Scalar Triple Product — Definition, Formula, Example

#Algebra
TL;DR
The scalar triple product of three vectors is $\vec{a} \cdot (\vec{b} \times \vec{c})$, a single number equal to the signed volume of the parallelepiped built on those vectors. This article covers the definition, the determinant formula, six worked examples, why a zero result means the vectors are coplanar, and the mistakes to avoid
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Bhanzu TeamLast updated on July 20, 202610 min read

What Is the Scalar Triple Product?

The scalar triple product of three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ is the number you get by taking the cross product of two of them and then the dot product with the third: $\vec{a} \cdot (\vec{b} \times \vec{c})$.

The result is a single scalar, and its absolute value is the volume of the parallelepiped (a slanted box) whose three edges from one corner are $\vec{a}$, $\vec{b}$, and $\vec{c}$. It is written compactly as $[\vec{a}\ \vec{b}\ \vec{c}]$, and it equals zero exactly when the three vectors lie in one plane.

The Determinant Formula

Writing out the cross product then the dot product works, but there is a cleaner route. If

$\vec{a} = (a_1, a_2, a_3)$

$\vec{b} = (b_1, b_2, b_3)$

$\vec{c} = (c_1, c_2, c_3)$

then the scalar triple product equals the determinant of the matrix whose rows are the three vectors:

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

Here is what each symbol in the formula stands for:

Symbol

Meaning

$\vec{a}, \vec{b}, \vec{c}$

The three vectors sharing a corner — the edges of the box

$a_1, a_2, a_3$

The $x$, $y$, $z$ components of $\vec{a}$ (likewise for $\vec{b}$ and $\vec{c}$)

$\vec{b} \times \vec{c}$

Cross product: a vector perpendicular to both $\vec{b}$ and $\vec{c}$

$\cdot$

Dot product: multiplies matching components and adds, giving a number

$[\vec{a}\ \vec{b}\ \vec{c}]$

Shorthand for the whole scalar triple product

The volume of the parallelepiped is the absolute value of this number, since a volume cannot be negative:

$$V = \left| \vec{a} \cdot (\vec{b} \times \vec{c}) \right|$$

The sign carries meaning too: a positive value means the three vectors form a right-handed set, a negative value a left-handed set. The size is the volume; the sign is the orientation.

Examples of the Scalar Triple Product

Example 1

Find $\vec{a} \cdot (\vec{b} \times \vec{c})$ for $\vec{a} = (1, 0, 0)$, $\vec{b} = (0, 1, 0)$, $\vec{c} = (0, 0, 1)$.

These are the three unit axes. Use the determinant.

$$\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 1$$

Final answer: the scalar triple product is 1 — the volume of the unit cube.

Example 2

Find the volume of the parallelepiped on $\vec{a} = (2, 0, 0)$, $\vec{b} = (0, 3, 0)$, $\vec{c} = (0, 0, 4)$. A student answers $2 + 3 + 4 = 9$. Where does that go wrong?

Adding the components treats the box like a sum of lengths, but volume is not a sum. The wrong answer of 9 has the wrong units entirely — it is a length, not a volume. Volume comes from the determinant.

$$\begin{vmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{vmatrix} = 2 \times 3 \times 4 = 24$$

Take the absolute value.

$V = |24| = 24$

Final answer: the volume is 24 cubic units. The fix was to multiply through the determinant, not add components — the box is $2 \times 3 \times 4$, which is exactly 24.

Example 3

Compute $\vec{a} \cdot (\vec{b} \times \vec{c})$ for $\vec{a} = (1, 2, 3)$, $\vec{b} = (0, 1, 4)$, $\vec{c} = (5, 6, 0)$.

Expand the determinant along the top row.

$$\begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{vmatrix}$$

$= 1(1 \cdot 0 - 4 \cdot 6) - 2(0 \cdot 0 - 4 \cdot 5) + 3(0 \cdot 6 - 1 \cdot 5)$

$= 1(0 - 24) - 2(0 - 20) + 3(0 - 5)$

$= -24 + 40 - 15$

$= 1$

Final answer: the scalar triple product is 1.

Example 4

Show that $\vec{a} = (1, 2, 3)$, $\vec{b} = (2, 4, 6)$, $\vec{c} = (1, 1, 1)$ are coplanar.

