# Vector product and vector product angles

1. Sep 8, 2014

### Bassa

Hello! I have a problem in my calculus based physics class regarding vectors. The problem says:

Vectors A and B have a scalar product -6.00 and their vector product has magnitude 9.00 what is the angle between these two vectors?

Here is how I approached it:

-6=|A||B|cos (theta)
9=|A||B|sin (theta)
tan (theta)= sin (theta)/cos (theta)
tan (theta)=9/-6=-56.31 degrees
since the sine is positive and cosine is negative the angle lies in the second quadrant.
180 degrees -56.31 degrees= 123.69 degrees which is approximately 124 degrees.

Now, why does the angle between the scalar product and the vector product of A and B give us the angle between A and B?

Thanks!

2. Sep 8, 2014

### BvU

Scalar product is a number. There is no angle between a number and a vector...

3. Sep 8, 2014

### Bassa

Then how would I explain how I got the right answer?

4. Sep 9, 2014

### BvU

|vector product| / |scalar product| gives you sin/cos, so the tangent. That is enough to extract the angle in $[0, \pi]$

5. Sep 9, 2014

### Bassa

Thanks! That clarifies a lot of things. ^-^