Scalar triple product

The scalar triple product of three 3D vectors is

$\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C}),$

a scalar. Geometrically, it is the signed volume of the parallelepiped spanned by $\mathbf{A},\mathbf{B},\mathbf{C}$. Positive if $(\mathbf{A},\mathbf{B},\mathbf{C})$ form a right-handed system, negative if left-handed, zero if the three vectors are coplanar.

Determinant formula

$\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ b_x& b_y& b_z\\ c_x& c_y& c_z\end{array}\right|.$

The determinant’s geometric meaning (signed volume) matches the scalar-triple-product’s geometric meaning. Determinants of matrices with rows of vectors are signed volumes — this is exactly why.

Cyclic invariance

The value is unchanged under cyclic permutation of the three vectors:

$\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})=\mathbf{B}\cdot (\mathbf{C}\times \mathbf{A})=\mathbf{C}\cdot (\mathbf{A}\times \mathbf{B}).$

Swapping any two of $\mathbf{A},\mathbf{B},\mathbf{C}$ flips the sign (matches the determinant’s row-swap rule).

A useful corollary: the dot and the cross can be swapped without changing the value,

$\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})=(\mathbf{A}\times \mathbf{B})\cdot \mathbf{C}.$

Coplanarity test

$\mathbf{A},\mathbf{B},\mathbf{C}$ are coplanar iff $\mathbf{A}\cdot (\mathbf{B}\times \mathbf{C})=0$. The parallelepiped collapses to a flat figure with zero volume.

This is one of the cleanest linear-algebra tests for coplanarity / linear dependence of three 3D vectors.

Right-handed vs left-handed

For the standard basis, $\hat{\mathbf{i}}\cdot (\hat{\mathbf{j}}\times \hat{\mathbf{k}})=1>0$ — right-handed. Negative sign would indicate a left-handed (mirror-image) system. Physical conventions about orientation (right-hand rule for cross products, counterclockwise = positive in 2D, outward normals on closed surfaces) all rest on consistent right-handedness.

Application: tetrahedron volume

A tetrahedron with vertices $P,Q,R,S$ has volume

$V=\frac{1}{6}|\overrightarrow{PQ}\cdot (\overrightarrow{PR}\times \overrightarrow{PS})|.$

The factor $1/6$ accounts for the tetrahedron being one-sixth of the spanning parallelepiped.