Vectors are coplanar when the parallelepiped on them is flat — zero volume. Compute the determinant.

$$\begin{vmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1 \end{vmatrix}$$

Notice row 2 is exactly twice row 1, so the determinant is 0.

$\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$

Final answer: the scalar triple product is 0, so the three vectors lie in one plane.

Example 5

Verify the cyclic property: for $\vec{a} = (1, 0, 2)$, $\vec{b} = (3, 1, 0)$, $\vec{c} = (0, 2, 1)$, check that $\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot (\vec{c} \times \vec{a})$.

Compute the first.

$$\begin{vmatrix} 1 & 0 & 2 \\ 3 & 1 & 0 \\ 0 & 2 & 1 \end{vmatrix} = 1(1 - 0) - 0(3 - 0) + 2(6 - 0) = 1 + 12 = 13$$

Now compute the second, rows $\vec{b}, \vec{c}, \vec{a}$.

$$\begin{vmatrix} 3 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{vmatrix} = 3(4 - 0) - 1(0 - 1) + 0 = 12 + 1 = 13$$

Final answer: both equal 13, confirming the cyclic property $\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot (\vec{c} \times \vec{a})$.

Example 6

Find the volume of a parallelepiped whose edges are $\vec{a} = (2, -1, 1)$, $\vec{b} = (1, 3, -2)$, $\vec{c} = (0, 1, 4)$.

Expand along the top row.

$$\begin{vmatrix} 2 & -1 & 1 \\ 1 & 3 & -2 \\ 0 & 1 & 4 \end{vmatrix}$$

$= 2(3 \cdot 4 - (-2) \cdot 1) - (-1)(1 \cdot 4 - (-2) \cdot 0) + 1(1 \cdot 1 - 3 \cdot 0)$

$= 2(12 + 2) + 1(4 - 0) + 1(1 - 0)$

$= 28 + 4 + 1$

$= 33$

Take the absolute value.

$V = |33| = 33$

Final answer: the volume is 33 cubic units.

Why "Volume" Is the Whole Idea

The scalar triple product exists because area and volume in higher dimensions need a way to be computed from coordinates alone, without a ruler. The reason $\vec{a} \cdot (\vec{b} \times \vec{c})$ is the volume comes straight from the two operations it chains.

  • The cross product gives the base. $\vec{b} \times \vec{c}$ is a vector perpendicular to the base parallelogram, and its length $|\vec{b} \times \vec{c}|$ equals that parallelogram's area.

  • The dot product gives the height. Dotting with $\vec{a}$ projects $\vec{a}$ onto that perpendicular direction, which is exactly the height of the box above its base.

  • Base area times height is volume. Multiply the two and you have volume — the same base-times-height rule you learned for a simple prism, now working for a slanted one.

This is why a zero result means the vectors are coplanar: if all three lie in one plane, the "box" is squashed flat, its height is zero, and so is its volume. Crystallographers use exactly this to compute the volume of a crystal's unit cell from its three edge vectors, and it underlies the change-of-variables factor in multivariable calculus — the Jacobian determinant that rescales volume when coordinates change.

At Bhanzu, trainers build this from the base-times-height picture first, so the determinant formula lands as a shortcut for a volume the student already understands. The two operations inside it are worth knowing on their own: the dot product that projects one vector onto another, and the cross product of two vectors that produces the perpendicular used for the base.

Properties of the Scalar Triple Product

These rules follow from the determinant form, and several of them let you skip work or catch an error before you finish computing.

  • Cyclic invariance. Rotating the vectors in a cycle leaves the value unchanged: $\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot (\vec{c} \times \vec{a}) = \vec{c} \cdot (\vec{a} \times \vec{b})$. In bracket form, $[\vec{a}\ \vec{b}\ \vec{c}] = [\vec{b}\ \vec{c}\ \vec{a}] = [\vec{c}\ \vec{a}\ \vec{b}]$.

  • Sign flips under a swap. Interchanging any two vectors reverses the sign but keeps the size: $[\vec{a}\ \vec{b}\ \vec{c}] = -[\vec{b}\ \vec{a}\ \vec{c}]$. This mirrors the rule that swapping two rows of a determinant negates it.

