Cross Products and Geometry in Space

The cross product of two vectors in $R^{3}$ is a vector perpendicular to both whose length is the area of the parallelogram they span. With the scalar triple product it measures volume, decides coplanarity, and gives closed-form distances from a point to a line or a plane.

Definition

For $u, v \in R^{3}$ the cross product is

u \times v = (u_{2} v_{3} - u_{3} v_{2}, u_{3} v_{1} - u_{1} v_{3}, u_{1} v_{2} - u_{2} v_{1}),

conveniently remembered as the formal determinant

u \times v = det (\begin{matrix} e_{1} & e_{2} & e_{3} \\ u_{1} & u_{2} & u_{3} \\ v_{1} & v_{2} & v_{3} \end{matrix}) .

Its two defining properties. $u \times v$ is orthogonal to both $u$ and $v$ , and

‖ u \times v ‖ = ‖ u ‖ ‖ v ‖ \sin θ,

the area of the parallelogram spanned by $u$ and $v$ . Its direction among the two perpendicular choices is fixed by the right-hand rule.

Algebraic rules. Anticommutative, $v \times u = - (u \times v)$ ; linear in each argument; and $u \times u = 0$ . It is not associative, and $u \times v = 0$ exactly when $u$ and $v$ are parallel or one is zero.

Lagrange's identity. $‖ u \times v ‖^{2} = ‖ u ‖^{2} ‖ v ‖^{2} - (u \cdot v)^{2}$ , which is $\sin^{2} = 1 - \cos^{2}$ written in vectors and is the easiest route to the area claim.

Scalar triple product. $u \cdot (v \times w) = det (\begin{matrix} u \\ v \\ w \end{matrix})$ , whose absolute value is the volume of the parallelepiped on $u$ , $v$ , $w$ . It is unchanged by cyclic permutation and changes sign under a swap, and it vanishes exactly when the three vectors are coplanar, which is the determinant's singularity test in geometric dress.

Distances. For a point $Q$ , a line through $A$ with direction $d$ , and a plane through $P_{0}$ with normal $n$ :

dist (Q, line) = \frac{‖ (Q - A) \times d ‖}{‖ d ‖}, dist (Q, plane) = \frac{| (Q - P_{0}) \cdot n |}{‖ n ‖} .

The first is an area divided by a base; the second is the length of a projection onto the normal.

Outer product. For $a \in R^{m}$ and $b \in R^{n}$ , $a b^{T}$ is the $m \times n$ matrix with entries $a_{i} b_{j}$ . It has rank 1 when both are nonzero, every row is a multiple of $b^{T}$ , and it is the building block of the singular value decomposition's expansion. It is unrelated to the cross product despite the similar name.

Assumptions and scope

The cross product is defined only in $R^{3}$ , and for $R^{7}$ by a different construction. The dot product, by contrast, is defined in every dimension.
The right-hand rule is a convention fixing which of the two normals is taken. Reversing it would negate every cross product consistently, and the area and volume magnitudes would be unaffected.
The cross product is not associative: $(u \times v) \times w$ and $u \times (v \times w)$ generally differ, so parentheses are never optional.
A vanishing cross product means the vectors are parallel or one is zero, which includes the case of two vectors pointing in opposite directions.
The outer product $a b^{T}$ and the cross product $a \times b$ are unrelated operations with unrelated outputs: a matrix and a vector respectively. The shared word product is the only thing they have in common.
Distance formulas assume the line or plane is given by a point together with a direction or a normal. A plane given by three points requires forming the normal as a cross product first.

Forms this is expressed in

The same content in several forms. Each makes something visible that the others leave implicit, so moving between them is part of understanding the topic rather than a presentation choice.

geometric

The parallelogram spanned, and the normal whose length is its area

The cross product drawn in space: $u$ and $v$ span a parallelogram, and $u \times v$ stands perpendicular to it.

Two facts the formula states are visible here. The direction is normal to the plane of $u$ and $v$ , which is why $(u \times v) \cdot u$ and $(u \times v) \cdot v$ both vanish. The magnitude $| u \times v | = | u | | v | \sin θ$ is the area of that parallelogram, so a cross product is small exactly when the vectors are nearly parallel and zero when they are.

Anticommutativity is the same picture with the normal reversed: $v \times u$ points the other way along the same line, because the right-hand rule follows the order of the factors.

This is one of the few places where three dimensions are the content rather than a flourish. There is no cross product in the plane, because a plane has no room for a normal direction.

Translates into: geometric

Worked material

Example

Standard basis products, and a plane through three points

The basis vectors cycle. $e_{1} \times e_{2} = e_{3}$ , $e_{2} \times e_{3} = e_{1}$ , $e_{3} \times e_{1} = e_{2}$ , and each reversed pair gives the negative. The cyclic pattern $1 \to 2 \to 3 \to 1$ is the same one governing the component formula, and checking a computed cross product against it catches a systematic sign error faster than checking orthogonality.

