Vector (Cross) Product

The vector (cross) product of two vectors is itself a vector, perpendicular to both, defined by

where $\theta$ ($0\le\theta\le\pi$) is the angle between them and $\hat n$ is the unit vector perpendicular to the plane of $\vec a,\vec b$, with direction given by the right-hand rule. In components it is most easily computed as a determinant:

Properties

• Anti-commutative: $\vec a\times\vec b=-(\vec b\times\vec a)$.
• Not associative in general: $\vec a\times(\vec b\times\vec c)\ne(\vec a\times\vec b)\times\vec c$.
• Distributive: $\vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c$; and $(\lambda\vec a)\times\vec b=\lambda(\vec a\times\vec b)$.
• Parallel test: $\vec a\parallel\vec b\iff\vec a\times\vec b=\vec 0$; in particular $\vec a\times\vec a=\vec 0$.
• Standard basis (right-handed): $\hat i\times\hat j=\hat k,\ \hat j\times\hat k=\hat i,\ \hat k\times\hat i=\hat j$ (reverse the order to flip the sign).
• Magnitude: $|\vec a\times\vec b|=|\vec a||\vec b|\sin\theta$, so $\sin\theta=\dfrac{|\vec a\times\vec b|}{|\vec a||\vec b|}$.

Areas, unit normals and the Lagrange identity

The magnitude of the cross product is a geometric area:

For a triangle with vertices $A,B,C$ take $\vec a=\overrightarrow{AB},\ \vec b=\overrightarrow{AC}$. A unit vector perpendicular to both $\vec a$ and $\vec b$ is $\hat n=\dfrac{\vec a\times\vec b}{|\vec a\times\vec b|}$ (the other choice is its negative). The Lagrange identity $|\vec a\times\vec b|^2+(\vec a\cdot\vec b)^2=|\vec a|^2|\vec b|^2$ links the two products and is handy when one of $\sin\theta,\cos\theta$ is known.

Scalar triple product

The scalar triple product of three vectors is the scalar

Geometrically its absolute value is the volume of the parallelepiped with the three vectors as coterminous edges (and a tetrahedron on the same edges has volume $\tfrac16\,|[\,\vec a\ \vec b\ \vec c\,]|$). The product is unchanged by a cyclic shift, $[\,\vec a\ \vec b\ \vec c\,]=[\,\vec b\ \vec c\ \vec a\,]=[\,\vec c\ \vec a\ \vec b\,]$, and the dot and cross may be swapped: $\vec a\cdot(\vec b\times\vec c)=(\vec a\times\vec b)\cdot\vec c$.

Coplanarity condition

Three vectors lie in one plane exactly when the parallelepiped they span is flat (zero volume):

Equivalently the $3\times3$ determinant of their components is zero. If any two of the three vectors are equal or parallel, the scalar triple product is automatically $0$.

Deeper Insight — cross product manufactures the perpendicular. The dot product collapses two vectors into a number; the cross product builds a brand-new vector at right angles to both, with length equal to the area they sweep out. That is exactly what you need for a normal to a plane, the axis of a rotation, or torque $\vec\tau=\vec r\times\vec F$ in physics. The determinant form is not a trick to memorise — it is the bookkeeping that keeps the $\hat i,\hat j,\hat k$ pattern straight, and the same determinant, with a dot in front, becomes the triple product (a volume). Carry one picture: cross $=$ area-and-perpendicular, triple product $=$ volume, and the algebra falls out.