Exercise 3.3 — Question 2

Topic: Scalar Triple Product and Volume Calculations

This exercise covers the scalar triple product of three vectors and its application to finding volumes of geometric solids.

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Key Concepts

Scalar Triple Product

For three vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ in space, the scalar triple product is defined as:

$\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$

The result is a scalar (a real number).

Determinant Form

If $\mathbf{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$, $\mathbf{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$, $\mathbf{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$, then:

$\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|$

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Volume Formulas

Volume of a Parallelepiped

The volume of a parallelepiped determined by three vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ from a common vertex is:

$V_{\text{parallelepiped}}=|\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})|$

Volume of a Tetrahedron

The volume of a tetrahedron determined by three vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ from a common vertex is:

$V_{\text{tetrahedron}}=\frac{1}{6}|\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})|$

This is $\frac{1}{6}$ of the parallelepiped volume.

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Worked Examples

Example 1: Volume of a Parallelepiped

Find the volume of the parallelepiped determined by: $\mathbf{a}=\hat{i}+2\hat{j}-\hat{k},\mathbf{b}=2\hat{i}-\hat{j}+\hat{k},\mathbf{c}=\hat{i}+\hat{j}+2\hat{k}$

Solution:

$\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=\left|\begin{array}{ccc}1& 2& -1\\ 2& -1& 1\\ 1& 1& 2\end{array}\right|$

Expanding along the first row: $=1[(-1)(2)-(1)(1)]-2[(2)(2)-(1)(1)]+(-1)[(2)(1)-(-1)(1)]$ $=1(-2-1)-2(4-1)+(-1)(2+1)$ $=-3-6-3=-12$

$V_{\text{parallelepiped}}=|-12|=12\text{ cubic units}$

Example 2: Volume of a Tetrahedron

Using the same vectors as above:

$V_{\text{tetrahedron}}=\frac{1}{6}|-12|=\frac{12}{6}=2\text{ cubic units}$

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Important Properties

• The scalar triple product is zero if and only if the three vectors are coplanar.
• The dot ($\cdot$) and cross ($\times$) are interchangeable: $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=(\mathbf{a}\times \mathbf{b})\cdot \mathbf{c}$
• Cyclic permutation preserves the value: $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=\mathbf{b}\cdot (\mathbf{c}\times \mathbf{a})=\mathbf{c}\cdot (\mathbf{a}\times \mathbf{b})$