3.6

Answer 13

u (dot) v = |u| * |v| cos(angle between)
-13 = 5.09*3.6 * cos(angle between)

Answer 14

So it its the absolute value of it??

Answer 15

ohhh, okok when we are dealing with vectors and we have the absolute value things like this: |v| it means to find the magnitude, aka lenght of the vector

remember triangles and when we did pythagorean theorem? a^2 + b^2 = c^2?

this is all finding the magnitude is, we are given the lenghts for side 'a' and side 'b' as givn in the vector

Answer 16

Okay I see how you got -13=5.09*3.6*cos now.. So what would you do after that???

Answer 17

now we isolate cos()

-13=5.09*3.6*cos
-13/(5.09*3.6) = cos(angel between)

then we do arc cosine, or inverse cosine
cos^-1(-13/(5.09*3.6) ) = angle between

Answer 18

135.1

Answer 19

^_^

Answer 20

correct

Answer 21

Thank You !! Lots!!! you don't know but this has been very helpful!