Q.Prove that $[\vec a-\vec b,\ \vec b-\vec c,\ \vec c-\vec a]=0$.

Step 1. Notice the sum.

Step 2. Apply the coplanarity criterion (Theorem 6.5). Three vectors $\vec p,\vec q,\vec r$ are coplanar iff there exist scalars $r,s,t$, not all zero, with $r\vec p+s\vec q+t\vec r=\vec 0$. Here $1\cdot(\vec a-\vec b)+1\cdot(\vec b-\vec c)+1\cdot(\vec c-\vec a)=\vec 0$ with all three coefficients equal to $1\ne0$ — so this criterion is satisfied automatically.

Step 3. Conclude coplanarity. $\vec a-\vec b,\ \vec b-\vec c,\ \vec c-\vec a$ are ALWAYS coplanar, for every choice of $\vec a,\vec b,\vec c$.

Step 4. Apply Theorem 6.4. Since they are coplanar, their scalar triple product is $0$:

(Cross-check by direct expansion: $(\vec a-\vec b)\cdot\big[(\vec b-\vec c)\times(\vec c-\vec a)\big]$ expands, after dropping every term of the form $\vec x\cdot(\vec x\times\vec y)=0$, to $\vec a\cdot(\vec b\times\vec c)-\vec b\cdot(\vec c\times\vec a)=[\vec a,\vec b,\vec c]-[\vec a,\vec b,\vec c]=0$, confirming the identity holds for every $\vec a,\vec b,\vec c$, not just special cases.)*

Notice the three vectors sum to zero, hence are automatically coplanar

• Trying to expand the full determinant symbolically instead of spotting the much faster linear-dependence shortcut
• Thinking the identity holds only for special $\vec a,\vec b,\vec c$ rather than universally