In short

The scalar triple product of three vectors \vec{a}, \vec{b}, \vec{c} is the number \vec{a} \cdot (\vec{b} \times \vec{c}), written [\vec{a}\;\vec{b}\;\vec{c}]. Geometrically, its absolute value is the volume of the parallelepiped formed by the three vectors. When this number is zero, the three vectors are coplanar — they all lie in one flat plane. Algebraically, it equals the determinant of the 3 \times 3 matrix whose rows are the components of the three vectors.

Take three sticks of different lengths and different directions. Tape their tails together at one corner. Now imagine building a box — not a cube, but a slanted box — by extending each stick into a full edge and filling in the six faces with parallelograms. The shape you get is called a parallelepiped (say it like "par-uh-lel-eh-PIE-ped"). It is the three-dimensional cousin of a parallelogram.

Here is the question: what is the volume of this box?

With a cube, volume is just side cubed. With a rectangular box, it is length times width times height. But a parallelepiped is slanted — its edges are not perpendicular — so none of those simple formulas apply. You need something that accounts for the angles between the edges, not just their lengths.

The answer turns out to be a single operation on the three edge vectors. It combines a cross product and a dot product in one step, producing a single number — a scalar. That number is called the scalar triple product, and it gives you the volume directly.

Building the formula

Suppose the three edges meeting at one corner of the parallelepiped are the vectors \vec{a}, \vec{b}, and \vec{c}.

Start with two of them, say \vec{b} and \vec{c}. Their cross product \vec{b} \times \vec{c} is a vector that is perpendicular to the parallelogram formed by \vec{b} and \vec{c}, and whose magnitude equals the area of that parallelogram:

|\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}|\sin\theta

where \theta is the angle between \vec{b} and \vec{c}.

Now you have the base of the parallelepiped (the parallelogram with area |\vec{b} \times \vec{c}|) and a direction perpendicular to it (\vec{b} \times \vec{c} points straight "up" from the base). The volume of the box is the base area times the perpendicular height — the component of \vec{a} in the direction of \vec{b} \times \vec{c}.

That perpendicular component is exactly what a dot product extracts. The projection of \vec{a} onto the direction of \vec{b} \times \vec{c} is

\text{height} = \frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|}

So the volume is

V = \text{base area} \times \text{height} = |\vec{b} \times \vec{c}| \times \frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|} = \vec{a} \cdot (\vec{b} \times \vec{c})

The magnitude cancels, and you are left with just the dot product of \vec{a} with the cross product of \vec{b} and \vec{c}. That product can be negative (if \vec{a} points "below" the base rather than "above" it), so the volume is the absolute value:

V = |\vec{a} \cdot (\vec{b} \times \vec{c})|
A parallelepiped built from three vectorsA three-dimensional parallelepiped with one corner at the origin. Three vectors a, b, and c form three edges from that corner. The base parallelogram is formed by b and c, and the height is the perpendicular component of a onto the direction of b cross c. b c a height base parallelogram (b × c gives its area)
The parallelepiped formed by three vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ from a common corner. The base is the parallelogram of $\vec{b}$ and $\vec{c}$ (area $= |\vec{b} \times \vec{c}|$), and the height is the perpendicular projection of $\vec{a}$ onto $\vec{b} \times \vec{c}$. The volume is their scalar triple product.

The definition

Scalar triple product

For three vectors \vec{a}, \vec{b}, \vec{c} in \mathbb{R}^3, the scalar triple product is

[\vec{a}\;\vec{b}\;\vec{c}] \;=\; \vec{a} \cdot (\vec{b} \times \vec{c})

If \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

[\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 volume of the parallelepiped formed by \vec{a}, \vec{b}, \vec{c} is |[\vec{a}\;\vec{b}\;\vec{c}]|.

Why it equals a determinant

This is worth seeing. Write out \vec{b} \times \vec{c} in component form:

\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = (b_2c_3 - b_3c_2)\hat{i} - (b_1c_3 - b_3c_1)\hat{j} + (b_1c_2 - b_2c_1)\hat{k}

Now take the dot product with \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}:

\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)

Look at the right side. That is exactly the cofactor expansion of a 3 \times 3 determinant along the first row — with a_1, a_2, a_3 as the first row, and the 2 \times 2 subdeterminants formed from the remaining rows. So

\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}

The scalar triple product is a determinant. This connection is one of the reasons determinants matter in geometry — a 3 \times 3 determinant is, quite literally, a volume.

Properties

The scalar triple product has several clean properties. Each follows from determinant properties.

