Problem 11 - Classical Dynamics of Particles and Systems

Question

Show that the triple scalar product \((\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C}\) can be written as $$(\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C}=\left|\begin{array}{lll} A_{1} & A_{2} & A_{3} \\ B_{1} & B_{2} & B_{3} \\ C_{1} & C_{2} & C_{3} \end{array}\right|$$ Show also that the product is unaffected by an interchange of the scalar and vector product operations or by a change in the order of \(\mathbf{A}, \mathbf{B}, \mathbf{C},\) as long as they are in cyclic order; that is, $$(\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C}=\mathbf{A} \cdot(\mathbf{B} \times \mathbf{C})=\mathbf{B} \cdot(\mathbf{C} \times \mathbf{A})=(\mathbf{C} \times \mathbf{A}) \cdot \mathbf{B}, \quad \text { etc. }$$ We may therefore use the notation ABC to denote the triple scalar product. Finally, give a geometric interpretation of \(\mathbf{A B C}\) by computing the volume of the parallelepiped defined by the three vectors \(\mathbf{A}, \mathbf{B}, \mathbf{C}\)

Step-by-Step Solution

Answer

The triple scalar product \((\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C}\) can be written as the determinant of the matrix containing the components of vectors \(\mathbf{A}, \mathbf{B},\) and \(\mathbf{C}\): \[(\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C} =\left|\begin{array}{lll} A_{1} & A_{2} & A_{3} \\ B_{1} & B_{2} & B_{3} \\ C_{1} & C_{2} & C_{3} \end{array}\right|\] The triple scalar product remains the same under interchanging the scalar and vector product operations, or when changing the order of the vectors involved, as long as they are in cyclic order. The triple scalar product represents the volume of the parallelepiped formed by the three vectors \(\mathbf{A}, \mathbf{B},\) and \(\mathbf{C}\) as its edges: \[V = |\mathbf{ABC}| = \left|\begin{array}{lll} A_{1} & A_{2} & A_{3} \\ B_{1} & B_{2} & B_{3} \\ C_{1} & C_{2} & C_{3} \end{array}\right|\]

Step 1: Calculate the cross product of \(\mathbf{A}\) and \(\mathbf{B}\)

Using the definition of cross product and the components of vectors \(\mathbf{A}\) and \(\mathbf{B}\), we can write: \[\mathbf{A} \times \mathbf{B} = \left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ A_{1} & A_{2} & A_{3} \\ B_{1} & B_{2} & B_{3} \end{array}\right|\] From this, we calculate each component of the cross product: \[\mathbf{A} \times \mathbf{B} =\left(\begin{array}{c} (A_{2}B_{3} - A_{3}B_{2})\hat{\mathbf{i}} \\ (A_{3}B_{1} - A_{1}B_{3})\hat{\mathbf{j}} \\ (A_{1}B_{2} - A_{2}B_{1})\hat{\mathbf{k}} \end{array}\right)\]

Step 2: Calculate the dot product of \((\mathbf{A} \times \mathbf{B})\) and \(\mathbf{C}\)

Using the cross product of \(\mathbf{A}\) and \(\mathbf{B}\) we computed in the previous step, we calculate the dot product with \(\mathbf{C}\): \[(\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C} =\left(\begin{array}{ccc} (A_{2}B_{3} - A_{3}B_{2})C_{1} \\ +(A_{3}B_{1} - A_{1}B_{3})C_{2} \\ +(A_{1}B_{2} - A_{2}B_{1})C_{3} \end{array}\right)\]

Step 3: Express the result as a determinant

We can rewrite the result from the previous step using the determinant notation: \[(\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C}=\left|\begin{array}{lll} A_{1} & A_{2} & A_{3} \\ B_{1} & B_{2} & B_{3} \\ C_{1} & C_{2} & C_{3} \end{array}\right|\]

Step 4: Show the invariance of the triple scalar product

Let's verify that the triple scalar product remains the same when we interchange the scalar and vector product operations or when we change the order of the vectors involved. We'll compute the triple scalar products \(\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})\) and \(\mathbf{B} \cdot (\mathbf{C} \times \mathbf{A})\) and show they are equal to the original expression: \[\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) = \left|\begin{array}{lll} A_{1} & A_{2} & A_{3} \\ B_{1} & B_{2} & B_{3} \\ C_{1} & C_{2} & C_{3} \end{array}\right| = (\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C}\] \[\mathbf{B} \cdot (\mathbf{C} \times \mathbf{A}) = \left|\begin{array}{lll} B_{1} & B_{2} & B_{3} \\ C_{1} & C_{2} & C_{3} \\ A_{1} & A_{2} & A_{3} \end{array}\right| = (\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C}\] Both expressions are equal to the original expression, confirming the invariance of the triple scalar product.

