- #1
Determinant of vector of AXB for 3-D
- Context: Undergrad
- Thread starter dpa
- Start date
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- Tags
- Determinant Vector
Click For Summary
Discussion Overview
The discussion revolves around the proof of the vector cross product formula a × b = |a||b|sin(θ), specifically focusing on the determinant representation of the cross product in three-dimensional space.
Discussion Character
- Exploratory, Technical explanation, Debate/contested
Main Points Raised
- Some participants propose that the expression involving a's represents the determinant, suggesting that it must be the whole thing squared.
- Another participant discusses the norm of a vector in R^n, indicating that it is the sum of the components squared, and challenges the use of this to prove the cross product formula.
- One participant mentions that while the cross product cannot be directly proven using the norm, it can be shown for a directional normal vector that a × b equals n̂ multiplied by the length, where the length is |a × b|.
Areas of Agreement / Disagreement
Participants express differing views on the applicability of the determinant and the norm in proving the cross product formula, indicating that multiple competing perspectives remain without a consensus.
Contextual Notes
There are limitations regarding the assumptions made about the determinant and the definitions of the norm, as well as unresolved mathematical steps in the proof process.
- #2
DonAntonio
- 606
- 2
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
- 4,817
- 134
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
- 146
- 0
Thank You.
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