Understanding Scalar Triple Product in Vector Algebra

The scalar triple product, also referred to as the mixed product, is a scalar quantity defined for three vectors in three-dimensional space by combining the dot product and cross product in a specific sequence. It is central to vector algebra and determinant-based problems.

Analytical Definition and Notation for the Scalar Triple Product

Given three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ in $\mathbb{R}^3$, the scalar triple product is defined as $\vec{a}\cdot(\vec{b}\times\vec{c})$, also notated as $[\vec{a} \; \vec{b} \;\vec{c}]$. This operation yields a real number and measures the oriented volume spanned by the three vectors.

If $[\vec{a} \; \vec{b} \;\vec{c}] = 0$, then the vectors are either coplanar or at least two of them are linearly dependent. The cyclic permutations $[\vec{a}\;\vec{b}\;\vec{c}] = [\vec{b}\;\vec{c}\;\vec{a}] = [\vec{c}\;\vec{a}\;\vec{b}]$ preserve the value, but swapping any two vectors changes the sign.

Determinant Representation of the Scalar Triple Product

Let the component forms be $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$. Then

$\displaystyle \vec{a}\cdot(\vec{b}\times\vec{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$

This determinant expands as:

$\vec{a}\cdot(\vec{b}\times\vec{c}) = a_1(b_2c_3-b_3c_2) - a_2(b_1c_3-b_3c_1) + a_3(b_1c_2-b_2c_1)$

Result: The scalar triple product is zero if and only if the vectors are coplanar.

Geometric Interpretation and Volume Associated with the Scalar Triple Product

The absolute value $|\vec{a}\cdot(\vec{b}\times\vec{c})|$ equals the volume of the parallelepiped formed by $\vec{a}$, $\vec{b}$, and $\vec{c}$ as coterminous edges. The direction of the cross product $\vec{b}\times\vec{c}$ is perpendicular to the plane formed by $\vec{b}$ and $\vec{c}$; its magnitude gives the area of the parallelogram with those sides. The scalar triple product then projects $\vec{a}$ along this normal to yield the volume.

If the scalar triple product is negative, the orientation of the vectors is left-handed with respect to the order taken, while a positive value corresponds to a right-handed system. For related geometric results, see Geometry of Complex Numbers.

Algebraic Structure and Properties of the Scalar Triple Product

The scalar triple product possesses several key algebraic properties. It is invariant under cyclic permutations and antisymmetric under exchange of any two vectors:

• Cyclic shift leaves the value unchanged
• Sign reverses upon swapping two vectors
• Zero if two vectors are equal or parallel
• Linearity in each vector individually

Explicitly, $[\vec{a}\;\vec{b}\;\vec{c}] = [\vec{b}\;\vec{c}\;\vec{a}] = [\vec{c}\;\vec{a}\;\vec{b}]$ and $[\vec{a}\;\vec{b}\;\vec{c}] = -[\vec{a}\;\vec{c}\;\vec{b}]$.

For any scalar $\lambda$, $[\lambda\vec{a}\;\vec{b}\;\vec{c}] = [\vec{a}\;\lambda\vec{b}\;\vec{c}] = [\vec{a}\;\vec{b}\;\lambda\vec{c}] = \lambda[\vec{a}\;\vec{b}\;\vec{c}]$.

If any two of the vectors are linearly dependent, the scalar triple product equals zero. Thus, coplanarity of three nonzero vectors is characterised by the vanishing of their scalar triple product.

For formal manipulations involving the triple product and other vector identities, refer to Vector Triple Product.

Derivation of the Scalar Triple Product Formula

Starting with general Cartesian vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$, compute $\vec{b}\times\vec{c}$ as a determinant with the standard basis. The dot product with $\vec{a}$ collects the coefficients, yielding the 3×3 determinant form.

This determinant representation gives both a compact computational tool and reveals the antisymmetric nature of the triple product. The equivalence of the determinant expansion and the definition $\vec{a}\cdot(\vec{b}\times\vec{c})$ establishes the correspondence for all vectors in $\mathbb{R}^3$.

For other scalar and vector products, see Scalar Product of Vectors and Dot Product vs Cross Product.

Test for Coplanarity Using the Scalar Triple Product

For three vectors to be coplanar, a necessary and sufficient condition is $[\vec{a}\;\vec{b}\;\vec{c}] = 0$. This is frequently examined in vector geometry problems, especially involving points or position vectors in space.

Given four points $O$, $A$, $B$, and $C$ with position vectors respectively $\vec{0}$, $\vec{a}$, $\vec{b}$, $\vec{c}$, the points $A$, $B$, $C$ are coplanar with $O$ if the triple product $[\vec{a}\;\vec{b}\;\vec{c}]=0$.

In coordinate geometry, parametric lines and planes involving coplanarity are frequently reduced to checking the vanishing of a scalar triple product, as illustrated in Coordinate Geometry.

Procedural Examples Involving the Scalar Triple Product

Example: For vectors $\vec{p} = \hat{i} - \hat{j} + \hat{k}$, $\vec{q} = 2\hat{i} + 3\hat{j} - \hat{k}$, $\vec{r} = -\hat{i} - \hat{j} + 5\hat{k}$, compute $[\vec{p}\;\vec{q}\;\vec{r}]$.

Substitute components: $\vec{p} = (1,-1,1)$, $\vec{q} = (2,3,-1)$, $\vec{r} = (-1,-1,5)$.

Evaluate determinant: $\begin{vmatrix} 1 & -1 & 1 \\ 2 & 3 & -1 \\ -1 & -1 & 5 \end{vmatrix} = 1[(3 \times 5) - (-1 \times -1)] - (-1)[(2 \times 5) - (-1 \times -1)] + 1[(2 \times -1) - (3 \times -1)]$

$= 1[15 - 1] - (-1)[10 - 1] + 1[-2 + 3] = 14 + 9 + 1 = 24$

Solution: The scalar triple product is $24$.

Example: Vectors $\vec{u} = \hat{i} + 2\hat{j} - 3\hat{k}$, $\vec{v} = 2\hat{i} - \hat{j} + 2\hat{k}$, $\vec{w} = 3\hat{i} + \hat{j} - \hat{k}$ are tested for coplanarity.

Substitute components to determinant:

$\begin{vmatrix} 1 & 2 & -3 \\ 2 & -1 & 2 \\ 3 & 1 & -1 \end{vmatrix} = 1[(-1)(-1) - 2 \times 1] - 2[2 \times -1 - 2 \times 3] + (-3)[2 \times 1 - (-1) \times 3]$

$= 1[1 - 2] - 2[-2 - 6] + (-3)[2 + 3] = -1 + 16 - 15 = 0$

Solution: The vectors are coplanar.

Example: Find $[\vec{a}\;\vec{b}\;\vec{c}]$ for $\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$, $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$, $\vec{c} = 3\hat{i} - \hat{j} + 2\hat{k}$.

Evaluate: $\begin{vmatrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix} = 2[(2\times2) - (-1\times-1)] - (-3)[1\times2 - (-1)\times3] + 4[1\times-1 - 2\times3]$

$= 2[4 - 1] + 3[2 + 3] + 4[-1 - 6] = 2\times3 + 3\times5 + 4\times(-7) = 6 + 15 - 28 = -7$

Solution: The scalar triple product is $-7$.

A comprehensive understanding of the scalar triple product and its properties enables efficient resolution of volume-related and coplanarity problems in three-dimensional geometry. For triangle solutions using vector methods, refer to Properties of Triangles.