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

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Homework Help Overview

The discussion revolves around finding the dot product of two vectors, u and v, given the constraints ||u+v||=1 and ||u-v||=5. The context suggests a focus on vector properties and relationships, particularly in Euclidean spaces.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss initial attempts to visualize the problem and apply the Pythagorean Theorem, noting that u and v are not orthogonal. There are considerations of squaring the norms and the challenges of dealing with multiple variables. Questions arise about the uniqueness of solutions and the dimensionality of the space involved.

Discussion Status

The conversation is ongoing, with participants exploring various interpretations of the problem. Some have suggested specific examples of vectors in different dimensions, while others question the implications of the constraints and the nature of the solutions.

Contextual Notes

There is a lack of clarity regarding the dimensionality of the space (R^1, R^2, or R^3), which influences the nature of the solutions. Participants express uncertainty about the implications of rotations and the scope of the problem.

kuahji
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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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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?
 
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.
 
How about R^1? u=3, v=-2. Is that good enough? If it's R^2, u=(3,0), v=(-2,0). Etc.
 

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