3.4 Q-6

Exercise 3.4 — Question 6

Topic: Scalar Triple Product, Volume of Parallelepiped/Tetrahedron, and Coplanar Vectors

---

Key Concepts Review

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 defined as:

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

In determinant form:

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

---

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:

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

---

Coplanar Vectors

Three vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ are coplanar if and only if their scalar triple product is zero:

$\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=0$

This means the volume of the parallelepiped formed by the three vectors is zero.

---

Worked Example (Q6 Type)

Problem: Given vectors $\mathbf{a}=\hat{i}+2\hat{j}-\hat{k}$, $\mathbf{b}=2\hat{i}-\hat{j}+3\hat{k}$, $\mathbf{c}=\hat{i}-3\hat{j}+2\hat{k}$:

(i) Find the volume of the parallelepiped. (ii) Find the volume of the tetrahedron. (iii) Determine whether the vectors are coplanar.

Solution:

Compute the scalar triple product using the determinant:

$\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=\left|\begin{array}{ccc}1& 2& -1\\ 2& -1& 3\\ 1& -3& 2\end{array}\right|$

Expanding along the first row:

$=1\left|\begin{array}{cc}-1& 3\\ -3& 2\end{array}\right|-2\left|\begin{array}{cc}2& 3\\ 1& 2\end{array}\right|+(-1)\left|\begin{array}{cc}2& -1\\ 1& -3\end{array}\right|$

$=1[(-1)(2)-(3)(-3)]-2[(2)(2)-(3)(1)]+(-1)[(2)(-3)-(-1)(1)]$

$=1[-2+9]-2[4-3]+(-1)[-6+1]$

$=1(7)-2(1)+(-1)(-5)$

$=7-2+5=10$

(i) Volume of parallelepiped $=|10|=10$ cubic units

(ii) Volume of tetrahedron $=\frac{1}{6}|10|=\frac{10}{6}=\frac{5}{3}$ cubic units

(iii) Since $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=10\ne 0$, the vectors are not coplanar.

---

Summary of Formulas