Exercise 3.3 — Question 16

Problem Statement

Using the scalar triple product, determine whether the vectors $\mathbf{a}=\hat{i}+2\hat{j}-\hat{k},\mathbf{b}=3\hat{i}-\hat{j}+2\hat{k},\mathbf{c}=5\hat{i}+3\hat{j}$ are coplanar.

---

Key Concepts

Scalar Triple Product

The scalar triple product of three vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ is defined as: $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$

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

Condition for Coplanarity

Three vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ are coplanar if and only if their scalar triple product is zero: $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=0$

Volume Interpretation

• Volume of a parallelepiped with edges $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$: $V=|\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})|$

• Volume of a tetrahedron with three edges $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ from a common vertex: $V=\frac{1}{6}|\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})|$

If the scalar triple product equals zero, the parallelepiped has zero volume, confirming the vectors are coplanar.

---

Solution

Given: $\mathbf{a}=\hat{i}+2\hat{j}-\hat{k},\mathbf{b}=3\hat{i}-\hat{j}+2\hat{k},\mathbf{c}=5\hat{i}+3\hat{j}+0\hat{k}$

Step 1: Write the scalar triple product in determinant form: $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=\left|\begin{array}{ccc}1& 2& -1\\ 3& -1& 2\\ 5& 3& 0\end{array}\right|$

Step 2: Expand along the first row: $=1\cdot \left|\begin{array}{cc}-1& 2\\ 3& 0\end{array}\right|-2\cdot \left|\begin{array}{cc}3& 2\\ 5& 0\end{array}\right|+(-1)\cdot \left|\begin{array}{cc}3& -1\\ 5& 3\end{array}\right|$

Step 3: Evaluate each $2\times 2$ determinant: $=1\cdot [(-1)(0)-(2)(3)]-2\cdot [(3)(0)-(2)(5)]+(-1)\cdot [(3)(3)-(-1)(5)]$ $=1\cdot [0-6]-2\cdot [0-10]+(-1)\cdot [9+5]$ $=1(-6)-2(-10)+(-1)(14)$ $=-6+20-14$ $=0$

Conclusion: Since $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=0$, the three vectors are coplanar.