1. Cyclic permutation preserves the value.

[\vec{a}\;\vec{b}\;\vec{c}] = [\vec{b}\;\vec{c}\;\vec{a}] = [\vec{c}\;\vec{a}\;\vec{b}]

Cycling the three vectors — moving the first to the end and shifting the rest forward — does not change the determinant. (In determinant language, two adjacent row-swaps bring you back to the same sign: swapping rows once flips the sign, swapping again flips it back.)

Equivalently, the "dot" and the "cross" can be interchanged:

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

This is sometimes written as: "the dot and cross are interchangeable in a scalar triple product." Geometrically, the volume of the parallelepiped does not depend on which face you call the base.

Cyclic permutation of the scalar triple productThree arrows forming a cycle between the expressions [a b c], [b c a], and [c a b], showing that all three are equal. A separate arrow to [b a c] with a minus sign shows that a non-cyclic swap flips the sign. [a b c] [b c a] [c a b] = = = −[b a c] swap
Cyclic permutations ($a \to b \to c \to a$) leave the scalar triple product unchanged. A non-cyclic swap (like swapping $\vec{a}$ and $\vec{b}$) flips the sign.

2. Swapping any two vectors flips the sign.

[\vec{a}\;\vec{b}\;\vec{c}] = -[\vec{b}\;\vec{a}\;\vec{c}] = -[\vec{a}\;\vec{c}\;\vec{b}]

Swapping two rows of a determinant flips its sign. So [\vec{a}\;\vec{b}\;\vec{c}] = -[\vec{b}\;\vec{a}\;\vec{c}]. The magnitude (and hence the volume) stays the same, but the sign changes. The sign tracks the orientation — whether the triple (\vec{a}, \vec{b}, \vec{c}) forms a right-handed or left-handed system.

3. Scalar factor pulls out.

[\lambda\vec{a}\;\vec{b}\;\vec{c}] = \lambda\,[\vec{a}\;\vec{b}\;\vec{c}]

Scaling one vector scales the volume proportionally. This follows from the determinant property that multiplying one row by \lambda multiplies the determinant by \lambda.

4. Linearity in each argument.

[(\vec{a} + \vec{a}')\;\vec{b}\;\vec{c}] = [\vec{a}\;\vec{b}\;\vec{c}] + [\vec{a}'\;\vec{b}\;\vec{c}]

The scalar triple product distributes over vector addition in any one slot. This is the multilinearity of the determinant.

5. If any two vectors are equal, the product is zero.

[\vec{a}\;\vec{a}\;\vec{c}] = 0

A determinant with two equal rows is zero. Geometrically, if two of your edge vectors are the same, the "parallelepiped" is flat — it has no volume.

The coplanarity test

This is the property that makes the scalar triple product indispensable. Three vectors \vec{a}, \vec{b}, \vec{c} are coplanar — they all lie in one flat plane — if and only if the parallelepiped they form has zero volume. And that happens exactly when

[\vec{a}\;\vec{b}\;\vec{c}] = 0

Turn this around: four points A, B, C, D are coplanar if and only if the three vectors \overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD} have scalar triple product zero.

Coplanar versus non-coplanar vectorsTwo diagrams side by side. On the left, three vectors lie in the same plane, giving a flat parallelepiped with zero volume. On the right, three vectors span three dimensions, forming a proper parallelepiped with nonzero volume. Coplanar: volume = 0 a b c all three in one plane Non-coplanar: volume ≠ 0 b c a a pokes out of the b-c plane
Left: when $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$, the "parallelepiped" is flat and has zero volume — $[\vec{a}\;\vec{b}\;\vec{c}] = 0$. Right: when $\vec{a}$ has a component perpendicular to the $\vec{b}$-$\vec{c}$ plane, the parallelepiped has genuine depth and nonzero volume.

Why does this work? If \vec{a}, \vec{b}, \vec{c} are coplanar, then \vec{a} lies in the plane spanned by \vec{b} and \vec{c}. But \vec{b} \times \vec{c} is perpendicular to that plane. So \vec{a} is perpendicular to \vec{b} \times \vec{c}, which means their dot product is zero. Conversely, if \vec{a} \cdot (\vec{b} \times \vec{c}) = 0, then \vec{a} has no component along \vec{b} \times \vec{c} — it lies entirely in the \vec{b}-\vec{c} plane.

