Vector Cross Product homework

physstudent1

This is the question:

Two vectors A and B have magnitude A = 3 and B = 3. Their vector product is A X B = -5k+2i. What is the angle Between A and B.

OK so I'll start with what I do know.

I do know that the cross product is the magnitude of A times magnitude of B times sin theta of B.
I end up with

3*3sinTHETA = 5.4 ( i got 5.4 from finding the magnitude with the components that they gave me )

eventually getting an angle of 37 degrees by dividing by 9 and using arcsin

im not sure what this angle is though..I think it is the angle of B but if it is how does that help me to find the angle between A and B ?

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andrevdh

Homework Helper
The $38^o$ angle is the actual angle between the two vectors $\vec{A},\ \vec{B}$.

physstudent1 said:
I do know that the cross product is the magnitude of A times magnitude of B times sin theta of B.
I end up with
This is not true.

The cross product is:
$$\vec A \times \vec B = \hat n |AB \sin \theta_{AB}|$$

where $\hat n$ is a unit vector with direction found by the right hand rule. It is orthogonal to the plane formed by $\vec A$ and $\vec B$.

The MAGNITUDE of the cross product however, can be written as:

$$|\vec A \times \vec B| = |AB \sin \theta_{AB}|$$

Notice that $\hat n$ disappears because it's magnitude is unity (equal to one).

Also note that $\theta_{AB}$ is the smaller angle between vectors $\vec A$ and $\vec B$.

Does that help?

Last edited:

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