Linear Algebra (Linearly Dependent)

[tweety24]

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

 

[Dick]

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.

 

[tweety24]

Thanks =)

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

c1 = -2c3 + 3c3

 

[Dick]

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?

 

[tweety24]

okay sweet, thank you! =)

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

 

[Dick]

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]?

 

[Mark44]

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.