Q.Let $\vec a=\hat i+\hat j+\hat k,\ \vec b=\hat i$ and $\vec c=c_1\hat i+c_2\hat j+c_3\hat k$. If $c_1=1$ and $c_2=2$, find $c_3$ such that $\vec a,\vec b$ and $\vec c$ are coplanar.

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 …