The Scalar Triple Product of Vectors

The Definition of the Scalar Triple Product

Scalar triple product

Let a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} be three vectors. The scalar triple product is defined as:

[a⃗,b⃗,c⃗]=(a⃗×b⃗)⋅c⃗[\vec{a}, \vec{b}, \vec{c}] = (\vec{a} \times \vec{b}) \cdot \vec{c}

The result of the scalar triple product is a scalar.

How to Compute It

The scalar triple product can be computed with a determinant:

The determinant form of the scalar triple product
[a⃗,b⃗,c⃗]=∣a1a2a3b1b2b3c1c2c3∣[\vec{a}, \vec{b}, \vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}

The Geometric Meaning

The geometric meaning of the scalar triple product

The absolute value of the scalar triple product is the volume of the parallelepiped with a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} as its edges:

V=∣[a⃗,b⃗,c⃗]∣V = |[\vec{a}, \vec{b}, \vec{c}]|

The Sign of the Volume

  • when [a⃗,b⃗,c⃗]>0[\vec{a}, \vec{b}, \vec{c}] > 0, the three vectors form a right-handed system
  • when [a⃗,b⃗,c⃗]<0[\vec{a}, \vec{b}, \vec{c}] < 0, the three vectors form a left-handed system
  • when [a⃗,b⃗,c⃗]=0[\vec{a}, \vec{b}, \vec{c}] = 0, the three vectors are coplanar

Properties

The basic properties of the scalar triple product

Invariance under cyclic permutation: [a⃗,b⃗,c⃗]=[b⃗,c⃗,a⃗]=[c⃗,a⃗,b⃗][\vec{a}, \vec{b}, \vec{c}] = [\vec{b}, \vec{c}, \vec{a}] = [\vec{c}, \vec{a}, \vec{b}]

Anti-commutativity: [a⃗,b⃗,c⃗]=−[b⃗,a⃗,c⃗][\vec{a}, \vec{b}, \vec{c}] = -[\vec{b}, \vec{a}, \vec{c}]

Linearity: [ka⃗+la⃗′,b⃗,c⃗]=k[a⃗,b⃗,c⃗]+l[a⃗′,b⃗,c⃗][k\vec{a} + l\vec{a}', \vec{b}, \vec{c}] = k[\vec{a}, \vec{b}, \vec{c}] + l[\vec{a}', \vec{b}, \vec{c}]

Applications

Testing for Coplanarity

The condition for three vectors to be coplanar
a⃗,b⃗,c⃗ are coplanar⇔[a⃗,b⃗,c⃗]=0\vec{a}, \vec{b}, \vec{c} \text{ are coplanar} \Leftrightarrow [\vec{a}, \vec{b}, \vec{c}] = 0

Computing a Volume

The volume of a tetrahedron

The volume of the tetrahedron with a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} as its edges is:

V=16∣[a⃗,b⃗,c⃗]∣V = \frac{1}{6} |[\vec{a}, \vec{b}, \vec{c}]|

Computing with a Determinant

The scalar triple product equals the value of the third-order determinant whose rows (or columns) are the three vectors.


Exercises

Exercise 1

Compute the scalar triple product [a⃗,b⃗,c⃗][\vec{a}, \vec{b}, \vec{c}], where a⃗=(1,0,0)\vec{a} = (1, 0, 0), b⃗=(0,1,0)\vec{b} = (0, 1, 0) and c⃗=(0,0,1)\vec{c} = (0, 0, 1).

the scalar triple productcomputing a determinantthe standard basis vectors

Approach:

Use the determinant to compute the scalar triple product.

Steps:

  1. [a⃗,b⃗,c⃗]=∣100010001∣=1[\vec{a}, \vec{b}, \vec{c}] = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 1

  2. Or compute it directly: a⃗×b⃗=(0,0,1)\vec{a} \times \vec{b} = (0, 0, 1) (a⃗×b⃗)⋅c⃗=(0,0,1)⋅(0,0,1)=1(\vec{a} \times \vec{b}) \cdot \vec{c} = (0, 0, 1) \cdot (0, 0, 1) = 1

Answer: [a⃗,b⃗,c⃗]=1[\vec{a}, \vec{b}, \vec{c}] = 1

Exercise 2

Decide whether the vectors a⃗=(1,1,0)\vec{a} = (1, 1, 0), b⃗=(1,0,1)\vec{b} = (1, 0, 1) and c⃗=(0,1,1)\vec{c} = (0, 1, 1) are coplanar.

the scalar triple producttesting coplanaritythe determinant

Approach:

Compute the scalar triple product; if it is zero, they are coplanar.

Steps:

  1. [a⃗,b⃗,c⃗]=∣110101011∣[\vec{a}, \vec{b}, \vec{c}] = \begin{vmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{vmatrix}

  2. Compute the determinant: Expanding along the first row: 1×∣0111∣−1×∣1101∣+0×∣1001∣1 \times \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} - 1 \times \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} + 0 \times \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} =1(0−1)−1(1−0)+0=−1−1=−2≠0= 1(0-1) - 1(1-0) + 0 = -1 - 1 = -2 \neq 0

Answer: the scalar triple product is not zero, so the three vectors are not coplanar.


Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
[a⃗,b⃗,c⃗][\vec{a}, \vec{b}, \vec{c}]Scalar triple productscalar triple productThe scalar triple product of the three vectors
VVVolumevolumeThe volume of the parallelepiped

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