Angle between vectors via scalar product vs vector product

[andylatham82]

Hello, I have a question about why I can't determine the angle between two vectors using their cross product.

Say there are two vectors in the XY-plane that we want to find the angle between:

A = -2.00i + 6.00j
B = 2.00i - 3.00j

The method to do this would be to work out the scalar product of the vectors, calculate the magnitude of each vector, and use these to determine the angle via this relationship:

A⋅B = AB cos ∅

In the example above, this gives a correct angle of 165°.

However, I feel like it should be possible to arrive at the same answer using a vector product method instead. So I tried calculating the vector product, and used it with the calculated magnitudes of the vectors and the following relationship:

AXB = AB sin Φ

However, using this method results in an angle of 15.3°.

I must be missing something in the way all of this works and wondered if anyone could provide me with the knowledge I'm missing!

Thanks!
Andy

 

[DrClaude]

It is because $\sin(180^\circ - \theta) = \sin(\theta)$. The arc sin on your calculator usually returns a value between -90° and 90°, so you have to check if the angle you get is the correct one, or if you have to take $180^\circ - \theta$.

 

[andylatham82]

It is because $\sin(180^\circ - \theta) = \sin(\theta)$. The arc sin on your calculator usually returns a value between -90° and 90°, so you have to check if the angle you get is the correct one, or if you have to take $180^\circ - \theta$.

Ahh of course! That's solved the mystery! Thanks a lot DrClaude :)

 

[Erland]

And since the angle between two vectors always lies between 0 and 180 degrees, but not always between -90 and 90 degrees, the scalar product is more suitable than the vector product for this type of problem.