Prove that vec a,vec b,vec c are coplanar if (vec a×vec…

Prove that $\vec a,\vec b,\vec c$ are coplanar if $(\vec a\times\vec b)+(\vec b\times\vec c)+(\vec c\times\vec a)=\vec 0$.

Explanation of Method

A vanishing cyclic sum of vector products forces coplanarity.

Step 1: Dot product with $\vec c$

\[ (\vec a\times\vec b)\cdot\vec c+\underbrace{(\vec b\times\vec c)\cdot\vec c}_{0}+\underbrace{(\vec c\times\vec a)\cdot\vec c}_{0}=0\ \Longrightarrow\ [\vec a,\vec b,\vec c]=0. \]

\[[\vec a,\vec b,\vec c]=0\ \Longrightarrow\ \vec a,\vec b,\vec c\ \text{coplanar}\qquad\blacksquare\]

(Taking the dot product with $\vec a$ or $\vec b$ gives the same.)

Step 2: What the condition really says (added)

\[ (\vec a-\vec c)\times(\vec b-\vec c)=\vec a\times\vec b-\vec a\times\vec c-\vec c\times\vec b=\vec a\times\vec b+\vec b\times\vec c+\vec c\times\vec a. \]

So the condition is $\overrightarrow{CA}\times\overrightarrow{CB}=\vec 0$: the tips $A,B,C$ of the three vectors are collinear. Equivalently, triangle $ABC$ has zero area. This is stronger than coplanarity, and it immediately gives coplanarity: all three vectors lie in the plane through $O$ and the line $ABC$.

$A$ $B$ $C$ $O$ $\vec a$ $\vec b$ $\vec c$

The condition means $(\vec a-\vec c)\times(\vec b-\vec c)=\vec 0$: the tips $A,B,C$ lie on one line (dashed), so $\vec a,\vec b,\vec c$ lie in the plane through $O$ and that line.

Result

\[\boxed{\,[\vec a,\vec b,\vec c]=0\ \Longrightarrow\ \vec a,\vec b,\vec c\ \text{coplanar}\qquad\blacksquare\,}\]

Remark

Note. The source's geometric paragraph says that the three cross products "cannot cancel" unless the vectors are coplanar. The identity above makes this precise. Coplanarity alone is not enough: $\vec a=\vec i$, $\vec b=\vec j$, $\vec c=\vec 0$ are coplanar, yet the sum is $\vec k\neq\vec 0$.