9.2 Cross product, scalar triple product, and vector geometry

From a scalar product to an oriented normal

The dot product measures how much two vectors point in the same direction. Its output is a scalar. In three-dimensional geometry we also need an operation that produces a vector perpendicular to two given directions. Such a vector specifies a normal to their plane; its length can record the area they span. The cross product supplies both pieces of information at once.

Throughout this note all vectors belong to $\mathbb R^3$. We use the standard right-handed orthonormal basis $\mathbf i=(1,0,0)$, $\mathbf j=(0,1,0)$, $\mathbf k=(0,0,1)$. The norm $\|\mathbf u\|$ denotes length, while $\mathbf0$ is the zero vector. The cross product introduced here is an operation in three dimensions; its coordinate formula is not a definition for arbitrary $\mathbb R^n$. Its order also matters, unlike the order of a dot product.

Definition and the coordinate rule

Definition

Cross product in three dimensions

For $\mathbf u=(u_1,u_2,u_3)$ and $\mathbf v=(v_1,v_2,v_3)$, define

$$\mathbf u\times\mathbf v =(u_2v_3-u_3v_2,\ u_3v_1-u_1v_3,\ u_1v_2-u_2v_1).$$

This is a vector, not a scalar. A convenient expansion mnemonic is

$$\mathbf u\times\mathbf v= \begin{vmatrix} \mathbf i&\mathbf j&\mathbf k\\ u_1&u_2&u_3\\ v_1&v_2&v_3 \end{vmatrix}.$$

Expand the first row with signs plus, minus, plus. In particular, the middle component is $-(u_1v_3-u_3v_1)$; forgetting that minus sign changes the answer. The displayed array with basis vectors is shorthand for the coordinate formula. The ordinary scalar determinant will appear later when all its entries are real numbers.

The formula gives $\mathbf i\times\mathbf j=\mathbf k$, $\mathbf j\times\mathbf k=\mathbf i$, and $\mathbf k\times\mathbf i=\mathbf j$. Each basis vector crossed with itself is $\mathbf0$. These identities fix the positive orientation used throughout.

Worked example

Compute a cross product and check its output

Take $\mathbf u=\mathbf i+2\mathbf k=(1,0,2)$ and $\mathbf v=2\mathbf i-3\mathbf j+\mathbf k=(2,-3,1)$. Then

$$\begin{aligned} \mathbf u\times\mathbf v &=(0\cdot1-2(-3),\ 2\cdot2-1\cdot1,\ 1(-3)-0\cdot2)\\ &=(6,3,-3). \end{aligned}$$

As a check, its dot products with the inputs are

$$(6,3,-3)\cdot(1,0,2)=6-6=0, \qquad (6,3,-3)\cdot(2,-3,1)=12-9-3=0.$$

Both vanish, as a normal vector should. This check can expose a sign error, although perpendicularity alone would not determine the required length or choose between opposite normal directions.

Algebraic rules and their proofs

Theorem

Cross-product algebra

For $\mathbf u,\mathbf v,\mathbf w\in\mathbb R^3$ and $t\in\mathbb R$,

$$\begin{aligned} \mathbf u\times\mathbf v&=-\mathbf v\times\mathbf u,\\ \mathbf u\times(\mathbf v+\mathbf w) &=\mathbf u\times\mathbf v+\mathbf u\times\mathbf w,\\ (t\mathbf u)\times\mathbf v &=\mathbf u\times(t\mathbf v)=t(\mathbf u\times\mathbf v). \end{aligned}$$

Swapping the inputs reverses each coordinate difference. For example, $v_2u_3-v_3u_2=-(u_2v_3-u_3v_2)$, and the other two coordinates behave identically. This proves anticommutativity. For distributivity, the first coordinate on the left expands to

$$u_2(v_3+w_3)-u_3(v_2+w_2) =(u_2v_3-u_3v_2)+(u_2w_3-u_3w_2).$$

The second and third coordinates have the same distributive expansion, proving the vector equality. A scalar multiplying either input factors out of every coordinate difference, proving the last rule even for $t=0$. Distributivity in the first input follows by reversing order, distributing in the second, and reversing order again. Thus the coordinate definition establishes the rules without assuming them from the geometric picture.

