Solutions to cross product, a x u = b

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SUMMARY

The discussion centers on the mathematical problem of finding solutions to the equation \( a \times u = b \) for vectors \( a, b \in \mathbb{R}^3 \) with \( a \neq 0 \). It is established that a solution exists if and only if the dot product \( a \cdot b = 0 \). The general solution is given by the formula \( u = -\frac{a \times b}{|a|^2} + \lambda a \), where \( \lambda \) is a real parameter. The discussion also confirms that the solution set is one-dimensional, derived through the vector triple product identity.

PREREQUISITES
  • Understanding of vector operations, specifically cross products and dot products.
  • Familiarity with vector spaces and linear algebra concepts.
  • Knowledge of the vector triple product identity.
  • Basic algebraic manipulation skills for solving vector equations.
NEXT STEPS
  • Study the properties of the vector triple product in detail.
  • Explore the implications of the dot product in determining orthogonality in vector spaces.
  • Learn about the geometric interpretation of cross products in three-dimensional space.
  • Investigate the concept of one-dimensional solution sets in linear algebra.
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Mathematicians, physics students, and anyone studying vector calculus or linear algebra who seeks to deepen their understanding of vector operations and their applications in solving equations.

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Studying outer product spaces at the moment and thought I'd quickly recap on the cross product when I stumbled across this problem which has me fairly stumped!

If a,b∈R^3 with a≠0 show that the equation a x u = b has a solution if and only if a.b = 0 and fi nd all the solutions in this case.

The answer is,

-((a x b)/|a|^2) + λ a , where λ is a real parameter.


The first part is trivial, but I have no idea how to get to the solution set. Could anybody shed any light on this matter? I would be very grateful.
 
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Can you show that a x [solutions] = b (for all λ)? Can you show that the set of solutions is one-dimensional?
It is possible to derive the answer if you work out the individual components of the equation.
 
First take the cross product of the equation with a:
[itex]a \times (a \times u = b)[/itex]

The vector triple product gives

[itex](a \cdot u) a -a^2 u = a \times b[/itex]

With a little algebra you can then get

[itex]u =-\frac{ a \times b}{a^2} + \frac{a\cdot u}{a^2} a[/itex]
 

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