Coplanarity of four points

To check whether four points A(x_1, y_1, z_1), B(x_2, y_2, z_2), C(x_3, y_3, z_3), D(x_4, y_4, z_4) are coplanar, compute:

[\overrightarrow{AB}\;\overrightarrow{AC}\;\overrightarrow{AD}] = \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \\ x_4 - x_1 & y_4 - y_1 & z_4 - z_1 \end{vmatrix}

If this determinant is zero, the four points lie in one plane.

Volume of a tetrahedron

A parallelepiped and a tetrahedron are related. A tetrahedron with vertices A, B, C, D occupies exactly \frac{1}{6} of the parallelepiped formed by the vectors \overrightarrow{AB}, \overrightarrow{AC}, \overrightarrow{AD}. (The factor \frac{1}{6} plays the same role as the \frac{1}{2} in the triangle area formula — a triangle is half its parallelogram, and a tetrahedron is one-sixth of its parallelepiped.)

\text{Volume of tetrahedron} = \frac{1}{6}\,|[\overrightarrow{AB}\;\overrightarrow{AC}\;\overrightarrow{AD}]|
A tetrahedron as one-sixth of a parallelepipedA parallelepiped shown as a transparent box, with a tetrahedron drawn inside it with solid edges. The tetrahedron uses the same three edges from one corner but has only four faces instead of six. A B C D V(tet) = ⅙ V(ppd)
The tetrahedron $ABCD$ (solid red) fits inside the parallelepiped built from the same three edges $\overrightarrow{AB}$, $\overrightarrow{AC}$, $\overrightarrow{AD}$ (dashed outline). Six copies of the tetrahedron fill the parallelepiped exactly, so the tetrahedron's volume is $\frac{1}{6}$ of the parallelepiped's.

This formula is used constantly in 3D geometry problems.

Computing one from start to finish

Example 1: Volume of a parallelepiped

Find the volume of the parallelepiped whose three co-terminal edges are \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}.

Step 1. Write the scalar triple product as a determinant.

[\vec{a}\;\vec{b}\;\vec{c}] = \begin{vmatrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix}

Why: the three vectors form the rows, in order. The scalar triple product is this determinant.

Step 2. Expand along the first row.

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

Why: cofactor expansion — each element of the first row multiplies the 2 \times 2 determinant left after deleting its row and column, with alternating signs +, -, +.

Step 3. Compute each 2 \times 2 determinant.

\begin{vmatrix} 2 & -1 \\ -1 & 2 \end{vmatrix} = 4 - 1 = 3, \qquad \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix} = 2 + 3 = 5, \qquad \begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} = -1 - 6 = -7

Why: a 2 \times 2 determinant is ad - bc. Careful with the signs — 1 \times 2 - (-1) \times 3 = 2 + 3 = 5.

Step 4. Combine.

[\vec{a}\;\vec{b}\;\vec{c}] = 2(3) + 3(5) + 4(-7) = 6 + 15 - 28 = -7

Why: the sign of -3 in the first row was handled by the cofactor sign pattern. The entry is -3, the cofactor sign is -, so the contribution is -(-3) \times 5 = +15.

Step 5. The volume is the absolute value.

V = |{-7}| = 7 \text{ cubic units}

Result: The parallelepiped has volume 7 cubic units.

Parallelepiped from the three vectors in Example 1A parallelepiped formed by vectors a, b, c with volume 7 cubic units. The three vectors extend from a common origin, and the box they form is shaded. b c a Volume = |−7| = 7
The parallelepiped built from $\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$, $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$, $\vec{c} = 3\hat{i} - \hat{j} + 2\hat{k}$. The determinant gives $-7$, so the volume is $7$ cubic units. The negative sign tells us the triple $(\vec{a}, \vec{b}, \vec{c})$ forms a left-handed system.

The negative sign in the scalar triple product has a geometric meaning: it tells you about orientation. A positive value means (\vec{a}, \vec{b}, \vec{c}) form a right-handed system (like the standard \hat{i}, \hat{j}, \hat{k}); a negative value means they form a left-handed system. The volume itself is always the absolute value.

Example 2: Testing four points for coplanarity

Are the four points A(1, 2, 3), B(4, 0, 1), C(2, 1, 5), D(7, -1, 3) coplanar?

Step 1. Compute the three vectors from A to the other points.

\overrightarrow{AB} = (4-1)\hat{i} + (0-2)\hat{j} + (1-3)\hat{k} = 3\hat{i} - 2\hat{j} - 2\hat{k}
\overrightarrow{AC} = (2-1)\hat{i} + (1-2)\hat{j} + (5-3)\hat{k} = \hat{i} - \hat{j} + 2\hat{k}
\overrightarrow{AD} = (7-1)\hat{i} + (-1-2)\hat{j} + (3-3)\hat{k} = 6\hat{i} - 3\hat{j} + 0\hat{k}

Why: four points are coplanar exactly when the three vectors connecting one of them to the other three have scalar triple product zero.