Each of $e_{i} \times e_{i}$ is zero, as it must be: a vector is parallel to itself, and the parallelogram on a single direction has no area.

A plane through three points. Take $P = (1, 0, 2)$ , $Q = (2, 1, 0)$ , $R = (0, 3, 1)$ .

Form two edge vectors from $P$ :

P Q = Q - P = (1, 1, - 2), P R = R - P = (- 1, 3, - 1) .

Their cross product is the normal:

n = P Q \times P R = (1 (- 1) - (- 2) (3), (- 2) (- 1) - 1 (- 1), 1 (3) - 1 (- 1)) = (5, 3, 4) .

Check. $n \cdot P Q = 5 + 3 - 8 = 0$ and $n \cdot P R = - 5 + 9 - 4 = 0$ .

The plane is $n \cdot (X - P) = 0$ , that is $5 (x - 1) + 3 y + 4 (z - 2) = 0$ , or

5 x + 3 y + 4 z = 13 .

Verify on all three points. $P$ : $5 + 0 + 8 = 13$ . $Q$ : $10 + 3 + 0 = 13$ . $R$ : $0 + 9 + 4 = 13$ .

Areas and volumes from the same data. $‖ n ‖^{2} = 25 + 9 + 16 = 50$ , so the parallelogram on $P Q$ and $P R$ has area $\sqrt{50} \approx 7.071$ and the triangle $P Q R$ has area $\sqrt{50} / 2 \approx 3.536$ .

Adding a fourth point $S = (4, 4, 4)$ , the displacement $P S = (3, 4, 2)$ gives the triple product

P Q \cdot (P R \times P S) = 35,

so the parallelepiped on the three edges has volume 35 and the tetrahedron $P Q R S$ has volume $35 / 6 \approx 5.833$ .

And the distance from $S$ to the plane. $P S \cdot n = 15 + 12 + 8 = 35$ , so

dist = \frac{35}{\sqrt{50}} \approx 4.950 .

The same number 35 appears as the triple product and as the numerator here, which is not a coincidence: volume equals base area times height, and $\sqrt{50} \times 4.950 = 35$ .

What one computation supplied. A normal, a plane equation, a triangle area, a tetrahedron volume and a point-to-plane distance all came from one cross product and one dot product.

Non-example

Five errors in space

Treating the cross product as commutative. Writing $u \times v = v \times u$ . For $u = (1, 2, 3)$ and $v = (4, 5, 6)$ the two are $(- 3, 6, - 3)$ and $(3, - 6, 3)$ , opposite vectors. Any normal computed in the wrong order points the wrong way, which reverses an orientation even though it leaves an area or a distance unchanged.

Dropping the parentheses. Writing $u \times v \times w$ . With $u = v = e_{1}$ and $w = e_{2}$ , $(e_{1} \times e_{1}) \times e_{2} = 0$ while $e_{1} \times (e_{1} \times e_{2}) = e_{1} \times e_{3} = - e_{2}$ . The expression has two different values, so it names nothing.

Getting the middle component's signs wrong. Computing the second component as $u_{1} v_{3} - u_{3} v_{1}$ instead of $u_{3} v_{1} - u_{1} v_{3}$ . The result is a vector that fails the orthogonality check, for the example above it would give $(- 3, - 6, - 3)$ , whose dot product with $u$ is $- 3 - 12 - 9 = - 24 \neq 0$ . Two dot products catch this immediately, which is why the check is part of the procedure.

Forgetting to normalise a distance. Reporting $| (Q - P_{0}) \cdot n |$ as the distance to a plane. That is the distance multiplied by $‖ n ‖$ . For $n = (2, - 1, 2)$ with $‖ n ‖ = 3$ and $(Q - P_{0}) \cdot n = 10$ , the distance is $10 / 3 \approx 3.33$ , not 10. The error is invisible whenever the normal happens to be a unit vector, which is exactly why it survives.

Confusing the cross product with the outer product. Expecting $u \times v$ and $u v^{T}$ to be related. The first is a vector in $R^{3}$ ; the second is a $3 \times 3$ matrix of rank 1, defined for vectors of any lengths including different ones. They share the word product and nothing else, and the outer product is the one appearing in the singular value decomposition.

What unites them. The first three mistake the cross product for something better behaved than it is, commutative, associative, symmetric in its indices. The last two drop a normalisation or conflate two operations. All five are caught by two cheap habits: check perpendicularity after every cross product, and check that a distance has the units of a length.

Common errors

Common misconception

The cross product behaves like the dot product with respect to order and association, so $u \times v = v \times u$ and parentheses may be dropped in $u \times v \times w$ .

Cross Products and Geometry in Space

Assumptions and scope

Forms this is expressed in

Worked material

Standard basis products, and a plane through three points

Five errors in space

Common errors

Related units

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