Precalculus

Analytic Geometry

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Triple Product

Scalar Triple Product

The scalar triple product (also called the mixed product) of three vectors a, b and c is defined as the dot product of the vector a and the cross product of the other two vectors b and c. As the name suggests, the scalar triple product is a scalar quantity. It is denoted by

\[\mathbf{a} \cdot \left({\mathbf{b} \times \mathbf{c}}\right) \text{ or } \left[{\mathbf{a}, \mathbf{b}, \mathbf{c}}\right]\]

In coordinate form, the scalar triple product is calculated using the determinant:

\[\mathbf{a} \cdot \left({\mathbf{b} \times \mathbf{c}}\right) = \left| {\begin{array}{*{20}{c}} {{X_1}} & {{Y_1}} & {{Z_1}}\\ {{X_2}} & {{Y_2}} & {{Z_2}}\\ {{X_3}} & {{Y_3}} & {{Z_3}} \end{array}} \right|,\]

where \(\mathbf{a} = \left({X_1, Y_1, Z_1}\right),\) \(\mathbf{b} = \left({X_2, Y_2, Z_2}\right),\) and \(\mathbf{c} = \left({X_3, Y_3, Z_3}\right).\)

Geometric Properties of the Scalar Triple Product

The volume of a parallelepiped defined by three vectors a, b and c is equal to the absolute value of the scalar triple product of these vectors:

\[V = \Vert{\mathbf{a} \cdot \left({\mathbf{b} \times \mathbf{c}}\right)}\Vert\]
Volume of a parallelepiped as a scalar triple product
Figure 1.

The volume of a pyramid spanned by three vectors a, b and c is expressed by the formula

\[V = \frac{1}{6}\Vert{\mathbf{a} \cdot \left({\mathbf{b} \times \mathbf{c}}\right)}\Vert\]
Volume of a pyramid spanned by three vectors
Figure 2.

Algebraic Properties of the Scalar Triple Product

The scalar triple product is invariant under a circular permutation of the three vectors:

\[\mathbf{a} \cdot \left({\mathbf{b} \times \mathbf{c}}\right) = \mathbf{b} \cdot \left({\mathbf{c} \times \mathbf{a}}\right) = \mathbf{c} \cdot \left({\mathbf{a} \times \mathbf{b}}\right)\]

Swapping any two vectors changes the sign of the scalar triple product:

\[\mathbf{a} \cdot \left({\mathbf{b} \times \mathbf{c}}\right) = - \mathbf{b} \cdot \left({\mathbf{a} \times \mathbf{c}}\right) = - \mathbf{c} \cdot \left({\mathbf{b} \times \mathbf{a}}\right) = - \mathbf{a} \cdot \left({\mathbf{c} \times \mathbf{b}}\right)\]

If the scalar triple product of vectors a, b and c is zero, then the three vectors are linearly dependent (coplanar), i.e. one of the vectors can be represented as a linear combination of the two other vectors:

\[\mathbf{c} = \lambda\mathbf{a} + \mu\mathbf{b},\]

where \(\lambda, \mu\) are real numbers.

A necessary and sufficient condition for three vectors \(\mathbf{a} = \left({X_1, Y_1, Z_1}\right),\) \(\mathbf{b} = \left({X_2, Y_2, Z_2}\right),\) and \(\mathbf{c} = \left({X_3, Y_3, Z_3}\right)\) to be coplanar is that the determinant whose rows are the coordinates of these vectors is equal to zero:

\[\left| {\begin{array}{*{20}{c}} {{X_1}} & {{Y_1}} & {{Z_1}}\\ {{X_2}} & {{Y_2}} & {{Z_2}}\\ {{X_3}} & {{Y_3}} & {{Z_3}} \end{array}} \right| = 0\]

If the scalar triple product of vectors a, b and c is nonzero, then these vectors are linearly independent.

Vector Triple Product

The vector triple product of three vectors a, b and c is defined as the cross product of the vector a with the cross product of the other two vectors b and c.

As the name suggests, the vector triple product is a vector quantity. It can be calculated by the formula

\[\mathbf{a} \times \left({\mathbf{b} \times \mathbf{c}}\right) = \left({\mathbf{a} \cdot \mathbf{c}}\right)\mathbf{b} - \left({\mathbf{a} \cdot \mathbf{b}}\right)\mathbf{c} = \left| {\begin{array}{*{20}{c}} \mathbf{b} & \mathbf{c}\\ \left({\mathbf{a} \cdot \mathbf{b}}\right) & \left({\mathbf{a} \cdot \mathbf{c}}\right) \end{array}} \right|\]

Solved Problems

Click or tap a problem to see the solution.