Counterexample mode

Cross products do not commute

The tempting claim $\mathbf u\times\mathbf v=\mathbf v\times\mathbf u$ fails for $\mathbf u=\mathbf i$, $\mathbf v=\mathbf j$: the two answers are $\mathbf k$ and $-\mathbf k$. The correct rule has a minus sign. Reversing order preserves area but reverses the oriented normal. Equality in both orders occurs only when the cross product is $\mathbf0$, since a real vector equal to its negative must be zero.

Perpendicularity, magnitude, and orientation

Theorem

Geometric meaning of the cross product

The vector $\mathbf u\times\mathbf v$ is perpendicular to both inputs. For nonzero inputs let $\theta\in[0,\pi]$ be their angle. Then

$$\|\mathbf u\times\mathbf v\|=\|\mathbf u\|\|\mathbf v\|\sin\theta.$$

If the vectors are not parallel, its direction is the right-handed normal from $\mathbf u$ toward $\mathbf v$. If either input is zero or the two nonzero inputs are parallel, their cross product is $\mathbf0$.

Perpendicularity follows directly by cancellation:

$$\begin{aligned} \mathbf u\cdot(\mathbf u\times\mathbf v) &=u_1u_2v_3-u_1u_3v_2+u_2u_3v_1-u_1u_2v_3 +u_1u_3v_2-u_2u_3v_1\\ &=0. \end{aligned}$$

Interchanging the inputs and using anticommutativity gives perpendicularity to $\mathbf v$ as well. To check the magnitude, expand squares to obtain

$$\begin{aligned} \|\mathbf u\times\mathbf v\|^2 &=(u_2v_3-u_3v_2)^2+(u_3v_1-u_1v_3)^2+(u_1v_2-u_2v_1)^2\\ &=(u_1^2+u_2^2+u_3^2)(v_1^2+v_2^2+v_3^2) -(u_1v_1+u_2v_2+u_3v_3)^2. \end{aligned}$$

In this identity the diagonal square terms cancel after subtraction; the remaining square terms and mixed terms are exactly those in the three squared coordinate differences. For nonzero inputs substitute $\mathbf u\cdot\mathbf v=\|\mathbf u\|\|\mathbf v\|\cos\theta$. Taking the nonnegative square root yields the claimed formula because $\sin\theta\ge0$ on the stated interval.

The coordinate convention selects the right-hand rule: curl the fingers of your right hand from the first vector toward the second through their angle; the thumb gives the normal direction. The basis case $\mathbf i\times\mathbf j=\mathbf k$ fixes which orientation is positive. For zero or parallel inputs there is no nonzero cross-product direction to choose. No angle is assigned to the zero vector. For nonzero inputs the magnitude vanishes exactly when $\theta=0$ or $\theta=\pi$.

Worked example

Read area and unit normals from the same calculation

For the preceding pair, $\mathbf u\times\mathbf v=(6,3,-3)$, so

$$\|\mathbf u\times\mathbf v\|=\sqrt{36+9+9}=3\sqrt6.$$

The parallelogram spanned by the two vectors has this area: its base length is $\|\mathbf u\|$ and its perpendicular height is $\|\mathbf v\|\sin\theta$. The triangle spanned by them has half the area, $3\sqrt6/2$. The ordered pair chooses unit normal

$$\widehat{\mathbf n}=\frac{(6,3,-3)}{3\sqrt6} =\frac{(2,1,-1)}{\sqrt6}.$$

The opposite vector is also a unit normal to the plane, but it corresponds to the reverse input order. Division by the cross-product norm is legitimate here because the computed norm is nonzero; a degenerate cross product cannot be normalized this way.

Scalar triple product and determinant

Definition

Scalar triple product

For three vectors in $\mathbb R^3$, their scalar triple product is

$$[\mathbf u,\mathbf v,\mathbf w]=(\mathbf u\times\mathbf v)\cdot\mathbf w.$$

The cross product is taken first. Its output is a vector, so the final dot product produces a real scalar.

Substituting the coordinate formula gives

$$\begin{aligned} [\mathbf u,\mathbf v,\mathbf w] &=(u_2v_3-u_3v_2)w_1-(u_1v_3-u_3v_1)w_2 +(u_1v_2-u_2v_1)w_3\\ &=\begin{vmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1&w_2&w_3\end{vmatrix}. \end{aligned}$$

The last equality is expansion of the determinant along its third row. Consequently cyclic permutations preserve the scalar triple product, whereas interchanging any two vectors changes its sign. A cyclic permutation is two row interchanges, so its two minus signs cancel. Row order must therefore match the order used in the geometric interpretation.

