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.
Quick Example
Let a=i^, b=j^, c=i^. Then a⋅c=1 and a⋅b=0, so
a×(b×c)=(1)j^−(0)i^=j^.
Checking directly: b×c=j^×i^=−k^, and i^×(−k^)=j^. The identity agrees.
The vector triple product identity goes beyond the core NCERT Class 12 Vector Algebra syllabus, but it is an important topic for JEE Advanced and select state CETs, building on the scalar and vector product foundations already laid in the NCERT curriculum. Students searching "BAC CAB rule vector triple product" should master the basic cross product and dot product first, since this identity is really just a compressed combination of both.
Apply the expansion u×(a×u)=∣u∣2a−(u⋅a)u to each of u=i^,j^,k^ and add.
✓Final answer
Sum =3a−[(i^⋅a)i^+(j^⋅a)j^+(k^⋅a)k^]=3a−a=2a. ■
Expand each of the three terms with the vector triple product formula (each unit vector has ∣u∣2=1); the sum (i^⋅a)i^+(j^⋅a)j^+(k^⋅a)k^ that's left over is exactly a itself, decomposed in the standard basis.
Step 1. Expand each term. Using u×(a×u)=(u⋅u)a−(u⋅a)u and ∣i^∣=∣j^∣=∣k^∣=1:
Step 3. Identify the bracketed sum. Writing a=a1i^+a2j^+a3k^: i^⋅a=a1,j^⋅a=a2,k^⋅a=a3, so the bracket is a1i^+a2j^+a3k^=a — this is just a decomposed in the standard basis.
Step 4. Substitute back. Sum =3a−a=2a.
✓Final answer
i^×(a×i^)+j^×(a×j^)+k^×(a×k^)=2a. ■
Vector triple product expansion applied to each basis vector, then recognise the leftover sum as a
Forgetting ∣i^∣=∣j^∣=∣k^∣=1 so the first term of each expansion is simply a
Not recognising (i^⋅a)i^+(j^⋅a)j^+(k^⋅a)k^ as just a