3.3 Q-3

Exercise 3.3 — Question 3

This question involves the scalar triple product of three vectors to find the volume of a parallelepiped or tetrahedron.

Key Concepts

Scalar Triple Product:

For three vectors $\mathbf{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$, $\mathbf{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$, and $\mathbf{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$, the scalar triple product is:

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

Volume of a Parallelepiped:

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

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

Volume of a Tetrahedron:

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

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

Method

1. Write the three vectors in component form.
2. Set up the $3\times 3$ determinant with the components of $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ as rows.
3. Evaluate the determinant to find the scalar triple product.
4. Take the absolute value to get the volume of the parallelepiped, or multiply by $\frac{1}{6}$ for the tetrahedron.