The Cross Product, Lines, and Planes

The dot product returns a number. The second vector product returns a vector: given $\vec{a}$ and $\vec{b}$, the cross product $\vec{a}\times \vec{b}$ is perpendicular to both, with length the area of the parallelogram they span. That single construction orients a plane in space, measures parallelogram area and parallelepiped volume, and supplies the direction of a line and the normal of a plane.

The cross product

A vector $\vec{c}=⟨c_1,c_2,c_3⟩$ perpendicular to both $\vec{a}$ and $\vec{b}$ must satisfy $\vec{a}\cdot \vec{c}=0$ and $\vec{b}\cdot \vec{c}=0$ — two equations in three unknowns, so solutions form a line through the origin. The cross product is the solution obtained by eliminating one unknown between the two equations.

The component pattern is hard to remember but easy to reconstruct as a symbolic $3\times 3$ determinant with the standard basis in the first row:

where a $2\times 2$ determinant is $\left|\begin{array}{cc}p& q\\ r& s\end{array}\right|=ps-qr$. For $\vec{a}=⟨1,3,4⟩$ and $\vec{b}=⟨2,7,-5⟩$,

Geometry of the cross product

The computation $(\vec{a}\times \vec{b})\cdot \vec{a}$ expands to a sum in which every term cancels, giving $0$; the same holds for $\vec{b}$. So $\vec{a}\times \vec{b}$ is perpendicular to the plane the two vectors span. Which of the two perpendicular directions it takes is set by the right-hand rule: curl the fingers of your right hand from $\vec{a}$ toward $\vec{b}$ through the angle between them, and your thumb points along $\vec{a}\times \vec{b}$.

Expanding $\Vert \vec{a}\times \vec{b}{\Vert }^2$ and using $\Vert \vec{a}{\Vert }^2\Vert \vec{b}{\Vert }^2-(\vec{a}\cdot \vec{b}{)}^2$ from the dot-product identity yields $\Vert \vec{a}{\Vert }^2\Vert \vec{b}{\Vert }^2(1-{\cos }^2\theta )$, whose square root is the claimed formula. Two consequences follow at once. First, since $\sin \theta =0$ exactly when the vectors are parallel, the cross product detects parallelism.

Second, $\Vert \vec{a}\Vert (\Vert \vec{b}\Vert \sin \theta )$ is base times height for the parallelogram with sides $\vec{a}$ and $\vec{b}$:

To find a vector perpendicular to the plane through $P(1,4,6)$, $Q(-2,5,-1)$, and $R(1,-1,1)$, cross two edge vectors: $\overrightarrow{PQ}=⟨-3,1,-7⟩$, $\overrightarrow{PR}=⟨0,-5,-5⟩$, and $\overrightarrow{PQ}\times \overrightarrow{PR}=⟨-40,-15,15⟩$. Half its length, $\frac{1}{2}\Vert ⟨-40,-15,15⟩\Vert$, is the area of triangle $PQR$.

Algebra of the cross product

The cross product does not obey the usual rules of multiplication: it is anticommutative and not associative, so order and grouping matter.

The last row introduces the scalar triple product$\vec{a}\cdot (\vec{b}\times \vec{c})$, computable as a determinant whose rows are the three vectors:

Its absolute value is the volume of the parallelepiped with edges $\vec{a},\vec{b},\vec{c}$: the base parallelogram has area $\Vert \vec{b}\times \vec{c}\Vert$, and $\Vert \vec{a}\Vert \left|\cos \theta \right|$ is the height, so $V=\left|\vec{a}\cdot (\vec{b}\times \vec{c})\right|$. A zero triple product means zero height: the three vectors are coplanar.

When a force $\vec{F}$ acts at a point with position vector $\vec{r}$, the torque about the origin is the cross product $\vec{\tau }=\vec{r}\times \vec{F}$, whose direction is the axis of rotation and whose magnitude $\Vert \vec{r}\Vert \Vert \vec{F}\Vert \sin \theta$ is the turning effect.

Lines in space

In the plane a line is fixed by a point and a slope. In space a slope is not enough; the direction is carried by a direction vector $\vec{v}$. A point $\vec{r}$ is on the line through ${\vec{r}}_0$ parallel to $\vec{v}$ exactly when $\vec{r}-{\vec{r}}_0$ is a multiple of $\vec{v}$, which gives the vector equation $\vec{r}={\vec{r}}_0+t\vec{v}$.

For the line through $A(2,4,-3)$ and $B(3,-1,1)$, take $\vec{v}=\overrightarrow{AB}=⟨1,-5,4⟩$ and the point $A$:

Setting $z=0$ finds where it pierces the $xy$-plane. Restricting the parameter to $0\le t\le 1$ describes just the segment from ${\vec{r}}_0$ to ${\vec{r}}_1$: $\vec{r}(t)=(1-t){\vec{r}}_0+t{\vec{r}}_1$.

Two lines in space that are neither parallel nor intersecting are skew: they lie in parallel planes and never meet. Parallel lines have proportional direction vectors; intersecting lines admit a common point. When both fail, the lines are skew.

Planes

A single direction does not orient a plane, but a single perpendicular does. A normal vector $\vec{n}$ is orthogonal to every vector lying in the plane. If ${\vec{r}}_0$ is a known point and $\vec{r}$ is arbitrary, then $\vec{r}-{\vec{r}}_0$ lies in the plane, so $\vec{n}\cdot (\vec{r}-{\vec{r}}_0)=0$.

The cross product supplies the normal when a plane is given by three points. For $P(1,3,2)$, $Q(3,-1,6)$, $R(5,2,0)$, the edge vectors $\overrightarrow{PQ}=⟨2,-4,4⟩$ and $\overrightarrow{PR}=⟨4,-1,-2⟩$ cross to $\vec{n}=⟨12,20,14⟩$, and the plane through $P$ is $12(x-1)+20(y-3)+14(z-2)=0$, i.e. $6x+10y+7z=50$.

Three further computations recur:

• Line meets plane. Substitute the line's parametric $x,y,z$ into the plane equation and solve for $t$. For $x=2+3t$, $y=-4t$, $z=5+t$ and the plane $4x+5y-2z=18$, the substitution gives $-10t=20$, so $t=-2$ and the point is $(-4,8,3)$.
• Angle between planes. The angle between two planes is the angle between their normals: $\cos \theta =({\vec{n}}_1\cdot {\vec{n}}_2)/(\Vert {\vec{n}}_1\Vert \Vert {\vec{n}}_2\Vert )$. Parallel planes have parallel normals. The third computation, the distance from a point to a plane, is again a scalar projection onto the normal.

Take any point $P_0$ on the plane; the distance is the length of the projection of $\overrightarrow{P_0{P}_1}$ onto $\vec{n}=⟨a,b,c⟩$. Writing that scalar projection out and using $a{x}_0+b{y}_0+c{z}_0=-d$ collapses it to the formula.