How to come up with a set of three linearly dependent vectors in a systematic way

Question

Give an example of three linearly dependent vectors in $\mathbb{R}^{3}$ such that none of the three is a multiple of another.

Three vectors that satisfy this property are the vectors :$\{(-1,2,1), (3,0,-1),(-5,4,3)\}$.

Now that's all fine and dandy, but I only got this solution by looking at the answer. Prior to looking at the solution I was trying to come up in some systematic way the necessary set of vectors. But I failed at that task. My question is how could I come up with a set of vectors in a systematic way instead of depending on blind luck and hoping to get a correct set?

Answer 1

Take two non-zero vectors $\vec u$ and $\vec v$ that are not multiples of one another

(i.e., there is no real number such that $\vec v = c\vec u$).

Take a linear combination of them $\vec w=a\vec u + b\vec v$, with $a,b\in\mathbb R\setminus\{0\}$.

Then $\vec u, \vec v$, and $\vec w$ are linearly dependent because $a\vec u + b\vec v - \vec w=\vec 0$,

but, if $\vec w=d\vec u$, then $\vec v=\dfrac{d-a}b \vec u$,

contradicting the choice of $\vec u$ and $\vec v$ as not multiples of each other,

and a similar argument shows that $\vec w$ is not a multiple of $\vec v$ either.

The solution you gave is an example of this,

with $\vec u =(-1,2,1), \vec v=(3,0,-1), a=2, $ and $b=-1$.

Answer 2

Just pick three planar $\mathbb{R}^3$-vectors such as $(1,0,0), (0,1,0)$ and $(1,1,0)$, or as an alternative, a closed path of three vectors in the euclidean space, non necessarily contained in the same 2-dimensional subspace of $\mathbb{R}^3$, or can take at least one of them equal to zero, if in your question you are allowing for a vector to be multiple of the zero vector.