Vectors ; specifically cross product application

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kapitanma
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Homework Statement



vector A = 1.5i + 6.7j - 7.4k
vector B= -8.2i + 6.5j + 2.3k

(f) What is the magnitude of the component of vector A perpendicular to the direction of vector B but in the plane of vector A and B.


The Attempt at a Solution



This part of the problem has me kinda stumped. My attempt at the solution was using the application of the cross product : C = ABsin(theta).

I calculated the angle between the two as 82.43 degrees, and realized that this formula gave me the same thing as simply taking the magnitude of the cross product vector which I calculated to be 107.207.

This shows me that my fundamental approach to this problem is incorrect, but I have no idea where to go with it. Any pointers would be appreciated.
 
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Magnitudes of vectors isn't all you need, you also need directions. It would be nice to find a vector that is perpendicular to B but in the plane of A and B, right? Then you could just find the magnitude of A along that direction. How about (AxB)xB? Can you see why that works?
 
Looking at a graph of AxB, I think I can see why that (AxB)xB would give me a vector perpendicular to B in the plane, and I'd just have to apply the dot product to get the component of A in the direction of B.
 
kapitanma said:
Looking at a graph of AxB, I think I can see why that (AxB)xB would give me a vector perpendicular to B in the plane, and I'd just have to apply the dot product to get the component of A in the direction of B.

Sure. (AxB)xB is perpendicular to B, and it's also perpendicular to AxB which is the normal to plane containing A and B. And, yes, from here you can use a dot product.
 
I successfully solved this problem, thanks for the nudge in the right direction.