How can the vector cross product be calculated when only magnitudes are given?

[glemkat]

1. Find axb: Given the magnitude of lal=3 and the magnitude of lbl=2.


2. Since I have the magnitudes, I thought maybe I could use the equation of axb=lal lbl sin theta.



3. I thought since I am trying to find axb that I could use 90 as the angle theta to find axb. I am thinking that is not a good approach because theta isn't given in the problem. I also thought that I could possibly find theta by the dot product but again, I don't have the actual points to find it. I did calculate the (3)(2)sin90 and that is equal to 6, but I don't think that is the right approach. If someone can give me pointers on how to possibly start this, I will be able to finish it up.

Any help would be greatly appreciated!

 

[benorin]

You do not have enough information to calculate the cross product uniquely. Re-read the problem: did they give any more information? did they ask for the answer in terms of something else? if available, check the answer in the back of the book...

Note: the correct formula is: | \vec{a}\times\vec{b}| =|\vec{a}|\cdot |\vec{b}|\sin\theta[/tex].

 

[glemkat]

You were right. I was missing the picture. I can use the formula that you gave above because the angle is 90 so I was right before.

Thanks for your help is suggesting that I was missing some information.