Theorem

Signed and ordinary volume

The scalar triple product is the signed volume of the parallelepiped spanned by the ordered vectors. Its ordinary volume is

$$V=|[\mathbf u,\mathbf v,\mathbf w]|.$$

The triple product is zero exactly when the three vectors lie in a common plane through the origin, including all degenerate cases.

If $\mathbf u\times\mathbf v\ne\mathbf0$, use its unit normal $\widehat{\mathbf n}$. The scalar $h=\widehat{\mathbf n}\cdot\mathbf w$ is the signed height above the base plane. Hence

$$[\mathbf u,\mathbf v,\mathbf w] =\|\mathbf u\times\mathbf v\|h.$$

Base area times signed height proves the signed-volume statement. Ordinary height is $|h|$, giving the absolute value. For nonzero $\mathbf w$, if $\alpha$ is its angle with the chosen normal, then $h=\|\mathbf w\|\cos\alpha$. Heights are negative when $\pi/2\lt\alpha\le\pi$. If $\mathbf w=\mathbf0$, height is zero directly, without defining this angle.

With a nondegenerate base, zero height means precisely that $\mathbf w$ is in its plane. If the base cross product is zero, its area and the triple product are both zero; the first two vectors span at most a line, and adding the third vector still leaves all three in a common plane. No unit normal needs to be formed in that case.

Concept lensGeometric

A determinant measures height as well as base area

Keep the first two rows fixed. Their cross product records the oriented base area. Taking its dot product with the third row retains only the third vector's normal component: moving that vector within the base plane does not change the volume. The determinant is thus one scalar combining area and signed height. Its absolute value forgets orientation, while the cross product itself retains the chosen normal direction. This also explains why changing the third vector's length need not change the volume: only its projection onto the normal direction contributes.

Worked example

A side-two cube in explicit coordinates

Choose cube vertices $A=(0,0,0)$, $B=(2,0,0)$, $D=(0,2,0)$, $C=(2,2,0)$, and $E=(0,0,2)$. Thus $ABCD$ is the base square and $E$ lies above $A$. Then

$$\overrightarrow{AB}=(2,0,0),\quad \overrightarrow{AD}=(0,2,0),\quad \overrightarrow{CE}=(-2,-2,2).$$

The first cross product is $(0,0,4)$, giving

$$(\overrightarrow{AB}\times\overrightarrow{AD})\cdot\overrightarrow{CE} =(0,0,4)\cdot(-2,-2,2)=8.$$

The horizontal components of $\overrightarrow{CE}$ do not contribute. Its signed vertical height is $2$, and the base area is $4$. Reversing $AB$ and $AD$ makes the signed answer $-8$, while the ordinary volume stays $8$.

Set up the geometry before computing

In a diagram, first turn directed segments into vectors by subtracting the initial point from the terminal point. The order is part of the data: $\overrightarrow{CE}=E-C$, whereas $\overrightarrow{EC}=C-E$. The base vectors $\overrightarrow{AB}$ and $\overrightarrow{AD}$ already share a starting point. The vector $\overrightarrow{CE}$ can be drawn with its tail at that same point without changing its components. The volume formula then concerns the parallelepiped spanned by these three vectors; it does not require all three original arrows to start at one cube vertex.

There is also a direct algebraic reason that the horizontal displacement in the cube example is harmless. Put $\mathbf n=\mathbf u\times\mathbf v$. For any real numbers $s,t$, perpendicularity gives

$$\mathbf n\cdot(\mathbf w+s\mathbf u+t\mathbf v) =\mathbf n\cdot\mathbf w+s\mathbf n\cdot\mathbf u+t\mathbf n\cdot\mathbf v =\mathbf n\cdot\mathbf w.$$

Thus sliding the third vector's endpoint parallel to the base leaves the signed height, and hence the triple product, unchanged. This conclusion uses the full vector identity and works for every choice of the real coefficients, not only for the particular coordinates of the cube.

Common mistake

The sloping edge length is not the height

In the cube calculation, $\|\overrightarrow{CE}\|=2\sqrt3$, but the height above the base plane is $2$. Multiplying the base area $4$ by the sloping length would give $8\sqrt3$, which is not the volume. The dot product with the unit normal performs the missing projection. Use the full length as height only when the third vector is perpendicular to the base; use an absolute value as well if ordinary rather than signed volume is requested. A negative signed answer indicates orientation, not a negative amount of space.