Example 1

Find the scalar triple product of vectors \(\mathbf{a}\left({1,2,3}\right),\) \(\mathbf{b}\left({1,-1,-1}\right),\) and \(\mathbf{b}\left({1,1,1}\right).\)

Example 2

Show that four points \(A\left({-1,2,1}\right),\) \(B\left({0,4,2}\right),\) \(C\left({-1,5,5}\right),\) and \(D\left({2,5,0}\right)\) lie in the same plane.

Example 3

Show that vectors \(\mathbf{a} = 2\mathbf{i} + 3\mathbf{k},\) \(\mathbf{b} = \mathbf{i} - \mathbf{j}\) and \(\mathbf{c} = 2\mathbf{j} + 3\mathbf{k}\) are coplanar and express vector \(\mathbf{c}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}.\)

Example 4

Given a pyramid with vertices \(A\left({2,1,0}\right),\) \(B\left({-1,5,0}\right),\) \(C\left({-1,1,0}\right),\) and \(D\left({0,0,6}\right).\) Calculate its volume and the height drawn to the face \(ABC.\)

Example 1.

Find the scalar triple product of vectors \(\mathbf{a}\left({1,2,3}\right),\) \(\mathbf{b}\left({1,-1,-1}\right),\) and \(\mathbf{b}\left({1,1,1}\right).\)

Solution.

We can calculate the triple product using the determinant. This yields:

\[\mathbf{a} \cdot \left({\mathbf{b} \times \mathbf{c}}\right) = \left| {\begin{array}{*{20}{c}} {{X_1}} & {{Y_1}} & {{Z_1}}\\ {{X_2}} & {{Y_2}} & {{Z_2}}\\ {{X_3}} & {{Y_3}} & {{Z_3}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 1 & 2 & 3\\ 1 & -1 & -1\\ 1 & 1 & 1 \end{array}} \right| = 1\cdot\left| {\begin{array}{*{20}{c}} -1 & -1\\ 1 & 1 \end{array}} \right| - 2\cdot\left| {\begin{array}{*{20}{c}} 1 & -1\\ 1 & 1 \end{array}} \right| + 3\cdot\left| {\begin{array}{*{20}{c}} 1 & -1\\ 1 & 1 \end{array}} \right| = 1\cdot0 - 2\cdot2 + 3\cdot2 = 2.\]

Example 2.

Show that four points \(A\left({-1,2,1}\right),\) \(B\left({0,4,2}\right),\) \(C\left({-1,5,5}\right),\) and \(D\left({2,5,0}\right)\) lie in the same plane.

Solution.

Compute the coordinates of vectors \(\mathbf{a} = \mathbf{AB},\) \(\mathbf{b} = \mathbf{AC},\) and \(\mathbf{c} = \mathbf{AD}:\)

\[\mathbf{a} = \mathbf{AB} = \left({1,2,1}\right),\;\mathbf{b} = \mathbf{AC} = \left({0,3,4}\right),\;\mathbf{c} = \mathbf{AD} = \left({3,3,-1}\right).\]

Check that these vectors are coplanar (this will mean that they lie in the same plane). Calculate the determinant:

\[\left| {\begin{array}{*{20}{c}} 1 & 2 & 1\\ 0 & 3 & 4\\ 3 & 3 & -1 \end{array}} \right| = 1\cdot\left| {\begin{array}{*{20}{c}} 3 & 4\\ 3 & -1 \end{array}} \right| + 3\cdot\left| {\begin{array}{*{20}{c}} 2 & 1\\ 3 & 4 \end{array}} \right| = 1\cdot\left({-15}\right) + 3\cdot5 = 0.\]

The determinant is zero. Hence, the vectors are coplanar.

Example 3.

Show that vectors \(\mathbf{a} = 2\mathbf{i} + 3\mathbf{k},\) \(\mathbf{b} = \mathbf{i} - \mathbf{j}\) and \(\mathbf{c} = 2\mathbf{j} + 3\mathbf{k}\) are coplanar and express vector \(\mathbf{c}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}.\)

Solution.

