Q.Find the volume of the parallelepiped whose coterminous edges are represented by the vectors $-6\hat i+14\hat j+10\hat k,\ 14\hat i-10\hat j-6\hat k$ and $2\hat i+4\hat j-2\hat k$.

Scalar Triple Product and Coplanarity

The scalar triple product of vectors $\vec a,\vec b,\vec c$ is

$[\,\vec a\ \vec b\ \vec c\,]=\vec a\cdot(\vec b\times\vec c)$, equal to the

determinant of their components. Geometrically its absolute value is the volume of the parallelepiped built on the three vectors, and it is unchanged under cyclic

permutation but changes sign under a swap.

Three vectors are coplanar exactly when this volume is zero:

$[\,\vec a\ \vec b\ \vec c\,]=0.$

This condition, written as a $3\times3$ determinant set to zero, is the standard way to

find an unknown that makes vectors coplanar. Related magnitudes such as

$|\vec b\times\vec c|$ (area of a face) and dot products $\vec a\cdot\vec b$ combine with

the triple product in identities like Lagrange's, letting one relate

$(\vec p\cdot\vec q)^2$ and $|\vec r\times\vec q|^2$ for vectors constrained to be

coplanar.

The scalar triple product and the coplanarity condition it gives are part of the NCERT/CBSE Class 12 Mathematics "Vector Algebra" chapter, matching "scalar triple product and coplanarity of vectors class 12 maths" searches. This determinant-based test is a frequently asked JEE Main and JEE Advanced vector-algebra question.