Vector. Is there an inverse of dot and cross product?

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  • #1
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Given the following cross product equation:
[itex]\vec{A}[/itex][itex]\times[/itex][itex]\vec{B}[/itex]=[itex]\vec{C}[/itex]
How to express [itex]\vec{A}[/itex] in term of [itex]\vec{B}[/itex] and [itex]\vec{C}[/itex] (or [itex]\vec{B}[/itex] in term of [itex]\vec{A}[/itex] and [itex]\vec{C}[/itex] ). I think the question I want to ask can also be rephrased as if one was told that a known vector when cross product with an unknown vector, yield a known vector, find the unknown vector.

Similarly,
Given the following dot product equation:
[itex]\vec{D}[/itex][itex]\bullet[/itex][itex]\vec{E}[/itex]=k
How to express [itex]\vec{D}[/itex] in term of [itex]\vec{E}[/itex] and k. Similarly also, the question I want to ask can be rephrased as if one was told that a known vector when dot product with an unknown vector, yield a known scalar, find the unknown vector.

My personal thought is, it can't be done. Never heard of a "division" in vector operation. But, maybe I am wrong.
 

Answers and Replies

  • #2
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That's an ill posed problem. Take the vector v = (0, 0, 1); I can find two vectors orthogonal to v, and so we have two vectors who's dot product with v is zero. Given "0v", there is no way to recover information about the vector that created the dot product.
 
  • #3
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No it can't be done as there are many different correct answers.
 
  • #4
Simon Bridge
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You mean, u.v=k, given u and k, find v?
Consider the geometry of this...

k=u.v.cos(A), where A is the angle between them.
Thus k/u = vcosA giving one equation and two unknowns (the magnitude of the vector and it's angle to u.)

So to solve the problem you also need the direction of the vector you want to find or it's magnitude.

It's similar with a cross product.
 
  • #5
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Each of these ultimately boil down to solving a large numerical equation (or system of equations). The answer will generally not be unique and its rather pointless to do so. Say we are given[tex]\vec{x}\cdot(3, -2, 1) = 6[/tex]then we are essentially solving[tex]3x_{1} - 2x_{2} + x_{3} = 6[/tex]for arbitrary numbers [itex]x_{1}, x_{2}, x_{3}[/itex]. By specifying restrictions on the form of [itex]\vec{x}[/itex], we can reduce the number of vectors satisfying the equation. Similarly, the cross product will reduce to essentially solving the system obtained by equating the components of the vectors.
 
  • #6
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That's an ill posed problem. Take the vector v = (0, 0, 1); I can find two vectors orthogonal to v, and so we have two vectors who's dot product with v is zero. Given "0v", there is no way to recover information about the vector that created the dot product.
No it can't be done as there are many different correct answers.

Each of these ultimately boil down to solving a large numerical equation (or system of equations). The answer will generally not be unique and its rather pointless to do so. Say we are given[tex]\vec{x}\cdot(3, -2, 1) = 6[/tex]then we are essentially solving[tex]3x_{1} - 2x_{2} + x_{3} = 6[/tex]for arbitrary numbers [itex]x_{1}, x_{2}, x_{3}[/itex]. By specifying restrictions on the form of [itex]\vec{x}[/itex], we can reduce the number of vectors satisfying the equation. Similarly, the cross product will reduce to essentially solving the system obtained by equating the components of the vectors.
Hence it is still do-able but the answer is never unique and there are infinitely many correct answer. Got it.

You mean, u.v=k, given u and k, find v?
Consider the geometry of this...

k=u.v.cos(A), where A is the angle between them.
Thus k/u = vcosA giving one equation and two unknowns (the magnitude of the vector and it's angle to u.)....
Yes this is exactly what I mean. I came into a situation where I want to reverse the process then I realize it doesn't seem to be reversible, thus I seek for confirmation.
 
  • #7
lurflurf
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As others have said there is too much freedom for and unconstrained unique solution.
You want given a vector v a vector V such that
v.V=1
The natural thing is to work with sets of vectors that span the space
Like thereciprocal_basis.
Thus (in three space) given
{v1,v2,v3}
we can define
{V1,V2,V3}
such that
v1.V1=1
v2.V1=0
v3.V1=0
v1.V2=0
v2.V2=1
v3.V2=0
v1.V3=0
v2.V3=0
v3.V3=1
 

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