I'm currently stuck with what I reckon to be a simple problem, but a little insight would be greatly appreciated!
Let $a, b, c, d \in \mathbb R^3$ and $dim(span(a - d, b - d, c - d)) \le 2$ (i.e. the vectors $a - d$, $b - d$, $c - d$ are linearly dependant). Prove that $a, b, c, d$ lie on an affine plane.
I'm currently struggling with the meaning of an affine plane (I know the definition) and whether or not vectors are meant to be their endpoints or the whole line leading from the origin to that endpoint. In the first case, proving that would be trivial, since we can prove that $d$ is linearly dependant on $a, b, c$, hence the point defined by d (in terms of the origin) lies within the same affine plane defined by the endpoints of $a, b, c$. But I suspect that's not the case. How should I approach? Thanks in advance!
The vectors $\,a-d,\, b-d,\, c-d\,$ being linearly dependent, one of them, say $\,a-d\,$ for example, is a linear combination of the two others:
$$ a-d = \lambda(b-d) + \mu(c-d) $$
for some $\lambda,\mu\in\mathbb R$. Hence
$$ a = d+\lambda(b-d)+\mu(c-d) \in d+\text{span}(b-d,c-d) $$
which is an affine subspace of ${\mathbb R}^3$. This subspace contains, in addition to $a$, also $\,d$ $\,$(if $\lambda =\mu=0$), $\,b$ $\,$(if $\lambda=1, \mu=0)$ , and $\,c$ $\,$(if $\lambda=0,\mu=1$).
Note that the affine subspace above is a plane if $\,b-d,\,c-d\,$ are linearly independent; or a straight line if they are linearly dependent and not both zero (in which case, however, there are infinite planes containing $\,a,b,c,d$); or a single point if $b-d=c-d=0$, since in the latter case we have $a=b=c=d$.