Cross Products and Geometry in Space

The dot product of two vectors is a number saying how much they agree. It cannot tell you which plane they span, or how much area they enclose.

In three dimensions a second product does both. The cross product returns a vector: its direction is perpendicular to both inputs, naming their plane, and its length is the area of the parallelogram they span. Dotting that against a third vector then gives a volume, and a volume of zero says the three vectors lie in a plane.

From those two operations come closed-form answers to questions that otherwise need calculus or coordinates chosen by hand: the distance from a point to a line, the distance from a point to a plane, the area of a triangle in space, the equation of a plane through three points. Each is one formula, and each is the determinant machinery from the previous units wearing geometric clothes.

Each component of $u\times v$ omits one coordinate and takes a $2\times 2$ determinant of the other two:

The middle component is the one that catches people out: its indices run $3,1$ then $1,3$ rather than $1,3$ then $3,1$. The pattern is cyclic, $1\to 2\to 3\to 1$, and writing it as the formal determinant with $e_1,e_2,e_3$ along the top row generates the alternating sign automatically, which is why that mnemonic is worth using.

Why only in ${\mathbb{R}}^3$. The construction needs a single vector perpendicular to both inputs. In ${\mathbb{R}}^3$ the space orthogonal to two independent vectors is one-dimensional, so up to length there are exactly two candidates and the right-hand rule picks one. In ${\mathbb{R}}^4$ that orthogonal space is two-dimensional and no single vector names it; in ${\mathbb{R}}^2$ there is no room at all. The dot product has no such restriction because its output is a scalar.

The right-hand rule is a convention. It fixes which of the two normals is taken, and nothing mathematical forces the choice. Adopting the opposite convention would negate every cross product and every triple product consistently, leaving all areas and volumes, which are absolute values, unchanged.

Not associative. $(u\times v)\times w$ and $u\times (v\times w)$ are generally different vectors. With $u=v=e_1$ and $w=e_2$: the first is $0\times e_2=0$, the second is $e_1\times (e_1\times e_2)=e_1\times e_3=-e_2$. So parentheses are never optional, and an expression written $u\times v\times w$ is not well defined.

Cross product against outer product. $u\times v$ is a vector in ${\mathbb{R}}^3$; $u{v}^{\mathsf{T}}$ is a $3\times 3$ matrix of rank 1. They share a word and nothing else. The outer product is what the singular value decomposition expands a matrix into, and it is defined for vectors of any two lengths, including different ones.

Cross product. Write the formal determinant with $e_1,e_2,e_3$ on the top row, $u$ on the second and $v$ on the third, and expand along the first row. Then check: $(u\times v)\cdot u$ and $(u\times v)\cdot v$ must both be zero. Two dot products, and they catch almost every sign error.

Area of a parallelogram on $u$ and $v$: $\Vert u\times v\Vert$. Area of a triangle on the same two edges: half that.

Volume of a parallelepiped on $u,v,w$: $|u\cdot (v\times w)|$, equal to $|det|$ of the matrix with those rows. Volume of a tetrahedron on the same three edges: one sixth of it.

Coplanarity test. The three vectors lie in a common plane exactly when $u\cdot (v\times w)=0$, equivalently when the determinant vanishes, the vectors are dependent, and the matrix is singular.

Distance from a point $Q$ to a line through $A$ with direction $d$:

The numerator is the area of the parallelogram on the displacement and the direction; dividing by the base length leaves the height.

Distance from a point $Q$ to a plane through $P_0$ with normal $n$:

This is the length of the projection of the displacement onto the normal. Forgetting to divide by $\Vert n\Vert$ is the standard error, and it is invisible when $n$ happens to be a unit vector.

Plane through three points $P$, $Q$, $R$. Form $n=(Q-P)\times (R-P)$, then the plane is $n\cdot (X-P)=0$. Expanding gives $n_{1}x+n_{2}y+n_{3}z=n\cdot P$. Verify by substituting all three points. Each must satisfy it.

Order matters everywhere. Swapping the arguments of a cross product negates it, so $(Q-A)\times d$ and $d\times (Q-A)$ differ by a sign. Distances are unaffected because they take a magnitude, but a normal vector's direction and a triple product's sign both flip, which matters whenever orientation is part of the answer.

Take $u=(1,2,3)$, $v=(4,5,6)$, $w=(2,-1,1)$.

