Exercise 3.4 — Q1: Scalar Triple Product, Volume & Coplanarity

Key Concepts

This exercise applies the scalar triple product to find volumes of 3D figures and test coplanarity of vectors.

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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:

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

Geometrically, it equals the signed volume of the parallelepiped formed by the three vectors.

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Volume of a Parallelepiped

If $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ are three coterminal edges of a parallelepiped, then:

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

Volume of a Tetrahedron

A tetrahedron has volume equal to $\frac{1}{6}$ of the parallelepiped formed by the same three vectors:

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

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Condition for Coplanarity

Three vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ are coplanar if and only if:

$\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=0⟺\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|=0$

This is because coplanar vectors form a parallelepiped of zero volume.

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Worked Example: Finding an Unknown for Coplanarity

Problem: Find the value of $d$ such that the vectors $\mathbf{a}=\hat{i}+2\hat{j}-3\hat{k},\mathbf{b}=3\hat{i}-\hat{j}+2\hat{k},\mathbf{c}=d\hat{i}+3\hat{j}+\hat{k}$ are coplanar.

Solution: Set the scalar triple product equal to zero:

$\left|\begin{array}{ccc}1& 2& -3\\ 3& -1& 2\\ d& 3& 1\end{array}\right|=0$

Expanding along the first row:

$1[(-1)(1)-(2)(3)]-2[(3)(1)-(2)(d)]+(-3)[(3)(3)-(-1)(d)]=0$

$1(-1-6)-2(3-2d)-3(9+d)=0$

$-7-6+4d-27-3d=0$

$d-40=0⟹d=40$

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Worked Example: Volume of a Parallelepiped

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

Solution:

$V=\left|\left|\begin{array}{ccc}2& 3& 1\\ 1& -1& 2\\ 3& 1& -1\end{array}\right|\right|$

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

$=|2(1-2)-3(-1-6)+1(1+3)|$

$=|2(-1)-3(-7)+4|$

$=|-2+21+4|=|23|=23\text{ cubic units}$

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