# Physics Vector Cross Product problem

1. Nov 22, 2012

### majormaaz

1. The problem statement, all variables and given/known data
Two vectors are given by A = -6 i + 5 j and B = 1 i + 4 j
Find A X B (answer only in terms of i, j, k)
Find the angle between A and B (answer is terms of degrees)

2. Relevant equations
All I was told was that if I set a 3x3 matrix like this:
i j k
-6 5 0
1 4 0
then AxB is the determinant

3. The attempt at a solution
I made the 3x3 matrix and found the determinant to be only -29 k, which I am told is correct.
I have absolutely no idea on how to approach the angle problem. If I may ask, can someone get me started in the right direction for that problem?

2. Nov 22, 2012

### FeynmanIsCool

Start by graphing it that may help. Think of sin,cosine.

Last edited: Nov 22, 2012
3. Nov 22, 2012

### FeynmanIsCool

i is along X axis
j is along y axis
k is along z axis (not relevant)

4. Nov 22, 2012

### Staff: Mentor

The dot product of two vectors is equal to the product of the vector magnitudes times the cosine of the angle between them. The magnitude of the cross product is equal to the product of the vector magnitudes times the sine of the angle between them. The direction of the cross product is perpendicular to the two vectors.

5. Nov 22, 2012

### majormaaz

Thanks for the info, but I'm just trying to understand cross products as it relates to this problem.

So you're basically saying that A X B = ABsin∅? Great! But in this case, I was given A and B as vectors. So would that mean that I would have to find the displacement between A and B, let's call it C, and use law of sines to get the angle? That seems like a bit of work.

6. Nov 22, 2012

### haruspex

You have already calculated A×B, so you can easily determine |A×B|, |A| and |B|. From those calculate sin(∅).

7. Nov 22, 2012

### Staff: Mentor

The dot product gives you the cosine of the angle. Once you know the cosine of the angle, you can get the sine. Or you can use the info that haruspex provided you. Both should give you the same result.