Prove analytically and geometrically that vec a-vec b , vec…

Prove analytically and geometrically that $\vec a-\vec b$, $\vec b-\vec c$, $\vec c-\vec a$ are coplanar for any vectors $\vec a,\vec b,\vec c$.

Step 1: Analytical proof

Let $\vec u=\vec a-\vec b$, $\vec v=\vec b-\vec c$, $\vec w=\vec c-\vec a$. Then

\[ \vec u+\vec v+\vec w=\vec 0, \] a non-trivial linear relation (all coefficients equal to $1$). Hence the vectors are linearly dependent, i.e. coplanar. In terms of the triple product: \[[\vec u,\vec v,\vec w]=[\vec u,\vec v,-\vec u-\vec v]=-[\vec u,\vec v,\vec u]-[\vec u,\vec v,\vec v]=0.\]

Determinant view: adding rows 2 and 3 to row 1 of $\det(\vec u,\vec v,\vec w)$ gives a zero row.

Step 2: Geometric proof

With $\vec a,\vec b,\vec c$ the position vectors of $A,B,C$: \[\vec a-\vec b=\overrightarrow{BA},\qquad \vec b-\vec c=\overrightarrow{CB},\qquad \vec c-\vec a=\overrightarrow{AC}.\]

These are the sides of triangle $ABC$ traversed once around ($B\to A\to C\to B$). They all lie in the plane $ABC$. If $A,B,C$ are collinear, the vectors are collinear, which is a special case of coplanar. $\blacksquare$

\[(\vec a-\vec b)+(\vec b-\vec c)+(\vec c-\vec a)=\vec 0\ \Longrightarrow\ [\vec a-\vec b,\ \vec b-\vec c,\ \vec c-\vec a]=0\]