# Vectors and division

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FS98
As far as I know, vectors can only be added, subtracted, or “multiplied” by dot or cross product.

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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Yes, you cannot divide a vector by another vector. Such division is meaningless. You can divide the magnitude of ##\vec F## by the magnitude of ##\vec a## to get a scalar quantity, namely the mass.

FS98
FS98
Yes, you cannot divide a vector by another vector. Such division is meaningless. You can divide the magnitude of ##\vec F## by the magnitude of ##\vec a## to get a scalar quantity, namely the mass.
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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Mentor
Or is there some way to get the answer by manipulated the vector form of the equation?
Not really. A vector equation is three equations (one for each component). So solving it that way would be three equations for one unknown.

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

FS98
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The great advantage that the Complex numbers have over vectors in R2 is that division by a non-zero element is always possible. The same can be said for the advantage of quaternions over vectors in R3.

Because division of complex numbers is defined, the derivative of a complex-valued function can be defined mimicking the definition of real-valued functions (limz->z0 ( f(z)-f(z0) )/(z-z0) ). The consequences are profound and beautiful.

So solving it that way would be three equations for one unknown.
In the case of F = ma each of the three equations yields the same m. In the case of M=r x F you have no unique solution for r or F. So in either case there is no practical use for a vector division operation.

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In the case of F = ma each of the three equations yields the same m. In the case of M=r x F you have no unique solution for r or F. So in either case there is no practical use for a vector division operation.
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.

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

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But is shows the amplitude of one vector, given the other two and the direction you want. That involves division - it allows you to solve and equation which the bare fact that "you can't divide vectors" would seem to forbid.

That involves division
Not of two vectors.

the bare fact that "you can't divide vectors"
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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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.
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.

Dale and A.T.
Gold Member
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.
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?
I understand that there is not an inverse operation for vector of scalar multiplication but the information can be found.