Triple Products

Beyond dot and cross, two third-order combinations appear in mechanics and electromagnetism: the scalar triple product (a volume) and the vector triple product (used in expanding cross-product cascades, especially for angular-momentum problems).

Concept

Scalar triple product: $[\vec a, \vec b, \vec c] = \vec a \cdot (\vec b \times \vec c) = \begin{vmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{vmatrix}.$ Geometrically, $|[\vec a,\vec b,\vec c]|$ equals the volume of the parallelepiped with edges $\vec a,\vec b,\vec c$. Cyclic permutations preserve the value; swapping two vectors flips the sign. The vectors are coplanar iff the scalar triple product is zero.

Vector triple product: $\vec a \times (\vec b \times \vec c) = (\vec a \cdot \vec c)\,\vec b - (\vec a \cdot \vec b)\,\vec c.$ Mnemonic: BAC$-$CAB. The result lies in the plane of $\vec b$ and $\vec c$.

Derivation

For the vector triple, set up a Cartesian frame with $\vec b$ along $\hat x$, $\vec c$ in the $xy$ plane. Then $\vec b \times \vec c$ is along $\hat z$, and crossing with any $\vec a$ produces a vector in the $xy$ plane, i.e. spanned by $\vec b$ and $\vec c$. Bilinearity forces the form $\alpha \vec b + \beta \vec c$. Determining $\alpha, \beta$ by matching simple cases (set $\vec a = \vec c$) yields BAC$-$CAB.

JEE Worked Example

Q. Show that $\vec a = \hat i + \hat j$, $\vec b = \hat j + \hat k$, $\vec c = \hat k + \hat i$ are non-coplanar and find the volume of the parallelepiped.

Solution. $[\vec a, \vec b, \vec c] = \begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix} = 1(1 - 0) - 1(0 - 1) + 0 = 1 + 1 = 2.$

Volume $= 2$ cubic units. Since it is non-zero, the vectors are non-coplanar.

Traps

• Forgetting the cyclic property: $\vec a\cdot(\vec b\times\vec c) = \vec b\cdot(\vec c\times\vec a) = \vec c\cdot(\vec a\times\vec b)$.
• Mis-remembering BAC$-$CAB as CAB$-$BAC.
• Writing $(\vec a\times\vec b)\times\vec c$ as if it equalled $\vec a\times(\vec b\times\vec c)$; the cross product is non-associative.
• Sign errors when expanding the $3\times 3$ determinant.

Key Takeaways

• Scalar triple = signed volume; zero $\Leftrightarrow$ coplanar.
• $\vec a\times(\vec b\times\vec c) = (\vec a\cdot\vec c)\vec b - (\vec a\cdot\vec b)\vec c$.
• Use cyclic symmetry of scalar triples to simplify before computing.