3.3 Q-11

Exercise 3.3 — Question 11

Problem

Find the volume of the parallelepiped (and tetrahedron) determined by the vectors:

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

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Key Concepts

Scalar Triple Product: The scalar triple product of three vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ is defined as:

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

In determinant form, if $\mathbf{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$, $\mathbf{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$, $\mathbf{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$:

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

Volume Formulas:

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

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Solution

Step 1: Set up the determinant.

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

Step 2: Expand along the first row.

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

Step 3: Evaluate each $2\times 2$ determinant.

$\left|\begin{array}{cc}-1& 3\\ 1& 2\end{array}\right|=(-1)(2)-(3)(1)=-2-3=-5$

$\left|\begin{array}{cc}2& 3\\ 3& 2\end{array}\right|=(2)(2)-(3)(3)=4-9=-5$

$\left|\begin{array}{cc}2& -1\\ 3& 1\end{array}\right|=(2)(1)-(-1)(3)=2+3=5$

Step 4: Combine.

$=1(-5)-2(-5)+(-1)(5)=-5+10-5=0$

Step 5: Interpret the result.

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

$V_{\text{parallelepiped}}=|0|=0$

$V_{\text{tetrahedron}}=\frac{1}{6}|0|=0$

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Summary of Formulas