Test Question: Vector Proof Help

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Homework Help Overview

The discussion revolves around a true or false question regarding the existence of a vector \( v \) in \( \mathbb{R}^3 \) such that the dot product \( v \cdot v \) equals the magnitude \( ||v|| \). Participants are exploring the implications of using unit vectors in this context.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to validate the statement by calculating the dot product of a unit vector and questioning the validity of using unit vectors in their proof. Other participants provide feedback on this reasoning and express uncertainty about the restrictions on vector types.

Discussion Status

The discussion is ongoing, with some participants affirming the correctness of the original poster's reasoning while others express doubt based on peer feedback. There is no explicit consensus, but the conversation is exploring the validity of using unit vectors in the proof.

Contextual Notes

There is mention of opposition from classmates regarding the use of unit vectors, which adds a layer of complexity to the discussion about the problem's constraints.

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.
 
Last edited:

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