Exercise 3.3 — Question 1

Scalar Triple Product and Volume of Parallelepiped

Key Concepts

Scalar Triple Product of three vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ is defined as: $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$

If the vectors are given in component form: $\mathbf{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k},\mathbf{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k},\mathbf{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$

then the scalar triple product in determinant form 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 Formulas

Volume of a Parallelepiped determined by three vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$: $V=|\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})|$

Volume of a Tetrahedron determined by three vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ from a common vertex: $V=\frac{1}{6}|\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})|$

Worked Example

Find the volume of the parallelepiped determined by: $\mathbf{a}=\hat{i}+2\hat{j}-\hat{k},\mathbf{b}=2\hat{i}-\hat{j}+3\hat{k},\mathbf{c}=\hat{i}+\hat{j}+\hat{k}$

Step 1: Write the scalar triple product as a determinant: $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=\left|\begin{array}{ccc}1& 2& -1\\ 2& -1& 3\\ 1& 1& 1\end{array}\right|$

Step 2: Expand along the first row: $=1\left|\begin{array}{cc}-1& 3\\ 1& 1\end{array}\right|-2\left|\begin{array}{cc}2& 3\\ 1& 1\end{array}\right|+(-1)\left|\begin{array}{cc}2& -1\\ 1& 1\end{array}\right|$ $=1(-1-3)-2(2-3)+(-1)(2+1)$ $=1(-4)-2(-1)+(-1)(3)$ $=-4+2-3=-5$

Step 3: Volume of parallelepiped: $V=|-5|=5\text{ cubic units}$

Volume of tetrahedron with same vectors: $V=\frac{1}{6}|-5|=\frac{5}{6}\text{ cubic units}$

Rectangular Coordinate System in Space

In 3D space, three mutually perpendicular axes — $x$, $y$, and $z$ — define the rectangular (Cartesian) coordinate system. Any point $P$ is represented as an ordered triple $(x,y,z)$.

The unit vectors along these axes are:

• $\hat{i}$ along the $x$-axis
• $\hat{j}$ along the $y$-axis
• $\hat{k}$ along the $z$-axis

A general vector is written as $\mathbf{v}=v_1\hat{i}+v_2\hat{j}+v_3\hat{k}$, with magnitude: $|\mathbf{v}|=\sqrt{v_1^2+v_2^2+v_3^2}$