Step 2. Set up the determinant.

[\overrightarrow{AB}\;\overrightarrow{AC}\;\overrightarrow{AD}] = \begin{vmatrix} 3 & -2 & -2 \\ 1 & -1 & 2 \\ 6 & -3 & 0 \end{vmatrix}

Why: the three vectors become the three rows of a 3 \times 3 determinant.

Step 3. Expand along the third row (the zero makes it efficient).

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

Why: expanding along the row with a zero entry saves one minor computation. The cofactor signs for the third row are +, -, +.

Step 4. Evaluate.

= 6(-4 - 2) + 3(6 + 2) + 0 = 6(-6) + 3(8) = -36 + 24 = -12

Result: The scalar triple product is -12 \neq 0, so the four points are not coplanar. They span a tetrahedron of volume \frac{1}{6}|{-12}| = 2 cubic units.

Four non-coplanar points forming a tetrahedronFour points A, B, C, D in 3D space that do not lie in one plane. They form a tetrahedron with volume 2 cubic units. A B C D Volume = 2 Not coplanar
The four points $A(1,2,3)$, $B(4,0,1)$, $C(2,1,5)$, $D(7,-1,3)$ form a genuine tetrahedron in 3D space. The scalar triple product is $-12$, which is not zero, confirming that no single plane contains all four points. The tetrahedron has volume $\frac{1}{6} \times 12 = 2$ cubic units.

If the determinant had come out to zero, you would have concluded that the four points are coplanar. The scalar triple product is a single-number test: zero means coplanar, nonzero means not coplanar.

Common confusions

Going deeper

If you came here to learn what the scalar triple product is, how to compute it, and what zero means, you have it — you can stop here. The sections below explore how the scalar triple product connects to orientation, how it behaves under coordinate transformations, and its role in the volume element of integration.

Orientation and the sign

The sign of [\vec{a}\;\vec{b}\;\vec{c}] carries geometric information. If you curl the fingers of your right hand from \vec{b} toward \vec{c} and your thumb points roughly along \vec{a}, the triple product is positive — the system is right-handed. If your thumb points opposite to \vec{a}, the triple product is negative — the system is left-handed.

The standard basis (\hat{i}, \hat{j}, \hat{k}) is right-handed, and

[\hat{i}\;\hat{j}\;\hat{k}] = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 1
Right-handed versus left-handed systemsTwo coordinate systems side by side. On the left, a right-handed system where curling fingers from i to j makes the thumb point along k, with positive scalar triple product. On the right, a left-handed system where the vectors are ordered differently, giving a negative scalar triple product. Right-handed: [a b c] > 0 b c a Left-handed: [a b c] < 0 b c a
Left: the triple $(\vec{a}, \vec{b}, \vec{c})$ forms a right-handed system — $\vec{a}$ points the same way as $\vec{b} \times \vec{c}$, so the scalar triple product is positive. Right: $\vec{a}$ points opposite to $\vec{b} \times \vec{c}$, making the scalar triple product negative.

Any set of three vectors that forms a right-handed system has a positive scalar triple product. This idea — that a determinant encodes orientation — shows up again in the Jacobian of coordinate transformations, where the sign tells you whether the transformation preserves or reverses orientation.

Connection to the cross product magnitude

There is a useful identity relating the scalar triple product to magnitudes:

[\vec{a}\;\vec{b}\;\vec{c}]^2 + |\vec{a} \times (\vec{b} \times \vec{c})|^2 = |\vec{a}|^2 \cdot |\vec{b} \times \vec{c}|^2

This is a three-dimensional analogue of the identity \cos^2\theta + \sin^2\theta = 1, applied to the angle between \vec{a} and \vec{b} \times \vec{c}. The scalar triple product captures the \cos\theta part (the projection), while the vector triple product (which you will meet in the next article) captures the \sin\theta part (the rejection).

The volume element

In multivariable calculus, when you change from Cartesian to curvilinear coordinates (spherical, cylindrical, etc.), the volume element dV picks up a factor that is exactly a scalar triple product of the new basis vectors — the Jacobian determinant. The same idea of "three vectors spanning a tiny parallelepiped" reappears in a more sophisticated form.

Where this leads next

The scalar triple product sits at a junction between several important ideas.