3.3 Q-14

Exercise 3.3 — Question 14

Problem

Show that the vectors $\mathbf{a}=\hat{i}-2\hat{j}+3\hat{k}$, $\mathbf{b}=2\hat{i}+\hat{j}-\hat{k}$, and $\mathbf{c}=\hat{j}+\hat{k}$ are coplanar.

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Key Concept: Coplanar Vectors

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

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

Geometrically, this means the volume of the parallelepiped formed by the three vectors is zero — they all lie in the same plane.

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Scalar Triple Product 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:

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

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Solution

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

Compute the scalar triple product:

$\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=\left|\begin{array}{ccc}1& -2& 3\\ 2& 1& -1\\ 0& 1& 1\end{array}\right|$

Expanding along the first row:

$=1\left|\begin{array}{cc}1& -1\\ 1& 1\end{array}\right|-(-2)\left|\begin{array}{cc}2& -1\\ 0& 1\end{array}\right|+3\left|\begin{array}{cc}2& 1\\ 0& 1\end{array}\right|$

$=1(1\cdot 1-(-1)\cdot 1)+2(2\cdot 1-(-1)\cdot 0)+3(2\cdot 1-1\cdot 0)$

$=1(1+1)+2(2+0)+3(2-0)$

$=1(2)+2(2)+3(2)$

$=2+4+6=12$

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Volume of Parallelepiped and Tetrahedron

The same scalar triple product gives geometric volumes:

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Summary

• Scalar triple product $=\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})$ evaluated as a $3\times 3$ determinant.
• Coplanarity condition: $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=0$
• Volume of parallelepiped: $|\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})|$
• Volume of tetrahedron: $\frac{1}{6}|\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})|$