Coplanarity of b×c, c×a, a×b

Three vectors are coplanar exactly when their scalar triple product is zero. We use the result

[b×c c×a a×b] = [a b c]².

Proof of the result

(c × a) × (a × b) = a[(c × a) · b] − b[(c × a) · a] = [a b c] a, using P × (Q × R) = Q(P · R) − R(P · Q) and the fact that c × a is perpendicular to a.

Then [b×c c×a a×b] = (b × c) · [(c × a) × (a × b)] = [a b c] (b × c) · a = [a b c]².

Coplanar case

If a, b, c are coplanar, [a b c] = 0, so [b×c c×a a×b] = 0² = 0. Hence b × c, c × a, a × b are coplanar.

(Geometrically: when a, b, c lie in one plane, all three cross products are perpendicular to that plane, so they are parallel to each other, hence coplanar.)

Non-coplanar case

If a, b, c are non-coplanar, [a b c] ≠ 0, so [b×c c×a a×b] = [a b c]² > 0, which is non-zero. Hence b × c, c × a, a × b are non-coplanar.

So b×c, c×a, a×b are coplanar if a, b, c are coplanar and non-coplanar if a, b, c are non-coplanar. (The condition works both ways, since [a b c]² = 0 only when [a b c] = 0.)