3.4 Scalar and Vector Triple Products

This section extends vector operations to triple products, with applications to volume, coplanarity, and real-world problems.

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Scalar Triple Product

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

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

The result is a scalar (a real number).

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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Properties of Scalar Triple Product

1. Cyclic Permutation (Dot and Cross are Interchangeable)

The scalar triple product is unchanged under cyclic permutations:

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

This also shows that the dot and cross can be interchanged:

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

2. Sign Change on Swap

Swapping any two vectors reverses the sign:

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

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Geometric Interpretation: Volume of a Parallelepiped

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

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

Why? The cross product $\mathbf{b}\times \mathbf{c}$ gives a vector perpendicular to the base with magnitude equal to the base area. The dot product with $\mathbf{a}$ then gives $|\mathbf{a}|\cos \theta$, which is the perpendicular height. So:

$V=\text{Base Area}\times \text{Height}=|\mathbf{b}\times \mathbf{c}|\cdot |\mathbf{a}|\cos \theta$

Volume of a Tetrahedron

A tetrahedron shares the same three coterminal edges but has $\frac{1}{6}$ the volume of the parallelepiped:

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

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Coplanar Vectors

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

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

This makes sense geometrically: if the vectors are coplanar, the parallelepiped they form has zero volume.

Coplanarity of Four Points

Four points $A$, $B$, $C$, $D$ are coplanar if and only if:

$\overrightarrow{AB}\cdot (\overrightarrow{AC}\times \overrightarrow{AD})=0$

Finding an Unknown for Coplanarity

To find an unknown scalar $d$ such that vectors are coplanar:

1. Write the $3\times 3$ determinant of the vectors' components.
2. Set the determinant equal to zero.
3. Solve for $d$.

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Vector Triple Product

The vector triple product $\mathbf{a}\times (\mathbf{b}\times \mathbf{c})$ is a vector (unlike the scalar triple product). It is expanded using the BAC-CAB rule (Lagrange's Identity):

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

Note: The vector triple product is not associative in general: $\mathbf{a}\times (\mathbf{b}\times \mathbf{c})\ne (\mathbf{a}\times \mathbf{b})\times \mathbf{c}$

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Summary Table

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Worked Example

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

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

Expanding along the first row: $=|1[(-1)(-1)-(1)(1)]-2[(2)(-1)-(1)(1)]+3[(2)(1)-(-1)(1)]|$ $=|1(1-1)-2(-2-1)+3(2+1)|$ $=|0+6+9|=15\text{ cubic units}$