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Scalar Triple Product — Definition, Formula & Examples

The scalar triple product of three vectors a\mathbf{a}, b\mathbf{b}, and c\mathbf{c} is the dot product of one vector with the cross product of the other two, written a⋅(b×c)\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}). It produces a single number (a scalar) whose absolute value equals the volume of the parallelepiped formed by the three vectors.

For vectors a=⟨a1,a2,a3⟩\mathbf{a} = \langle a_1, a_2, a_3 \rangle, b=⟨b1,b2,b3⟩\mathbf{b} = \langle b_1, b_2, b_3 \rangle, and c=⟨c1,c2,c3⟩\mathbf{c} = \langle c_1, c_2, c_3 \rangle in R3\mathbb{R}^3, the scalar triple product is defined as the determinant of the 3×33 \times 3 matrix whose rows (or columns) are the components of a\mathbf{a}, b\mathbf{b}, and c\mathbf{c}. It equals zero if and only if the three vectors are coplanar.

Key Formula

a⋅(b×c)=∣a1a2a3b1b2b3c1c2c3∣\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}
Where:
  • a\mathbf{a} = First vector with components $a_1, a_2, a_3$
  • b\mathbf{b} = Second vector with components $b_1, b_2, b_3$
  • c\mathbf{c} = Third vector with components $c_1, c_2, c_3$

How It Works

You can compute the scalar triple product by evaluating a 3×33 \times 3 determinant rather than first finding the cross product and then dotting. Place each vector as a row of a matrix and take the determinant. A positive result means a\mathbf{a}, b\mathbf{b}, c\mathbf{c} form a right-handed system; a negative result means left-handed. Cyclically permuting the vectors (a⋅(b×c)=b⋅(c×a)=c⋅(a×b)\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b})) leaves the value unchanged, while swapping any two vectors flips the sign.

Worked Example

Problem: Find the scalar triple product of a=⟨1,0,2⟩\mathbf{a} = \langle 1, 0, 2 \rangle, b=⟨3,1,0⟩\mathbf{b} = \langle 3, 1, 0 \rangle, and c=⟨0,2,4⟩\mathbf{c} = \langle 0, 2, 4 \rangle, and state the volume of the parallelepiped they span.
Set up the determinant: Write the components as rows of a 3×3 matrix.
∣102310024∣\begin{vmatrix} 1 & 0 & 2 \\ 3 & 1 & 0 \\ 0 & 2 & 4 \end{vmatrix}
Expand along the first row: Use cofactor expansion.
1(1⋅4−0⋅2)−0(3⋅4−0⋅0)+2(3⋅2−1⋅0)=1(4)−0+2(6)=161(1 \cdot 4 - 0 \cdot 2) - 0(3 \cdot 4 - 0 \cdot 0) + 2(3 \cdot 2 - 1 \cdot 0) = 1(4) - 0 + 2(6) = 16
Find the volume: The volume of the parallelepiped is the absolute value of the scalar triple product.
V=∣16∣=16V = |16| = 16
Answer: The scalar triple product is 1616, and the volume of the parallelepiped is 1616 cubic units.

Why It Matters

The scalar triple product appears whenever you need to compute volumes in multivariable calculus, such as finding the volume of a tetrahedron (one-sixth the absolute value of the scalar triple product). It is also essential in physics for determining whether three force or velocity vectors are coplanar.

Common Mistakes

Mistake: Placing the parentheses incorrectly as (a⋅b)×c(\mathbf{a} \cdot \mathbf{b}) \times \mathbf{c}.
Correction: The dot product a⋅b\mathbf{a} \cdot \mathbf{b} returns a scalar, and you cannot take the cross product of a scalar with a vector. Always compute the cross product first: a⋅(b×c)\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}).