Use the cross product and dot product to find the angel of these two vectors?

Question

I am getting confused with my calculations. I dont know whether I did this right. Please help me out.

A = (10m)i + (20m)j

B = (-15m)i + (-20m)j

Cross product (vector product):

A x B = [(Ax)i + (Ay)j] x [(Bx)i + (By)j]

=[(Ax)i x (Bx)i] + [(Ax)i x (By)j] + [(Ay)j x (Bx)i] + [(Ay)j x (By)j]

=0 + k + (-k) + 0

=(Ax x By)k -(Ay x Bx)k

=(10 x (-20))k - (20 x (-15)k

=-200k - (-300)k = 100k

so...

B x A = -100k

A x B = (AyBz - AzBy)i + (AzBx - AxBz)j + (AxBy - AyBx)k =-400

Magnitude:

IAI = squareRoot(10^2 + 20^2 + 100^2) = 102

IBI = squareRoot((-15)^2 + ((-20)^2 + (-100)^2 =103

sin theta = A x B / (|A||B|) = -400 / (102 x 103) = -0.04

Dot product (scalar product):

note: (*) is dot or multiply

A * B = (Ax)i(Bx)j + (Ay)i(By)j

= (10)(-15) + (20)(-20)

= -550

IAI = squareRoot(10^2 + 20^2 + 0^2) =22.4

IBI = squareRoot((-15)^2 + ((-20)^2 + 0^2 =25.0

cos theta = A * B / (|A||B|) = -550 / (22.4 x 25.0) = -0.98

And from that then on, I'm stuck. Is this correct? Pleaseee help!

Answer 1

AXB = A x B sintheta

100 = root(500) x root(625) sintheta  so theta = 10.3 degrees

A.B = A . B costheta

550 = root(500) x root(625) x costheta   theta = 10.3 deg

you almost had it. Don't include 100^2 in modulus of A or B. Once you have the sine, or cosine, take inverse sine or cosine to find theta. calculator in degree mode.