- #1

FS98

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Does this mean that you couldn’t divide f-> by a-> to get m using the vector form of Newton’s second law? This would require dividing a vector by a vector, which seems to not be allowed.

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- #1

FS98

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Does this mean that you couldn’t divide f-> by a-> to get m using the vector form of Newton’s second law? This would require dividing a vector by a vector, which seems to not be allowed.

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- #3

FS98

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So can you not solve for mass with the vector form of Newton’s second law? Only the scalar form?

Or is there some way to get the answer by manipulating the vector form of the equation?

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- #4

Dale

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Not really. A vector equation is three equations (one for each component). So solving it that way would be three equations for one unknown.Or is there some way to get the answer by manipulated the vector form of the equation?

Basically, this type of operation won’t work on two arbitrary vectors, only on colinear vectors.

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Because division of complex numbers is defined, the derivative of a complex-valued function can be defined mimicking the definition of real-valued functions (lim

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A.T.

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In the case ofSo solving it that way would be three equations for one unknown.

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sophiecentaur

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Unless you choose the direction of the unknown vector. So you can predict the necessary Current in a given Wire to produce a given Force in a Given Field. But that process has reduced things to a scalar operation I guess.In the case ofF= maeach of the three equations yields the same m. In the case ofM=rxFyou have no unique solution forrorF. So in either case there is no practical use for a vector division operation.

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A.T.

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That's 3 inputs, not a division operation on 2 vectors....given Wire to produce a given Force in a Given Field.

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sophiecentaur

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A.T.

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Not of two vectors.That involves division

You can define a "vector division" to be whatever you want it to be. Whether its useful or sensible to call it that, is another question.the bare fact that "you can't divide vectors"

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To make a claim to the term "division", it should be the multiplicative inverse of some type of vector multiplication with a multiplicative identity. The dot and cross products do not have a multiplicative identity, so you would need to start with some new definition of vector multiplication.You can define a "vector division" to be whatever you want it to be. Whether its useful or sensible to call it that, is another question.

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sophiecentaur

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What would one call the process of finding the current if you knew the force and the field (plus the angle of the wire)? If it's not called division then what do you call it?You can define a "vector division" to be whatever you want it to be. Whether its useful or sensible to call it that, is another question.

I understand that there is not an inverse operation for vector of scalar multiplication but the information can be found.

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As you stated earlier, you can just call it "solving for the current". The solution may involve some sort of division by non-vectors, but I would object to calling it vector division. The term "vector division" is just too strong to apply here.What would one call the process of finding the current if you knew the force and the field (plus the angle of the wire)? If it's not called division then what do you call it?

The solution to a particular problem can be found, but that is not the same as having a guaranteed ability to divide by a non-zero vector in all situations.I understand that there is not an inverse operation for vector of scalar multiplication but the information can be found.

- #14

sophiecentaur

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