Scalar triple product

The scalar triple product of three vectors $\vec{a}, \vec{b}, \vec{c}$ is $[\vec{a}\;\vec{b}\;\vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c}).$ It is a single number (a scalar). Geometrically, its absolute value equals the volume of the parallelepiped whose three edges (from a common vertex) are $\vec{a}, \vec{b}, \vec{c}$. The sign tells us about the orientation of the three vectors (right-handed vs left-handed).

Determinant form

In components, $[\vec a\;\vec b\;\vec c] = \det \begin{pmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{pmatrix}.$

This follows directly: $\vec{b}\times\vec{c}$ expands as a "$\hat i, \hat j, \hat k$" determinant, then dotting with $\vec{a}$ replaces those unit vectors by $a_1, a_2, a_3$ , yielding the $3\times 3$ determinant.

Properties

• Cyclic invariance: $[\vec a\;\vec b\;\vec c] = [\vec b\;\vec c\;\vec a] = [\vec c\;\vec a\;\vec b]$. (Cyclically permuting the order leaves the value unchanged.)
• Transposition flips sign: swapping any two vectors changes the sign: $[\vec a\;\vec b\;\vec c] = -[\vec b\;\vec a\;\vec c]$.
• Linear in each argument: $[\lambda\vec a + \mu\vec a'\;\vec b\;\vec c] = \lambda[\vec a\;\vec b\;\vec c] + \mu[\vec a'\;\vec b\;\vec c]$.
• Dot-cross interchange: $\vec a\cdot(\vec b\times\vec c) = (\vec a\times\vec b)\cdot\vec c$. The dot and cross can swap without changing value.
• Vanishes for proportional vectors: if any two of $\vec a, \vec b, \vec c$ are parallel, $[\vec a\;\vec b\;\vec c] = 0$.

Coplanarity test

The triple product measures volume. Three vectors span no volume (i.e. are coplanar) iff the parallelepiped is flat, i.e. $[\vec a\;\vec b\;\vec c] = 0 \iff \vec a, \vec b, \vec c \text{ are coplanar}.$

This is one of the most useful tests in 3D geometry. To check whether four points $A, B, C, D$ are coplanar: form the three vectors $\vec{AB}, \vec{AC}, \vec{AD}$ and compute $[\vec{AB}\;\vec{AC}\;\vec{AD}]$. Zero $\Rightarrow$ coplanar.

Volume of a tetrahedron

The tetrahedron with three edges $\vec a, \vec b, \vec c$ from a common vertex has volume $V_{\text{tet}} = \frac{1}{6}\left|[\vec a\;\vec b\;\vec c]\right|.$

(Tetrahedron = $1/6$ of the parallelepiped.)

Worked examples

Example 1. Find $[\vec a\;\vec b\;\vec c]$ where $\vec a = \hat i + \hat j$, $\vec b = \hat j + \hat k$, $\vec c = \hat k + \hat i$.

$\det \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix} = 1(1 - 0) - 1(0 - 1) + 0 = 1 + 1 = 2$.

Example 2. Are the vectors $\vec a = \hat i - 2\hat j + \hat k$, $\vec b = 2\hat i + \hat j + \hat k$, $\vec c = 4\hat i - 3\hat j + 3\hat k$ coplanar?

$\det \begin{pmatrix} 1 & -2 & 1 \\ 2 & 1 & 1 \\ 4 & -3 & 3 \end{pmatrix} = 1(3 + 3) - (-2)(6 - 4) + 1(-6 - 4) = 6 + 4 - 10 = 0$. Yes, coplanar.

Example 3. Find the volume of the parallelepiped with edges $\hat i, \hat i + \hat j, \hat i + \hat j + \hat k$.

$\det \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix} = 1$. Volume $= 1$.

Example 4. Find $\lambda$ so that $\vec a = \hat i + 2\hat j + 3\hat k$, $\vec b = 2\hat i + 3\hat j + 4\hat k$, $\vec c = \hat i + \lambda \hat j + 5\hat k$ are coplanar.

