8.04e - The scalar triple product

A signed measure of volume

A box has base area $6$ square units and perpendicular height $4$ units. Push its top sideways so that the three edges from one corner are now $(2,0,0)$, $(1,3,0)$ and $(7,7,4)$. Can such a slanted box still have volume $24$, and which calculation sees through the slant? The scalar triple product measures the perpendicular height, wherever the top ends up.

The parallelepiped is the three-dimensional counterpart of a parallelogram: start with three edge vectors at one vertex and complete the six parallelogram faces. It may be slanted; its edges need not be perpendicular. Three coplanar edges give a flattened figure with zero volume.

The definition is not in the formula booklet: learn it. Parentheses make the order of the operations clear. The expression $(\mathbf{a}\cdot\mathbf{b})\times\mathbf{c}$ would ask for a vector product involving a scalar, so it is not the same operation.

To see the geometry, use the parallelogram on $\mathbf{b},\mathbf{c}$ as a base. If $\mathbf{n}=\mathbf{b}\times\mathbf{c}\ne\mathbf{0}$, its area is $\lvert\mathbf{n}\rvert$ and its unit normal is $\hat{\mathbf{n}}=\mathbf{n}/\lvert\mathbf{n}\rvert$. The perpendicular height from the opposite face is $h=\lvert\mathbf{a}\cdot\hat{\mathbf{n}}\rvert$. Therefore

[DIAGRAM: asset_name: A scalar triple product measures base times perpendicular height; asset_slug: h245_8_04_triple_height; caption: The slant changes the edge length, while the perpendicular height remains 4; description: Exact three-dimensional diagram of the parallelepiped with edges a=(2,0,0), b=(1,3,0), c=(7,7,4) from O. Its base is the blue and lavender parallelogram spanned by a and b, of area 6. A dashed sage perpendicular from the upper vertex c to (7,7,0) has label h=4. The clay slanted edge c is labelled separately. A side panel states base area=6, perpendicular height=4, volume=24.]

The sign of $T$ records orientation: a negative projection means the third dimension points to the other side of the chosen base normal. Volume uses the magnitude. Reversing two inputs changes the sign; cycling all three preserves it:

You can verify the cyclic identity by expanding components: both sides contain $a_1b_2c_3+a_2b_3c_1+a_3b_1c_2-a_1b_3c_2-a_2b_1c_3-a_3b_2c_1$. The cyclic identity and sign-change rule are not in the formula booklet: learn them. For a volume, any order of the same three edges gives the same magnitude.

The sign is useful information, but it is not a negative physical volume.

Volumes of parallelepipeds

Use the three edges issuing from one vertex; calculate one vector product and one scalar product, then take the magnitude.

The geometric argument gives a compact rule. It also covers a degenerate base: if the two base edges are parallel, their product and the three-dimensional volume are both zero.

Taking the magnitude at the end preserves the cancellation inside the scalar product. Replacing every component by its absolute value would generally change the volume.

Worked example 2 changes one feature: the third edge is sheared by adding $2\mathbf{a}+2\mathbf{b}$. Its endpoint moves sideways in the base directions.

More generally, adding any linear combination of two edges to the third leaves the triple product unchanged: each added triple product repeats an edge and is zero. This shear consequence is not in the formula booklet; derive it using distribution and the zero self-product.

The method remains just as short when no edge lies along an axis.

Tetrahedra and fixed-volume planes

A tetrahedron has four vertices and four triangular faces. Choose any vertex as a common start for three edge displacements.

A triangular base has half the area of the parallelogram on the same two edges. A pyramid on that triangle has volume one third of its base area times its perpendicular height. Together these factors give $(1/2)(1/3)=1/6$ of the parallelepiped volume. The pyramid result $V=Ah/3$ is assumed geometry and is not printed in the formula booklet; an examination question may supply it.

[DIAGRAM: asset_name: A tetrahedron occupies one sixth of the associated parallelepiped volume; asset_slug: h245_8_04_tetrahedron; caption: Half the parallelogram base and one third of base times height give the factor 1/6; description: Exact three-dimensional diagram of tetrahedron ABCD with A=(1,0,1), B=(3,1,1), C=(1,2,2) and D=(2,0,4). Three edges from A are labelled AB=(2,1,0), AC=(0,2,1), AD=(1,0,3). The triangular base ABC is pale blue, the remaining tetrahedron faces are pale clay, and dashed grey edges complete the associated parallelepiped. A side panel shows base triangle area=sqrt(21)/2, scalar triple product=13 and tetrahedron volume=13/6.]