Maximize a determinant under a length constraint

Worked example

Find both the bound and the equality case

Let $\mathbf a=(1,2,3)$, $\mathbf b=(3,2,1)$, and $\mathbf c=(x,y,z)$ with $\|\mathbf c\|=\sqrt6$. We have

$$\mathbf a\times\mathbf b=(-4,8,-4),\qquad \|\mathbf a\times\mathbf b\|=4\sqrt6.$$

Writing the determinant as a triple product gives

$$D=\begin{vmatrix}1&2&3\\3&2&1\\x&y&z\end{vmatrix} =(-4,8,-4)\cdot\mathbf c.$$

If $\alpha$ is the angle between the nonzero normal $(-4,8,-4)$ and $\mathbf c$, then $D=24\cos\alpha$. Therefore $|D|\le24$. The maximum signed value $D=24$ occurs when the vectors point in the same direction. Enforcing the length constraint gives

$$\mathbf c=\frac{\sqrt6}{4\sqrt6}(-4,8,-4)=(-1,2,-1).$$

This vector has squared norm $1+4+1=6$, and substitution gives $4+16+4=24$. Its negative gives $D=-24$. Thus the maximum absolute value is $24$, attained at exactly $\mathbf c=\pm(-1,2,-1)$. A bound without an attaining vector would not establish that the maximum is reached.

Quick checks

Checkpoint

What type of object is a cross product, and what type is a scalar triple product?

Follow the output of each operation.

Solution · Answer

The cross product is a vector in $\mathbb R^3$. Dotting it with the third vector produces a real scalar.

Checkpoint

Does a zero cross product imply that one input is zero?

Consider two nonzero vectors on the same line.

Solution · Answer

No. For example, $\mathbf i\times(2\mathbf i)=\mathbf0$. Both inputs are nonzero and parallel. The cross product also vanishes if an input is zero.

Checkpoint

What changes when the first two rows of a triple-product determinant are exchanged?

Distinguish signed volume from ordinary volume.

Solution · Answer

The signed triple product changes sign. Its absolute value, and therefore ordinary volume, remains the same.

Exercises

1. Recompute the cross product of $(1,0,2)$ and $(2,-3,1)$ in reverse order.
2. Use distributivity to compute $(\mathbf i+\mathbf j)\times\mathbf k$.
3. For $\mathbf u=(1,0,2)$ and $\mathbf v=(2,-3,1)$, find the area of their parallelogram and both unit normals to their plane.
4. In the specified cube, replace $\overrightarrow{CE}$ by $\overrightarrow{EC}$. Find the signed triple product and ordinary volume.
5. Prove that adding $t\mathbf u$ to the third vector leaves $[\mathbf u,\mathbf v,\mathbf w]$ unchanged for every real $t$.
6. Under the determinant example's constraint, find the minimum signed value, the maximum signed value, and every vector attaining the maximum absolute value.

Guided solutions

Solution · Model solution 1

Anticommutativity gives $(-6,-3,3)$, the negative of the original answer.

Solution · Model solution 2

$\mathbf i\times\mathbf k+\mathbf j\times\mathbf k =-\mathbf j+\mathbf i=(1,-1,0)$.

Solution · Model solution 3

The area is $\|(6,3,-3)\|=3\sqrt6$. Dividing by this nonzero norm gives unit normals $\pm(2,1,-1)/\sqrt6$.

Solution · Model solution 4

$\overrightarrow{EC}=(2,2,-2)$, so $(0,0,4)\cdot(2,2,-2)=-8$. The ordinary volume is $|-8|=8$.

Solution · Model solution 5

Dot-product distributivity gives $[\mathbf u,\mathbf v,\mathbf w+t\mathbf u] =[\mathbf u,\mathbf v,\mathbf w]+t(\mathbf u\times\mathbf v)\cdot\mathbf u$. The last term is zero by perpendicularity, including a degenerate base.

Solution · Model solution 6

The range is $-24\le D\le24$. The minimum occurs at $(1,-2,1)$, the maximum at $(-1,2,-1)$, and these two vectors are exactly the maximizers of $|D|$ because $|\cos\alpha|=1$ requires parallel directions.