The cross product. Expanding the formal determinant along the top row:

Check by orthogonality. $(u\times v)\cdot u=-3+12-9=0$ and $(u\times v)\cdot v=-12+30-18=0$. Both vanish, so the result really is perpendicular to the plane of $u$ and $v$.

Check by anticommutativity. $v\times u=(3,-6,3)=-(u\times v)$, which is what swapping the rows of the determinant does.

The area. $\Vert u\times v{\Vert }^2=9+36+9=54$, so the parallelogram on $u$ and $v$ has area $\sqrt{54}=3\sqrt{6}\approx 7.348$, and the triangle on those edges has area $\approx 3.674$.

Cross-check by Lagrange's identity. $\Vert u{\Vert }^2=14$, $\Vert v{\Vert }^2=77$, $u\cdot v=4+10+18=32$, so

And by the angle. $\cos \theta =32/\sqrt{14\cdot 77}\approx 0.9746$, so $\sin \theta \approx 0.2238$ and $\Vert u\Vert \Vert v\Vert \sin \theta \approx 3.7417\times 8.7750\times 0.2238\approx 7.348$. Three independent routes to the same area.

The volume. First $v\times w=(5(1)-6(-1),6(2)-4(1),4(-1)-5(2))=(11,8,-14)$, so

Check against the determinant.

The parallelepiped has volume $|-15|=15$, and the tetrahedron on the same three edges has volume $15/6=2.5$. The negative sign says $u$, $v$, $w$ form a left-handed frame.

Cyclic invariance. $v\cdot (w\times u)=-15$ and $w\cdot (u\times v)=-15$ as well, rotating the three vectors leaves the value alone, since it permutes the determinant's rows cyclically, an even permutation.

A coplanar case. Replace $w$ by $u+v=(5,7,9)$. Then

because the three vectors are dependent by construction. The third lies in the plane of the first two. Zero volume, zero determinant, dependent columns and coplanar vectors are the same fact stated four ways.

Theorem (Lagrange's identity). For $u,v\in {\mathbb{R}}^3$,

Proof. Both sides expand into polynomials in the six components. The left side is $(u_2{v}_3-u_3{v}_2{)}^2+(u_3{v}_1-u_1{v}_3{)}^2+(u_1{v}_2-u_2{v}_1{)}^2$; the right side is $\left(\sum _i{u}_i^2\right)\left(\sum _j{v}_j^2\right)-{\left(\sum _i{u}_i{v}_i\right)}^2$. Expanding the right side, the terms $u_i^2{v}_i^2$ cancel against the diagonal of the squared dot product, leaving , which regroups into exactly the three squares on the left. $◼$

Corollary (the area). Writing $u\cdot v=\Vert u\Vert \Vert v\Vert \cos \theta$,

and taking the nonnegative root gives $\Vert u\times v\Vert =\Vert u\Vert \Vert v\Vert \sin \theta$, the parallelogram's base times its height. So the area claim is not a separate definition but a consequence of the component formula.

Corollary. $u\times v=0$ exactly when $\sin \theta =0$ or one vector is zero, that is, exactly when $u$ and $v$ are parallel.

Theorem (triple product as determinant).

Proof. Expanding the determinant along its first row gives $u_1(v_2{w}_3-v_3{w}_2)-u_2(v_1{w}_3-v_3{w}_1)+u_3(v_1{w}_2-v_2{w}_1)$, and the three brackets are precisely the components of $v\times w$, with the middle sign absorbed by the component's own index order. $◼$

Corollaries, all inherited from the determinant unit. The triple product is unchanged by a cyclic permutation of $u,v,w$, a cyclic row permutation is even, and changes sign under a transposition. It vanishes exactly when the rows are dependent, which is what coplanarity means. Its absolute value is the volume of the parallelepiped, because the determinant is the volume scaling factor applied to the unit cube.

Why this unifies the two readings. The determinant unit established that $|det|$ scales volume; this theorem says the same number is reachable by an area-times-height argument. The agreement is why a single computation answers both a question about independence and a question about geometry.

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 $P$:

Their cross product is the normal:

Check. $n\cdot PQ=5+3-8=0$ and $n\cdot PR=-5+9-4=0$.

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

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. $\Vert n{\Vert }^2=25+9+16=50$, so the parallelogram on $PQ$ and $PR$ has area $\sqrt{50}\approx 7.071$ and the triangle $PQR$ has area $\sqrt{50}/2\approx 3.536$.

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

so the parallelepiped on the three edges has volume 35 and the tetrahedron $PQRS$ has volume $35/6\approx 5.833$.

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

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.