  • Dot and cross are interchangeable. The two operators can be swapped without changing the value: $\vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c}$. Only the cyclic order of the vectors matters, not which pair is crossed.

  • Zero means coplanar. The product is 0 if and only if the three vectors lie in one plane — the box is flat, so its volume is zero.

  • Repeated or parallel vectors give zero. If any two of the vectors are equal or parallel, the product is 0, because two proportional rows make a determinant vanish.

  • Scalars factor out. Scaling one vector scales the whole product: $[k\vec{a}\ \vec{b}\ \vec{c}] = k[\vec{a}\ \vec{b}\ \vec{c}]$.

Where the Scalar Triple Product Goes Sideways

Mistake 1: Mixing up the dot and cross operators

Where it slips in: reading $\vec{a} \cdot (\vec{b} \times \vec{c})$ and computing the cross product last.

Don't do this: try $(\vec{a} \cdot \vec{b}) \times \vec{c}$, which is meaningless — you cannot cross a scalar with a vector.

The correct way: the inner operation is the cross product $\vec{b} \times \vec{c}$ (vector result), and the outer is the dot product with $\vec{a}$ (number result). The memorizer who recalls "triple product" without the operator order often reaches for the wrong one first. The determinant formula sidesteps this entirely.

Mistake 2: Forgetting the absolute value for volume

Where it slips in: a determinant that comes out negative.

Don't do this: report a volume of $-33$.

The correct way: volume is $\left|\vec{a} \cdot (\vec{b} \times \vec{c})\right|$. A negative sign only tells you the orientation is left-handed; the volume is the size, so take the absolute value.

Mistake 3: Treating the components as something to add

Where it slips in: rushing a box-volume problem with axis-aligned vectors.

Don't do this: add the components, as in Example 2's wrong path.

The correct way: run the determinant. Adding gives a length, not a volume, and the units alone show the answer is wrong.

Conclusion

  • The scalar triple product is $\vec{a} \cdot (\vec{b} \times \vec{c})$, a single number equal to the signed volume of the parallelepiped on the three vectors.

  • It equals the 3×3 determinant with the vectors as rows, and the volume is its absolute value.

  • A result of zero means the three vectors are coplanar — the box is flat.

  • The × is the cross product and the · is the dot product; keeping the two operators straight is half the battle.

Work through the six examples above, then try building your own vectors and predicting whether the determinant will be zero before you compute it. To deepen vector algebra with a teacher, explore Bhanzu's algebra tutor, get help with algebra, or browse math classes online.

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Frequently Asked Questions

Is the scalar triple product a vector or a number?
A number. The cross product inside produces a vector, but the outer dot product collapses it to a single scalar — hence "scalar" triple product.
What does it mean when the scalar triple product is zero?
The three vectors are coplanar. The parallelepiped they span is flat, so its volume is zero.
Why can I compute it as a determinant?
Because the definition $\vec{a} \cdot (\vec{b} \times \vec{c})$ expands exactly to the 3×3 determinant with $\vec{a}$, $\vec{b}$, $\vec{c}$ as rows. It is the same value, faster to compute.
Does the order of the vectors matter?
Cyclic swaps ($\vec{a}, \vec{b}, \vec{c} \to \vec{b}, \vec{c}, \vec{a}$) keep the value the same. A single swap of two vectors flips the sign but not the size.
How is this different from the dot product or the cross product alone?
The dot product of two vectors gives a number (related to angle), the cross product of two vectors gives a perpendicular vector (related to area), and the scalar triple product chains both across three vectors to give a volume.
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Bhanzu Team
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Bhanzu’s editorial team, known as Team Bhanzu, is made up of experienced educators, curriculum experts, content strategists, and fact-checkers dedicated to making math simple and engaging for learners worldwide. Every article and resource is carefully researched, thoughtfully structured, and rigorously reviewed to ensure accuracy, clarity, and real-world relevance. We understand that building strong math foundations can raise questions for students and parents alike. That’s why Team Bhanzu focuses on delivering practical insights, concept-driven explanations, and trustworthy guidance-empowering learners to develop confidence, speed, and a lifelong love for mathematics.
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