Prove that any four vectors vec a,vec b,vec c,vec d in space…

Prove that any four vectors $\vec a,\vec b,\vec c,\vec d$ in space are linearly dependent: there exist $\alpha,\beta,\gamma,\delta$, not all zero, with $\alpha\vec a+\beta\vec b+\gamma\vec c+\delta\vec d=\vec 0$.

Step 1: Case: $\vec a,\vec b,\vec c$ dependent (coplanar)

This includes the case where one of them is $\vec 0$. Then $\alpha\vec a+\beta\vec b+\gamma\vec c=\vec 0$ with not all of $\alpha,\beta,\gamma$ zero, and $\delta=0$ completes the relation.

Step 2: Case: $\vec a,\vec b,\vec c$ not coplanar

Then $[\vec a,\vec b,\vec c]\neq0$ and they form a basis, so $\vec d=\alpha'\vec a+\beta'\vec b+\gamma'\vec c$ with (Cramer's rule)

\[ \alpha'=\frac{[\vec d,\vec b,\vec c]}{[\vec a,\vec b,\vec c]},\qquad \beta'=\frac{[\vec a,\vec d,\vec c]}{[\vec a,\vec b,\vec c]},\qquad \gamma'=\frac{[\vec a,\vec b,\vec d]}{[\vec a,\vec b,\vec c]}. \]

The relation $\alpha'\vec a+\beta'\vec b+\gamma'\vec c-\vec d=\vec 0$ has $\delta=-1\neq0$.

Step 3: One formula for all cases (added)

\[[\vec b,\vec c,\vec d]\,\vec a-[\vec a,\vec c,\vec d]\,\vec b+[\vec a,\vec b,\vec d]\,\vec c-[\vec a,\vec b,\vec c]\,\vec d=\vec 0\]

This identity holds for all four vectors. If some coefficient is non-zero, it is the required relation. If all four triple products are zero, all triples are coplanar, and the first case applies to $\vec a,\vec b,\vec c$. In three-dimensional space at most three vectors can be independent. $\blacksquare$