## Trouble with Cross Product

1. The problem statement, all variables and given/known data
Vector A = 3.5i + 1.8j and vector B = 1.7i + 4.8j . Find the components of A x B:

2. Relevant equations
AxB = AB sin(theta)

3. The attempt at a solution
Since vector A=3.5i+1.8j and B=1.7i+4.8j, I translated that into vectors. So, A is 3.94 @ 27.77*, and B is 5.09 @ 70.56*. This means that the angle between A and B is 42.79*.

Using the AxB formula, I have 13.62. However, the problem (it's on WebAssign) wants the i, j, and k components.

I tried the AxB formula with the individual i and j components, and I got 4.04 for the i direction, and 5.87 for the j direction. However, they're both wrong, and I have no idea how to find out the k direction...

Help!

-Angel.
 k is the third dimesion (or 3rd component in this case). A and B are 2 dimensions.
 I understand that the i, j, and k components are the x, y, and z directions, respectively. My frustration is that I believe I'm following the correct formula, but I still get the wrong answer. Where is my reasoning flawed? I have tried the following: AxB = (3.94)(5.09)sin(42.79) = 13.62, however, WebAssign wants the answer in components. So, I tried this: AxB = (3.5i)(1.7i)sin(42.79) = 4.04i and then AxB = (1.8)(4.8)sin(42.79) = 5.87j When I entered these two (out of three) answers, they were both marked wrong.

## Trouble with Cross Product

The formula you try to use is for the magnitude of the vector axb, but the question asks for the vector itself.

Do you know the way to calculate a cross product as if it were a determinant?

axb=|i j k; a_i a_j a_k; b_i b_j b_k|
 When you cross two vectors the resultant vector will be orthogonal to both original vectors. Keep that in mind.
 You can multiply by components, but: i x i=0 (same for j x j) i x j =k j x i = -k Try this. Newer mind the sin. Multiply the two vectors as you'll do for two binomials.