Friday
April 29, 2016

# Homework Help: physics (cross product vectors)

Posted by steve on Tuesday, January 23, 2007 at 11:30am.

Two vectors are lying in the xz plane: A=2.00i + 3.00k and B= -9.00i + 2.00k.

a) What is the value of AxB?

Check my solution: I found through calculations that the value is 23j.

b) What is the magnitude |AxB|?

I could do ABsin theta, but I have no angle between. So am confused here.

C) What is the ganle between the two vectors.

Here I could do the:
23
---- = sin theta.
AB

Thank you!

(2.00i + 3.00k)(-9.00i + 2.00k)

ixi=0
ixk= -j
kxi=j

from that, 4(-j) + 27(-j)

magnitude: (sqrt (4^2 + 27^2))

angle between vectors?
arc sin=magnitudeabove/magnitudeA*magnitudeB

I'm still somewhat confused. Does 4(-j) + 27(-j) equal -31j?

also, if you can, please elaborate a bit more on proper steps to conceptualize prob(the rest).

yes, I got -31j

if the cross product is equal to magnitude A * magnitude B * sinTheta, then

crossproduct/(magnitudeA*MagnitudeB)= sinTheta.

Solve for Theta.