Two vectors forming right handed set?

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The discussion revolves around the confusion regarding the concept of a "right handed set" of vectors, typically associated with three vectors, and its application to two vectors. Participants highlight the need to check if the two vectors are perpendicular and discuss the implications of not forming a right handed set, suggesting that the second vector may need to be multiplied by -1. The scalar triple product is mentioned as a method to determine the orientation of vectors, although its application to only two vectors is questioned. Clarification is sought on whether the vectors exist within a two-dimensional vector space. This conversation emphasizes the complexities of vector orientation and dimensionality in physics and mathematics.
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Homework Statement
Do the two vectors form a right handed set
Relevant Equations
Vectors
I am confused by a question. I thought "right handed set" only applied to sets of three vectors. However I have been given 2 vectors and asked "check whether they are perpendicular to each other and if they form a right handed set. If they don't form a right handed set, the second vector must be multiplied by -1".
 
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Can you post the question exactly as was given to you? Please include any accompanying diagrams. Thanks.
 
Andrew Tom said:
Homework Statement:: Do the two vectors form a right handed set
Relevant Equations:: Vectors

I am confused by a question. I thought "right handed set" only applied to sets of three vectors. However I have been given 2 vectors and asked "check whether they are perpendicular to each other and if they form a right handed set. If they don't form a right handed set, the second vector must be multiplied by -1".
to check if vectors form a right handed system, we can use the scalar triple product. Please give the full problem statement when asking for help.
 
MidgetDwarf said:
to check if vectors form a right handed system, we can use the scalar triple product. Please give the full problem statement when asking for help.
That's very difficult using only two vectors.
 
Do the vectors belong to a two-dimensional vector space?
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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