Step 5: Geometric interpretation of the triple scalar product

The triple scalar product represents the volume of the parallelepiped formed by the three vectors \(\mathbf{A}, \mathbf{B},\) and \(\mathbf{C}\) as its edges. It can be expressed as the absolute value of the determinant of a matrix containing the components of the vectors: \[V = |\mathbf{ABC}| = \left|\begin{array}{lll} A_{1} & A_{2} & A_{3} \\ B_{1} & B_{2} & B_{3} \\ C_{1} & C_{2} & C_{3} \end{array}\right|\] The volume is independent of the order of the vectors, and interchanging the scalar and vector product operations or changing the order of the vectors will not affect the volume of the parallelepiped, as demonstrated in Step 4.

Key Concepts

Cross Product

The cross product is a binary operation applied to two vectors in three-dimensional space. It results in another vector that is perpendicular to the plane formed by the original vectors. To compute \(\mathbf{A} \times \mathbf{B}\), where \(\mathbf{A}\) and \(\mathbf{B}\) are vectors, you can use the determinant of a matrix, which includes unit vectors \(\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}\) and the components of \(\mathbf{A}\) and \(\mathbf{B}\):

\[\mathbf{A} \times \mathbf{B} = \left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ A_{1} & A_{2} & A_{3} \ B_{1} & B_{2} & B_{3} \end{array}\right|.\]

The cross product formula gives the vector components by calculating the determinant along each axis:

• **i-component**: \(A_{2}B_{3} - A_{3}B_{2}\)
• **j-component**: \(A_{3}B_{1} - A_{1}B_{3}\)
• **k-component**: \(A_{1}B_{2} - A_{2}B_{1}\).
  

The resulting vector, \(\mathbf{A} \times \mathbf{B}\), represents the direction that is normal to both \(\mathbf{A}\) and \(\mathbf{B}\).

Dot Product

The dot product is another fundamental operation for vectors, but unlike the cross product, it results in a scalar rather than a vector. For two vectors \(\mathbf{X}\) and \(\mathbf{Y}\), the dot product is found by multiplying corresponding components and summing the result:\(\mathbf{X} \cdot \mathbf{Y} = X_1Y_1 + X_2Y_2 + X_3Y_3\).

This operation measures how much of one vector goes in the direction of another. In the context of the triple scalar product \((\mathbf{A} \times \mathbf{B}) \cdot \mathbf{C}\), we have already found \(\mathbf{A} \times \mathbf{B}\) which is then dotted with \(\mathbf{C}\), yielding:

\[((A_{2}B_{3} - A_{3}B_{2})C_{1} + (A_{3}B_{1} - A_{1}B_{3})C_{2} + (A_{1}B_{2} - A_{2}B_{1})C_{3})\]

This resulting scalar represents the volume of a parallelepiped, as we will see further.

Determinant

The mathematical tool that brings everything together in the triple scalar product is the determinant, specifically of a 3x3 matrix. This structure involves the vectors involved:

\[\left| \begin{array}{lll} A_{1} & A_{2} & A_{3} \ B_{1} & B_{2} & B_{3} \ C_{1} & C_{2} & C_{3} \end{array} \right|.\]

The determinant of this matrix gives the same result as the dot product of the cross product integrated with the third vector. It is calculated by expanding along a row (commonly the first):

\[A_{1}(B_{2}C_{3} - B_{3}C_{2}) - A_{2}(B_{1}C_{3} - B_{3}C_{1}) + A_{3}(B_{1}C_{2} - B_{2}C_{1}).\]

This concise expression can demonstrate the volume enclosed by the vectors \(\mathbf{A}\), \(\mathbf{B}\), and \(\mathbf{C}\). Determinants make it easy to see how the three-dimensional interaction of vectors works and more importantly, does not change with cyclic permutations of how these vectors come into play.

Geometric Interpretation

The geometric interpretation of the triple scalar product is an outstanding application, visualizing it as the volume of a parallelepiped formed by three vectors.

The triple scalar product \(\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})\) or any of its equivalent forms gives us this volume precisely. By considering the absolute value of the determinant mentioned earlier:

\[V = |\mathbf{ABC}| = \left| \begin{array}{lll} A_{1} & A_{2} & A_{3} \ B_{1} & B_{2} & B_{3} \ C_{1} & C_{2} & C_{3} \end{array} \right|,\]

this volume remains the same regardless of the vector order because of the cyclical property of determinants. In a more visual sense, if \(\mathbf{A}\), \(\mathbf{B}\), and \(\mathbf{C}\) represent the edges of a box, this value describes the box’s volume. It underscores how vector algebra is essential in connecting geometric visualization with quantitative analytics.