Linear Algebra (Linearly Dependent)

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Homework Help Overview

The problem involves determining the conditions under which the vectors a = [2, 2, 2], b = [3, 0, 1], and an unknown vector c are linearly dependent. The original poster is exploring the concept of linear dependence and the use of the dot product and triple product in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use the triple product to derive a relationship involving c, leading to the equation 2c1 + 4c2 - 6c3 = 0. Some participants suggest expressing c in terms of parameters c2 and c3, while others explore alternative methods to find c without relying on the triple product.

Discussion Status

Participants are actively discussing various approaches to express the vector c in terms of other parameters. There is a suggestion to consider c as a linear combination of vectors a and b, indicating a productive direction in the exploration of linear dependence.

Contextual Notes

There is an implicit assumption that the vectors a and b are linearly independent, which influences the discussion about the form of vector c. The participants are navigating through the constraints of the problem without reaching a definitive conclusion.

tweety24
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Homework Statement



a = [2, 2, 2]
b = [3, 0, 1]

Find all vectors c so that the vectors a, b, c are linearly dependent.


Homework Equations



(a x b) . c
( . = dot product)...is this how I'm supposed to get started?

linearly dependent would mean make it equal to zero right?



The Attempt at a Solution



when i use the triple product i get 2c1 + 4c2 - 6c3 = 0
i don't know where to go from there
 
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That's one way to express it. The set of all vectors c=[c1,c2,c3] such that 2c1 + 4c2 - 6c3 = 0. If you want to go one step further you solve your equation for, say c1. Then you could express c in terms of just the two parameters c2 and c3. If you think about it a little bit, you could probably find a way to express c in terms of two parameters without even going through the triple product.
 
Thanks =)

So if I was going to solve for c1 would it be something like

c1 = -2c3 + 3c3
 
What do you mean 'something like'?? Sure! So c=[-2c3+3c3,c2,c3] for any choice of c2 and c3. Now can you find a way to bypass the triple product?
 
okay sweet, thank you! =)

Not really, I'm not sure how it would work the other way without the triple product.
 
tweety24 said:
okay sweet, thank you! =)

Not really, I'm not sure how it would work the other way without the triple product.

How about c=s*[2, 2, 2]+t*[3, 0, 1]?
 
Dick said:
How about c=s*[2, 2, 2]+t*[3, 0, 1]?
What Dick is suggesting here is the simplest, most straightforward approach. If you want to find all vectors c so that {a, b, c} is a linearly dependent set, and you can tell by inspection that a and b are linearly independent, then c must be a linear combination of a and b. This is exactly what Dick's equation represents.
 

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