Linear span of given vectors expresses a plane in $\mathbb R^3$?

Given vectors $$(1, 3, 5), (-2 , -6, -10)$$ and $$(2, 6 , 10)$$ determine whether the linear span of the above is a plane in $$\mathbb R^3$$.

The vectors are linearly dependent nd hence do not form a basis and it is known that the set of linearly dependent vectors in $$R^2$$ are collinear.

So based on the above can it be said that linearly dependent vectors in $$R^3$$ will form a plane.

• You are right. The span a plane. – Kavi Rama Murthy Apr 18 at 7:42
• Thanks for the confirmation – tendo Apr 18 at 7:49
• Your final sentence is incorrect. For example, the vectors $(1,0,0)$, $(2,0,0)$ and $(3,0,0)$ are linearly dependent, but their span is a line, not a plane. You need another condition to ensure that the span is a plane. – amd Apr 18 at 18:18
• @amd But these vectors would only cover one dimension i.e. x axis? – tendo Apr 18 at 18:40
• Exactly. It can’t be said that linearly-dependent vectors in $\mathbb R^3$ will form a plane, contrary to your last sentence. They might, but you need other conditions besides linear dependence. – amd Apr 18 at 18:47

Your intuition is correct. It can be more rigorously stated, considering that the equation of a plane in $$\mathbb R^3$$ is :
$$Ax + By + Cz = 0$$