- #1
Determinant of vector of AXB for 3-D
- Thread starter dpa
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The discussion focuses on proving the equation aXb=|a||b|sin(theta) in the context of vector mathematics. Participants clarify that the determinant is involved in the proof, suggesting it relates to the squared magnitude of vectors. The norm of a vector in R^n is defined as the sum of the components squared, which is relevant to the proof. It is noted that while this approach cannot directly prove the expression for AXB, it can establish a relationship involving a directional normal vector. Overall, the conversation emphasizes the mathematical foundations necessary for understanding the cross product in three-dimensional space.
- #2
DonAntonio
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dpa said:Hi all,
This is a beginning step in proving aXb=|a||b|sin(theta)
thank you
Assuming that expression with a's is the determinant, then yes: it must be the whole thing squared.
DonAntonio
- #3
- #4
chiro
Homework Helper
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Hey dpa.
These brackets refer to the norm and the norm of a vector in R^n is simply is the sum of the components squared. [This norm is simply the ||x-0|| for some vector x]
Also you can't use this to prove what AXB is, but you can prove that for some directional normal vector n^ then you can prove that AXB = nhat*length where length is |AXB|.
These brackets refer to the norm and the norm of a vector in R^n is simply is the sum of the components squared. [This norm is simply the ||x-0|| for some vector x]
Also you can't use this to prove what AXB is, but you can prove that for some directional normal vector n^ then you can prove that AXB = nhat*length where length is |AXB|.
- #5
dpa
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Thank You.
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