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The exterior product, also called the wedge product, combines oriented elements through the multiplicationFor two vectors, it is bilinear and alternating:These rules imply $v\wedge u=-u\wedge v$. They distinguish the exterior product from both the symmetric dot product and the unconstrained tensor product.^[ref]

In an oriented orthonormal plane, write $u=(a,b)$ and $v=(c,d)$. Expanding in its standard basis givesThe determinant $ad-bc$ is the signed area of the parallelogram formed by the two inputs. Its absolute value is the ordinary area. Reversing the input order reverses the sign; parallel inputs produce zero.

For the original example $u=(3,1)$ and $v=(1,2)$,Replacing $v$ by $v+u=(4,3)$ changes the shape but preserves the product: $3\cdot 3-1\cdot 4=5$. Both parallelograms have the same oriented area element. If $v=2u$, the exterior product is zero, since $u\wedge (2u)=2(u\wedge u)=0$.

The same rule adds directions in higher dimensions. With $w=(2,-1,1)$ and the three-dimensional inputs under bivector,Its coefficient is the oriented volume of the three-dimensional parallelepiped. Repeating or linearly depending on earlier inputs collapses the volume, giving zero.

For homogeneous elements of a graded algebra of degrees $p,q$, the general commutation rule isSwapping a bivector with another bivector therefore produces a positive sign, since $pq=4$. It is the number of individual vector swaps that controls the sign, not a rule that every pair of objects must anticommute.

A parallelepiped is a polyhedron with six parallelogram faces in three opposite pairs. Three generating vectors determine its edges; the absolute value of their determinant gives its volume in orthonormal Cartesian coordinates.

Three vectors form a basis of ${\mathbb{R}}^3$ exactly when they are linearly independent. The determinant with those vectors as columns is their scalar triple product, so the condition isGeometrically their parallelepiped must have nonzero volume.

Subtract the first vertex from the remaining three to obtain edge vectors $a=(1,2,-2)$, $b=(0,1,3)$, $c=(-1,-1,0)$. Their scalar triple product isA tetrahedron occupies one sixth of the corresponding parallelepiped, so its volume isThe absolute value removes the orientation sign of the determinant.

The parallelepiped volume is the absolute value of the scalar triple product of its edge vectors, equivalently the absolute determinant of the matrix with those columns. Expansion along the first row givesTherefore

Multiplying a three-dimensional matrix by a real $\lambda$ multiplies its determinant by ${\lambda }^3$, whereas multiplicativity of the determinant gives $\det (A^5)=(\det A{)}^5$. Therefore equality of the two parallelepiped volumes meansBoth real signs are allowed:A negative value reverses the oriented volume while preserving the ordinary volume. If a positive scale factor is intended, select the positive sign.

Write $\times$ for the paper's $\wedge$. In a right-handed Cartesian coordinate system, the cross product isIt is perpendicular to both vectors, has norm $|\mathbf{a}||\mathbf{b}|\sin \varphi$, where $\varphi$ is their angle, and has direction given by the right-hand rule when nonzero. Its norm is the area of the parallelogram spanned by the two vectors.

If $\mathbf{b}\times \mathbf{c}\ne 0$, use that parallelogram as the base of the parallelepiped. Its perpendicular height is $|\mathbf{a}\cdot \mathbf{n}|$, where $\mathbf{n}=(\mathbf{b}\times \mathbf{c})/|\mathbf{b}\times \mathbf{c}|$. Thus its volume isIf $\mathbf{b}\times \mathbf{c}=0$, both the volume and this scalar triple product vanish. If the cross product is nonzero, the scalar triple product vanishes exactly when $\mathbf{a}$ lies in the plane spanned by $\mathbf{b},\mathbf{c}$. If it is zero, $\mathbf{b},\mathbf{c}$ are linearly dependent, and adjoining $\mathbf{a}$ still spans at most a plane. Consequently the three vectors are coplanar if and only if $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=0$, including cases involving a zero vector.

The volume of a parallelepiped is the absolute scalar triple product of its edge vectors. HereThe absolute value removes the orientation of the chosen edge ordering.

With $s=\sin {\theta }_0$ and $c=\cos {\theta }_0$, conversion from spherical coordinates gives $\mathbf{a}=r_0(s,0,c)$, $\mathbf{b}=r_0(-s/2,\sqrt{3}s/2,c)$ and $\mathbf{c}=r_0(-s/2,-\sqrt{3}s/2,c)$. The volume of the parallelepiped is the absolute value of their scalar triple product:The absolute value matters when the polar angle is in the lower hemisphere. At the poles or equator the three edge vectors are linearly dependent and the volume vanishes.

The volume of a parallelepiped is the absolute scalar triple product of its three edge vectors. Since $\overrightarrow{OP}\times \overrightarrow{OQ}=(0,0,3)$,The factor $1/6$ appropriate to a tetrahedron must not be applied to this parallelepiped.

The position vectors end at the midpoints of $BC$, $CA$ and $AB$, respectively. The drawing places $E$, $F$ and $G$ on those edges and shows the arrows from the origin. The volume of their parallelepiped is the absolute scalar triple product:Equivalently, this is the absolute determinant of the matrix whose columns are the three vectors. Since it is nonzero, the vectors are linearly independent. Three linearly independent vectors in the three-dimensional vector space ${\mathbb{R}}^3$ form a basis.

If the three vectors lie in one plane through the origin and $\mathbf{u},\mathbf{v}$ are independent, their cross product is a normal vector to that plane, so its dot product with $\mathbf{w}$ is zero. If $\mathbf{u},\mathbf{v}$ are dependent, their cross product is already zero. Thus coplanarity always forces the scalar triple product to vanish.

Taking the contrapositive proves that a nonzero scalar triple product rules out coplanarity. Geometrically its absolute value is the nonzero volume of the spanned parallelepiped.

The volume of the parallelepiped is the absolute value of the scalar triple product:

Replacing row one by $R_1-R_2-R_3-R_4$ does not change the determinant: adding a multiple of another row preserves a determinant. It produces first row $(-2,0,0,0)$, whose cofactor minor is the three-dimensional identity matrix. Thus the determinant is $-2$.

The columns are homogeneous coordinate vectors $(1,v_i{)}^{\top }$ of the four vertices. Subtracting the first column from each other column preserves the determinant and reduces its magnitude to the absolute determinant of the three edge vectors based at $v_1$. That absolute determinant is the parallelepiped volume. A tetrahedron occupies one sixth of that parallelepiped, so $V=|-2|/3!=1/3$ in cubic coordinate units.

A rectangular parallelepiped is a parallelepiped whose three edge directions at each vertex are mutually perpendicular. Its faces are rectangles; if its edge lengths are $a,b,c$, its volume is $abc$. A cube is the special case $a=b=c$.