Cross product of 2 vectors of same magnitude

AI Thread Summary
The discussion focuses on calculating the cross product of two vectors A and B, both with magnitude M, forming a 30-degree angle. The cross product formula involves the magnitudes of the vectors and the sine of the angle between them. For the given angle, the calculation leads to the result of (M^2) * sin(30°), which simplifies to (M^2)/2. The importance of understanding the relationship between vector orientation and the resulting cross product is emphasized. Ultimately, the cross product for the specified vectors is confirmed to be (M^2)/2.
Okokya
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Homework Statement


Vectors A and B both have magnitude M. Joined at the tails, they create a 30' angle. What is A x B in terms of M?

Homework Equations

The Attempt at a Solution


0? OR M^2? Sqrt(3)M/3?
 
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Remember that the cross product of two vectors is related to their magnitudes and the sine of the angle between them. Do you remember this?

If they were parallel, the answer would be zero, since the cross product gives you the perpendicular component of one vector along the other. If they were perpendicular, the answer would be M^2.
 
samnorris93 said:
Remember that the cross product of two vectors is related to their magnitudes and the sine of the angle between them. Do you remember this?

If they were parallel, the answer would be zero, since the cross product gives you the perpendicular component of one vector along the other. If they were perpendicular, the answer would be M^2.

So for theta 30', ab sin 30 = (M^2)/2 ?
 
Okokya said:
So for theta 30', ab sin 30 = (M^2)/2 ?

Yep!
 
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