Question

Vector A and B have a Scalar product of -10 and their vector product has a magnitude of 8.5. What is the angle between these two vectors?

Answer 1

The scalar product of two vectors

$$({\vec A}\cdot{\vec B})=AB\cos\theta$$

The magnitude of vector product of two vectors

$$|({\vec A}\times{\vec B})|=AB\sin\theta$$

So

$$\tan\theta=\frac{|({\vec A}\times{\vec B})|}{({\vec A}\cdot{\vec B})}$$

$$\tan\theta=\frac{8.5}{-10}=-0.85$$

Finally, the angle between vectors A and B

$$\theta=\tan^{-1}(-0.85)=139.6^{\circ}$$