$\det = 0$: $\begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 1 & \lambda & 5 \end{vmatrix} = 1(15 - 4\lambda) - 2(10 - 4) + 3(2\lambda - 3) = 15 - 4\lambda - 12 + 6\lambda - 9 = 2\lambda - 6 = 0 \Rightarrow \lambda = 3$.

Example 5. Volume of the tetrahedron with vertices $A = (0, 0, 0)$, $B = (1, 0, 0)$, $C = (0, 1, 0)$, $D = (0, 0, 1)$.

$\vec{AB} = (1, 0, 0)$, $\vec{AC} = (0, 1, 0)$, $\vec{AD} = (0, 0, 1)$. Triple product $= 1$ (identity matrix). Volume $= 1/6$.

Example 6. Show that the four points $A = (1, 0, 0)$, $B = (0, 1, 0)$, $C = (0, 0, 1)$, $D = (2, 1, 1)$ are coplanar.

$\vec{AB} = (-1, 1, 0)$, $\vec{AC} = (-1, 0, 1)$, $\vec{AD} = (1, 1, 1)$.

$\det \begin{pmatrix} -1 & 1 & 0 \\ -1 & 0 & 1 \\ 1 & 1 & 1 \end{pmatrix} = -1(0 - 1) - 1(-1 - 1) + 0 = 1 + 2 = 3$.

Not zero , so they are not coplanar. The example shows that you must compute, not guess.

Try it yourself

1. Compute $[\hat i\;\hat j\;\hat k]$.
2. Compute $[\vec a\;\vec b\;\vec c]$ if $\vec a = \hat i + 2\hat j + 3\hat k$, $\vec b = 2\hat i - \hat j + \hat k$, $\vec c = \hat i - \hat j + 2\hat k$.
3. Find $\lambda$ so that $\hat i + \hat j$, $\hat i + \hat k$, $\lambda \hat i + \hat j + \hat k$ are coplanar.
4. Find the volume of the parallelepiped with edges $\hat i + \hat j$, $2\hat j - \hat k$, $\hat i + \hat k$.
5. Volume of the tetrahedron with vertices $(1, 1, 1)$, $(2, 1, 3)$, $(3, 2, 2)$, $(3, 3, 4)$.
6. Show $[\vec a\;\vec b\;\vec c] = [\vec b\;\vec c\;\vec a]$ from determinant properties.
7. If $\vec a, \vec b, \vec c$ are mutually perpendicular unit vectors, find $[\vec a\;\vec b\;\vec c]$.
8. Are $\vec a = 2\hat i - \hat j + \hat k$, $\vec b = \hat i - 3\hat j - 5\hat k$, $\vec c = 3\hat i - 4\hat j - 4\hat k$ coplanar?
9. Find a unit vector perpendicular to both $\vec b = \hat i + \hat j$ and $\vec c = \hat i - \hat j + \hat k$, and verify it's orthogonal to both.
10. Find the volume of the parallelepiped spanned by $\hat i, \hat j, \hat i + \hat j + \hat k$.
11. Show that points $(1, 1, 1)$, $(2, 3, 5)$, $(3, 4, 6)$, $(4, 7, 11)$ are coplanar.
12. If $[\vec a\;\vec b\;\vec c] = 5$, find $[\vec b\;\vec c\;\vec a]$, $[\vec b\;\vec a\;\vec c]$.
13. The volume of a parallelepiped is $24$. If two edges are $\vec a = 2\hat i + \hat j$ and $\vec b = \hat j + 3\hat k$, find a possible third edge.
14. Show $\vec a\cdot(\vec b\times\vec c) = \vec b\cdot(\vec c\times\vec a)$.

Pitfalls and tricks

• Always take absolute value for volume , a negative triple product just means left-handed orientation.
• Coplanarity test: triple product $= 0$. Always.
• Cyclic permutation preserves; transposition flips. Memorise these symmetries.
• For four points coplanar, use vectors from one chosen vertex, then compute triple product.
• Tetrahedron volume: don't forget the $1/6$ factor.