Q.If the volume of the parallelepiped with a×b,b×c,c×a as coterminous edges is 8 cubic units, then the volume of the parallelepiped with (a×b)×(b×c),(b×c)×(c×a) and (c×a)×(a×b) as coterminous edges is,
When you cross three vectors together as a×(b×c), the result is again a vector — this is the vector triple product. The remarkable thing is that it can always be rewritten without any cross products at all, using only dot products.
The Key Identity (the "BAC − CAB" rule)
a×(b×c)=(a⋅c)b−(a⋅b)c
A memorable way to recall it: the answer is B times (A dot C) minus C times (A dot B) — "BAC minus CAB". The two survivors are b and c (the vectors inside the inner bracket); each is scaled by a dot product involving the outside vector a.
Why the Result Lies in the Plane of b and c
The inner product b×c is perpendicular to the plane containing b and c. Crossing a with that perpendicular swings the result back into the b–c plane. So the answer must be a combination λb+μc — and the identity tells you exactly what λ and μ are.
Order Matters — the Product Is Not Associative
The brackets are not decoration. In general,
a×(b×c)=(a×b)×c.
The left side lies in the plane of b,c; the right side lies in the plane of a,b. Its own expansion is
(a×b)×c=(a⋅c)b−(b⋅c)a,
which is a different vector. Always keep the parentheses where they are given.
Watch out
A frequent slip is to "cancel" and write a×(b×c) as some multiple of a. It is not — the surviving vectors are b and c, never the outer vector. …
The given volume 8 is already [a,b,c]2 (by the standard identity); applying that same squaring identity a second time — now to the vectors a×b,b×c,c×a themselves — squares 8 again.
Step 1. Identify the first volume. Volume with a×b,b×c,c×a as edges is [a×b,b×c,c×a]=[a,b,c]2=8. …
Multiplying by 8 instead of squaring it (giving 64 vs. the wrong 8×8=64... actually both give 64 here, but conceptually: mistaking the relationship for volume×8 rather than volume2, which would diverge for other start …