Because the cross product of two vectors is a vector, it is possible to combine the dot product and the cross product. The dot product of a vector with the cross product of two other vectors is called the triple scalar product because the result is a scalar.
Definition
The
triple scalar product of vectors
$\text{u},$$\text{v},$ and
$\text{w}$ is
$\text{u}\xb7\left(\text{v}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{w}\right).$
Calculating a triple scalar product
The triple scalar product of vectors
$\text{u}={u}_{1}\text{i}+{u}_{2}\text{j}+{u}_{3}\text{k},$$\text{v}={v}_{1}\text{i}+{v}_{2}\text{j}+{v}_{3}\text{k},$ and
$\text{w}={w}_{1}\text{i}+{w}_{2}\text{j}+{w}_{3}\text{k}$ is the determinant of the
$3\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}3$ matrix formed by the components of the vectors:
Let
$\text{u}=\u27e81,3,5\u27e9,\text{v}=\u27e82,\mathrm{-1},0\u27e9\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{w}=\u27e8\mathrm{-3},0,\mathrm{-1}\u27e9.$ Calculate the triple scalar product
$\text{u}\xb7\left(\text{v}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{w}\right).$
Calculate the triple scalar product
$\text{a}\xb7\left(\text{b}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{c}\right),$ where
$\text{a}=\u27e82,\mathrm{-4},1\u27e9,$$\text{b}=\u27e80,3,\mathrm{-1}\u27e9,$ and
$\text{c}=\u27e85,\mathrm{-3},3\u27e9.$
When we create a matrix from three vectors, we must be careful about the order in which we list the vectors. If we list them in a matrix in one order and then rearrange the rows, the absolute value of the determinant remains unchanged. However, each time two rows switch places, the determinant changes sign:
Rearranging vectors in the triple products is equivalent to reordering the rows in the matrix of the determinant. Let
$\text{u}={u}_{1}\text{i}+{u}_{2}\text{j}+{u}_{3}\text{k},$$\text{v}={v}_{1}\text{i}+{v}_{2}\text{j}+{v}_{3}\text{k},$ and
$\text{w}={w}_{1}\text{i}+{w}_{2}\text{j}+{w}_{3}\text{k}.$ Applying
[link] , we have
We can obtain the determinant for calculating
$\text{u}\xb7\left(\text{w}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}\right)$ by switching the bottom two rows of
$\text{u}\xb7\left(\text{v}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{w}\right).$ Therefore,
$\text{u}\xb7\left(\text{v}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{w}\right)=\text{\u2212}\text{u}\xb7\left(\text{w}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}\right).$
Following this reasoning and exploring the different ways we can interchange variables in the triple scalar product lead to the following identities:
Let
$\text{u}$ and
$\text{v}$ be two vectors in standard position. If
$\text{u}$ and
$\text{v}$ are not scalar multiples of each other, then these vectors form adjacent sides of a parallelogram. We saw in
[link] that the area of this parallelogram is
$\Vert \text{u}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{v}\Vert .$ Now suppose we add a third vector
$\text{w}$ that does not lie in the same plane as
$\text{u}$ and
$\text{v}$ but still shares the same initial point. Then these vectors form three edges of a
parallelepiped , a three-dimensional prism with six faces that are each parallelograms, as shown in
[link] . The volume of this prism is the product of the figure’s height and the area of its base. The triple scalar product of
$\text{u},\text{v},$ and
$\text{w}$ provides a simple method for calculating the volume of the parallelepiped defined by these vectors.
Volume of a parallelepiped
The volume of a parallelepiped with adjacent edges given by the vectors
$\text{u},\text{v},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{w}$ is the absolute value of the triple scalar product:
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