Q.If $\vec a=\hat i-2\hat j+3\hat k,\ \vec b=2\hat i+\hat j-2\hat k,\ \vec c=3\hat i+2\hat j+\hat k$, find

The Vector Triple Product

When you cross three vectors together as $\vec{a}\times(\vec{b}\times\vec{c})$, 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 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 $\vec{b}$ and $\vec{c}$ (the vectors inside the inner bracket); each is scaled by a dot product involving the outside vector $\vec{a}$.

Why the Result Lies in the Plane of $\vec{b}$ and $\vec{c}$

The inner product $\vec{b}\times\vec{c}$ is perpendicular to the plane containing $\vec{b}$ and $\vec{c}$. Crossing $\vec{a}$ with that perpendicular swings the result back into the $\vec{b}$–$\vec{c}$ plane. So the answer must be a combination $\lambda\vec{b}+\mu\vec{c}$ — and the identity tells you exactly what $\lambda$ and $\mu$ are.

Order Matters — the Product Is Not Associative

The brackets are not decoration. In general,

The left side lies in the plane of $\vec{b},\vec{c}$; the right side lies in the plane of $\vec{a},\vec{b}$. Its own expansion is

which is a different vector. Always keep the parentheses where they are given.

Quick Example

Let $\vec{a}=\hat{i}$, $\vec{b}=\hat{j}$, $\vec{c}=\hat{i}$. Then $\vec{a}\cdot\vec{c}=1$ and $\vec{a}\cdot\vec{b}=0$, so

Checking directly: $\vec{b}\times\vec{c}=\hat{j}\times\hat{i}=-\hat{k}$, and $\hat{i}\times(-\hat{k})=\hat{j}$. 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.