The coordinates of the vectors \(\mathbf{a},\) \(\mathbf{b},\) and \(\mathbf{c}\) are equal

\[\mathbf{a} = 2\mathbf{i} + 3\mathbf{k} = \left({2,0,3}\right),\;\mathbf{b} = \mathbf{i} - \mathbf{j} = \left({1,-1,0}\right),\;\mathbf{c} = 2\mathbf{j} + 3\mathbf{k} = \left({0,2,3}\right).\]

If vectors are coplanar then the determinant composed of these vectors is equal to zero. Let's check it out:

\[\left| {\begin{array}{*{20}{c}} {{X_1}} & {{Y_1}} & {{Z_1}}\\ {{X_2}} & {{Y_2}} & {{Z_2}}\\ {{X_3}} & {{Y_3}} & {{Z_3}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 2 & 0 & 3\\ 1 & -1 & 0\\ 0 & 2 & 3 \end{array}} \right| = 2\cdot\left| {\begin{array}{*{20}{c}} -1 & 0\\ 2 & 3 \end{array}} \right| + 3\cdot\left| {\begin{array}{*{20}{c}} 1 & -1\\ 0 & 2 \end{array}} \right| = 2\cdot\left({-3}\right) + 3\cdot2 = 0.\]

Since the three vectors are coplanar, vector \(\mathbf{c}\) can be represented as

\[\mathbf{c} = \lambda\mathbf{a} + \mu\mathbf{b},\]

where \(\lambda, \mu\) are some numbers. Find these numbers by equating the corresponding coordinates:

\[\left[ {\begin{array}{*{20}{c}} 0\\ 2\\ 3 \end{array}} \right] = \lambda\left[ {\begin{array}{*{20}{c}} 2\\ 0\\ 3 \end{array}} \right] + \mu\left[ {\begin{array}{*{20}{c}} 1\\ -1\\ 0 \end{array}} \right], \Rightarrow \left\{ {\begin{array}{*{20}{l}} 0 = 2\lambda + \mu\\ 2 = - \mu\\ 3 = 3\lambda \end{array}} \right., \Rightarrow \left\{ {\begin{array}{*{20}{l}} \lambda = 1\\ \mu = -2 \end{array}} \right..\]

Thus vector \(\mathbf{c}\) can be written as

\[\mathbf{c} = \mathbf{a} - 2\mathbf{b}.\]

Example 4.

Given a pyramid with vertices \(A\left({2,1,0}\right),\) \(B\left({-1,5,0}\right),\) \(C\left({-1,1,0}\right),\) and \(D\left({0,0,6}\right).\) Calculate its volume and the height drawn to the face \(ABC.\)

Solution.

A pyramid given by four points
Figure 3.

Determine the coordinates of the vectors \(\mathbf{a} = \mathbf{AB},\) \(\mathbf{b} = \mathbf{AC}\) and \(\mathbf{c} = \mathbf{AD}\) that span the pyramid:

\[\mathbf{a} = \mathbf{AB} = \left({-3,4,0}\right),\;\mathbf{b} = \mathbf{AC} = \left({-3,0,0}\right),\;\mathbf{c} = \mathbf{AD} = \left({-2,1,6}\right).\]

Calculate the volume of the pyramid using the triple product:

\[\mathbf{a} \cdot \left({\mathbf{b} \times \mathbf{c}}\right) = \left| {\begin{array}{*{20}{c}} -3 & 4 & 0\\ -3 & 0 & 0\\ -2 & 1 & 6 \end{array}} \right| = 6\cdot\left| {\begin{array}{*{20}{c}} -3 & 4\\ -3 & 0 \end{array}} \right| = 6\cdot12 = 72.\]

Hence, \(V = \frac{1}{6}\Vert{\mathbf{a} \cdot \left({\mathbf{b} \times \mathbf{c}}\right)}\Vert = 12.\)

We now find the area of the base of \(ABC\) using the cross product of vectors \(\mathbf{a}\) and \(\mathbf{b}.\)

\[\mathbf{a} \times \mathbf{b} = \left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ -3 & 4 & 0\\ -3 & 0 & 0 \end{array}} \right| = \mathbf{i}\cdot\left| {\begin{array}{*{20}{c}} 4 & 4\\ 0 & 0 \end{array}} \right| - \mathbf{j}\cdot\left| {\begin{array}{*{20}{c}} -3 & 0\\ -3 & 0 \end{array}} \right| + \mathbf{k}\cdot\left| {\begin{array}{*{20}{c}} -3 & 4\\ -3 & 0 \end{array}} \right| = \mathbf{i}\cdot0 - \mathbf{j}\cdot0+\mathbf{k}\cdot12 = 12\mathbf{k}.\]

The area of the base is \(A = \frac{1}{2}\Vert{\mathbf{a} \times \mathbf{b}}\Vert = \frac{1}{2}\cdot12=6.\)

Since the volume of a pyramid is determined by the formula

\[V = \frac{1}{3}A\cdot h,\]

the height of the pyramid is equal

\[h = \frac{3V}{A} = \frac{3\cdot12}{6} = 6.\]