Test Question: Vector Proof Help

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SUMMARY

The discussion centers on the mathematical proof regarding the existence of a vector \( v \) in \( \mathbb{R}^3 \) such that \( v \cdot v = ||v|| \). Participants confirm that the statement is true, particularly for unit vectors. The example provided demonstrates that for a unit vector \( v = \frac{1}{\sqrt{14}}(1, 2, 3) \), the dot product \( v \cdot v \) equals 1, which matches the magnitude \( ||v|| \). The conclusion is that unit vectors can indeed be used in this proof, and the equation holds true.

PREREQUISITES
  • Understanding of vector notation and operations in \( \mathbb{R}^3 \)
  • Familiarity with the concepts of dot product and vector magnitude
  • Knowledge of unit vectors and their properties
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the properties of unit vectors in vector spaces
  • Learn about vector dot products and their geometric interpretations
  • Explore the implications of vector norms and magnitudes in higher dimensions
  • Investigate common misconceptions in vector mathematics
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Students studying linear algebra, mathematics enthusiasts, and anyone seeking clarity on vector properties and proofs in \( \mathbb{R}^3 \).

Dahaka14
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Sorry for notation guys, but I don't know how to use LaTex.

Homework Statement


True or false, there exists a vector v (or set of vectors) in R3 such that v*v = ||v|| (v dot v equals the magnitude of v).


Homework Equations


At first I thought this was false, but then I considered an arbitrary unit vector v=(1/(14)/\(-1/2)*(1,2,3)...in words: one over the square root of fourteen times the vector one, two, three (the unit vector of (1,2,3)).


The Attempt at a Solution



Taking v*v, u get (1*1)/14 + (2*2)/14 + (3*3)/14 = 1/14 + 4/14 + 9/14 = 14/14 = 1, which is trivial for a unit vector. Also, the magnitude is just the square root of this answer, since the components are already squared for dotting itself, which is one; again, trivial. Am I correct or can I not consider a unit vector?
 
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You just demonstrated that the answer it true. Why are you doubting yourself?
 
I am doubting myself because I had strong opposition from 3 classmates that for some reason you can't use a unit vector for the proof.
 
Well, we can't really tell you if it said that you can't use unit vectors. Regardless, it's clearly true for any unit vector because, using your notation, the equation says ||v||^2 = v*v = ||v||.

Edit: Yeah, I'm really out of it today.
 
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