Scalar Triple Product

Definition. For three vectors $\bar a,\bar b,\bar c$ (order matters), the scalar triple product is

$\bar a\cdot(\bar b\times\bar c)$, written $[\bar a\ \bar b\ \bar c]$; for $\bar a=a_1\hat\imath+a_2\hat\jmath +a_3\hat k$ etc.,

It is also called the box product.

Properties (from properties of determinants).

(1) A cyclic change of the three vectors does not change the value: $[\bar a\ \bar b\ \bar c]=[\bar c\ \bar a\ \bar b]=[\bar b\ \bar c\ \bar a]$ (each cyclic shift is two row-interchanges, which cancel).

(2) A single interchange of any two vectors flips the sign: $[\bar a\ \bar b\ \bar c]=-[\bar b\ \bar a\ \bar c]=-[\bar c\ \bar b\ \bar a]=-[\bar a\ \bar c\ \bar b]$.

(3) The product is zero exactly when the three vectors are coplanar -- in particular when one of them is

$\bar 0$, or any two of them are collinear (a determinant with a repeated or dependent row is zero).

(4) Dot and cross may be interchanged without changing the value: $\bar a\cdot(\bar b\times\bar c)=(\bar a\times\bar b)\cdot\bar c$ (from property (1) plus the commutativity of the dot product).

Theorem 7 (Volume of a parallelepiped). The volume of the parallelepiped with coterminus edges $\bar a,\bar b,\bar c$ (i.e. $\overrightarrow{OA}=\bar a,\overrightarrow{OB}=\bar b,\overrightarrow{OC}=\bar c$) is

$|[\bar a\ \bar b\ \bar c]|$. Sketch: the base parallelogram $OBDC$ has area $|\bar b\times\bar c|$; the

height is the scalar projection of $\bar a$ onto the direction of $\bar b\times\bar c$, namely $\dfrac{\bar a\cdot(\bar b\times\bar c)}{|\bar b\times\bar c|}$; multiplying base $\times$ height gives volume $=\bar a\cdot(\bar b\times\bar c)=[\bar a\ \bar b\ \bar c]$.

Theorem 8 (Volume of a tetrahedron). The volume of the tetrahedron with coterminus edges $\bar a,\bar b,\bar c$ is $\tfrac16\left|[\bar a\ \bar b\ \bar c]\right|$, since a tetrahedron's volume is $\tfrac13$(base

area)(height) $=\tfrac13\cdot\tfrac12|\bar b\times\bar c|\cdot$(same height as the parallelepiped)

$=\tfrac16|[\bar a\ \bar b\ \bar c]|$.

Coplanarity of four points. Four points $A(\bar a),B(\bar b),C(\bar c),D(\bar d)$ are coplanar iff

$\overrightarrow{AB}\cdot(\overrightarrow{AC}\times\overrightarrow{AD})=0$.

Worked examples.

• $\bar a\cdot(\bar b\times\bar c)$ is computed directly as a $3\times3$ determinant for given component vectors, and if the result is $0$, the three vectors are declared coplanar.
• The volume of a parallelepiped with given coterminus-edge vectors is the absolute value of that same determinant.
• A vector orthogonal to a given vector $\bar a$ and coplanar with two other given vectors $\bar b,\bar c$ is produced by the vector triple product $\bar a\times(\bar b\times\bar c)=(\bar a\cdot\bar c)\bar b-(\bar a\cdot\bar b)\bar c$, which is automatically a combination of $\bar b,\bar c$ (hence coplanar with them) and …