Let $ \mathbb{K} $ be a field and, $ a=\left(\begin{array}{l}{a_{1}} \\ {a_{2}} \\ {a_{3}}\end{array}\right), b=\left(\begin{array}{l}{b_{1}} \\ {b_{2}} \\ {b_{3}}\end{array}\right), c=\left(\begin{array}{c}{c_{1}} \\ {c_{2}} \\ {c_{3}}\end{array}\right), d=\left(\begin{array}{c}{d_{1}} \\ {d_{2}} \\ {d_{3}}\end{array}\right) \in \mathbb{K}^{3} $
Show that $a,b,c,d$ are in an affine plane if and only if
$det\left(\begin{array}{llll} {a_{1}} & {b_{1}} & {c_{1}} & {d_{1}} \\ {a_{2}} & {b_{2}} & {c_{2}} & {d_{2}} \\ {a_{3}} & {b_{3}} & {c_{3}} & {d_{3}} \\ {1} & {1} & {1} & {1} \end{array}\right)=0$
How can I show this? Thanks in advance!
The determinent is $\det\left(\begin{array}{ccc} b_1-a_1 & & \\ b_2-a_2& \cdots & \cdots \\ b_3-a_3& & \end{array}\right)=0$. Then the linear subspace $U=\operatorname{span}(b-a,c-a,d-a)$ has rank 0,1 or 2.
I think affine plane is defined to be $p+V=\{p+v\mid v\in V\}$, where the point $p$ is in affine space and $V$ is a linear vector (sub)space. So if $\operatorname{rank}(U)=2$, then $a+U$ is a two dimension affine plane and $a,b,c,d\in a+U$. If $\operatorname{rank}(U)\le 1$, $a+U$ is a affine line or point, which can contain in a plane that pass the origin $(0,0,0)$.