Find u.v with Given Constraints: ||u+v||=1 and ||u-v||=5

In summary, the problem is to find u.v given that ||u+v||=1 and ||u-v||=5. The Pythagorean Theorem cannot be used since u and v are not orthogonal. Squaring both equations leads to a system with two variables, making it difficult to find a unique solution. However, any solution can be found by rotating u and v by any angle. In this case, u=3 and v=-2 could be a solution for R^1, while u=(3,0) and v=(-2,0) could be a solution for R^2.
  • #1
kuahji
394
2
Find u.v given that ||u+v||=1 & ||u-v||=5.

The first thing I did was drawn a simple picture, it became apparent that u & v wouldn't be orthogonal. So then the Pythagorean Theorem wouldn't work. Next I moved on to squaring both sides
||u+v||[tex]^{2}[/tex]=||u||[tex]^{2}[/tex]+2(u.v)+||u||[tex]^{2}[/tex]
However here again, I didn't seem to be getting anywhere because I can't do anything with the middle term. I also tried squaring the other equation, solving for ||u|| & substituting it into the other equation. But that still left me with two variables in one equations. So I'm kinda lost about what to actually do in this problem.
 
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  • #2
There isn't any unique solution. If you find a u and v that satisfy that then you can always rotate u and v by any angle and they will still satisfy that relation. Are you just supposed to find ANY u and v?
 
  • #3
Yes any solution. We had a similar one earlier that had infinitely many solutions. But it was specified that it was R^2 or R^3. Here it does not say. We haven't learned about rotations yet, that is the next section.
 
  • #4
How about R^1? u=3, v=-2. Is that good enough? If it's R^2, u=(3,0), v=(-2,0). Etc.
 

Related to Find u.v with Given Constraints: ||u+v||=1 and ||u-v||=5

What is linear algebra?

Linear algebra is a branch of mathematics that deals with the study of linear equations and their solutions. It involves the use of algebraic techniques to solve problems related to linear systems of equations, vector spaces, matrices, and linear transformations.

What are the basic concepts in linear algebra?

Some of the basic concepts in linear algebra include matrices, vectors, systems of linear equations, linear transformations, and eigenvalues and eigenvectors.

What are the applications of linear algebra?

Linear algebra has a wide range of applications in various fields such as physics, engineering, computer science, economics, and statistics. It is used for solving problems involving linear systems, optimization, data analysis, and computer graphics.

What is the difference between a vector and a matrix?

A vector is a one-dimensional array of numbers, while a matrix is a two-dimensional array of numbers. Vectors are used to represent magnitude and direction, while matrices are used to represent linear transformations and systems of linear equations.

What are eigenvalues and eigenvectors?

Eigenvalues and eigenvectors are important concepts in linear algebra that are used to analyze linear transformations. Eigenvalues are scalar values that represent the scaling factor of an eigenvector when it is transformed by a linear transformation. Eigenvectors are non-zero vectors that remain in the same direction after being transformed by a linear transformation.

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