Angle Between Vectors

1. Feb 3, 2009

raptik

1. The problem statement, all variables and given/known data
If A = 1i + 2j + 3k and B = 1i + 2k, and if C = A X B, then the angle between the vector A and the vector C is:

2. Relevant equations
AxB = ((a2b3-a3b2)i + (a3b1-a1b3)j + (a1b2-a2b3)k)
A·B = ABcosθ = AiBi + AjBj + AkBk

3. The attempt at a solution
I got C = 4i + j -2k and A = 1i + 2j + 3k
A·C = 4i + 2j - 6k with a magnitude of 7.48
then I used the magnitude of A and C individually to get 3.74 and 4.58 respectively.
The I used the equation cos-1(7.48/(3.74 x 4.58)) = θ
θ = 64.11 degrees. The actual answer is 90 degrees. What am I doing wrong?

2. Feb 3, 2009

Nabeshin

This is not meant to be a numerical question! Do you know what physical (spacial) property the cross product of a two vectors has with respect to the original vectors?

3. Feb 3, 2009

raptik

I suppose that the cross product is perpendicular to the plane of the initial vectors considering that the answer is 90 degrees. I was not fully aware of this, but if this is the case then I'll keep it in mind.