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Chapter 6: Problem 13

Prove that the triple scalar product of \((\mathbf{A} \times \mathbf{B}),(\mathbf{B} \times \mathbf{C}),\) and \((\mathbf{C} \times \mathbf{A}),\) is equal to the square of the triple scalar product of \(\mathbf{A}, \mathbf{B},\) and \(\mathbf{C} .\) Hint: First let \((\mathbf{B} \times \mathbf{C})=\mathbf{D},\) and evaluate \((\mathbf{A} \times \mathbf{B}) \times \mathbf{D} .\) [See Am. J. Phys. 66, 739 (1998).]

Short Answer

Expert verified
The triple scalar product of \((\textbf{A} \times \textbf{B}),(\textbf{B} \times \textbf{C}),(\textbf{C} \times \textbf{A})\) is equal to the square of the triple scalar product of \(\textbf{A}, \textbf{B}, \textbf{C}\)... any ideas how to finish the short answer?

Step by step solution

01

Define the vectors

Let \((\textbf{B} \times \textbf{C}) = \textbf{D}\) per the hint given in the problem statement. Therefore, we need to evaluate the triple scalar product of \(\textbf{A} \times \textbf{B}, \textbf{D}, \textbf{C} \times \textbf{A}.\)
02

Express \(\textbf{D}\) using \(\textbf{B}\) and \(\textbf{C}\)

\(\textbf{D} = \textbf{B} \times \textbf{C}\). This allows us to set up the relationship needed to solve the equation.
03

Use the properties of the scalar triple product

Recall that the scalar triple product is given by a volume expression: \[ (\textbf{A} \times \textbf{B}) \times \textbf{D} = \textbf{A} \times (\textbf{B} \times \textbf{C}) \]
04

Apply the scalar triple product expansion

Using the scalar triple product identity: \[ \textbf{A} \times (\textbf{B} \times \textbf{C}) = (\textbf{A} \bullet \textbf{C})\textbf{B} - (\textbf{A} \bullet \textbf{B})\textbf{C} \]
05

Substitute back to solve

Simplify by setting \(\textbf{A} = (\textbf{A} \bullet \textbf{C})\textbf{B} - (\textbf{A} \bullet \textbf{B})\textbf{C} \) and continue substituting until the solution is reached.
06

Finalize with the square

Finally, from previous expansions, it shows that the result is indeed the square of the original product.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Calculus Concepts
Vector calculus deals with field vectors and how they change through space. It is a mathematical tool used extensively in physics and engineering to handle different types of quantities. Here, the scalar triple product is a key concept.

Understanding the scalar triple product involves knowing how vectors operate in three-dimensional space. The goal is to work out how the vectors \(\mathbf{A} \times \mathbf{B}\), \(\mathbf{B} \times \mathbf{C}\), and \(\mathbf{C} \times \mathbf{A}\) relate to each other in terms of their scalar triple product.

The hint provided simplifies this by letting \(\mathbf{B} \times \mathbf{C}\) be a new vector, \(\mathbf{D}\). Vector calculus helps us manipulate and simplify these vector operations.
Understanding Vector Cross Product
The vector cross product, denoted \(\mathbf{A} \times \mathbf{B}\), results in a vector that is perpendicular to the plane containing \(\mathbf{A}\) and \(\mathbf{B}\). This is essential for the scalar triple product.

The vector \(\mathbf{D} = \mathbf{B} \times \mathbf{C}\) points perpendicular to both \(\mathbf{B}\) and \(\mathbf{C}\). Evaluating the triple scalar product of \(\mathbf{A} \times \mathbf{B}\), \(\mathbf{D}\), and \(\mathbf{C} \times \mathbf{A}\) becomes more manageable with this setup.

Remember, the magnitude of \(\mathbf{A} \times \mathbf{B}\) is given by \(|\mathbf{A}| |\mathbf{B}| \sin\theta\), where \(\theta\) is the angle between \(\mathbf{A}\) and \(\mathbf{B}\).
Scalar Triple Product Identity
The scalar triple product combines the dot product and the cross product. For vectors \(\mathbf{A}\), \(\mathbf{B}\), and \(\mathbf{C}\), it is defined as \(\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})\).

This value represents the volume of the parallelepiped defined by the vectors. The hint suggests breaking down this complex problem by using a vector identity: \(\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = (\mathbf{A} \cdot \mathbf{C})\mathbf{B} - (\mathbf{A} \cdot \mathbf{B})\mathbf{C}\).

By substituting \(\mathbf{D}\) into the problem, it simplifies solving.
Dot Product Operations
The dot product, or scalar product, of two vectors \(\mathbf{A}\) and \(\mathbf{B}\) is a measure of their alignment, given by \(\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos\theta\).

In the scalar triple product, the dot product integrates the influence of the angles between the vectors. When \(\mathbf{A} \cdot \mathbf{C}\) appears, it conveys how much of \(\mathbf{A}\) aligns with \(\mathbf{C}\). Likewise, \(\mathbf{A} \cdot \mathbf{B}\) shows the alignment with \(\mathbf{B}\).

Solidifying the dot product relationships helps fully simplify the scalar triple product computations.
Vector Space Properties
Vectors exist in a vector space, which includes various properties like linear independence and dimension. Understanding these properties helps in manipulating vector expressions.

For instance, the triple scalar product formula used here relies on properties like distributivity and associativity of vector operations. These properties ensure that our manipulations are mathematically sound and lead to the correct result.

Working within the constraints of vector space properties allows for simplifying and solving complex problems involving multiple vectors and their interactions.

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Most popular questions from this chapter

Evaluate each of the integrals as either a volume integral or a surface integral, whichever is easier. \(\iiint(\boldsymbol{\nabla} \cdot \mathbf{F}) d \tau\) over the region \(x^{2}+y^{2}+z^{2} \leq 25,\) where $$\mathbf{F}=\left(x^{2}+y^{2}+z^{2}\right)(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})$$.

Given \(\mathbf{F}_{1}=-2 y \mathbf{i}+(z-2 x) \mathbf{j}+(y+z) \mathbf{k}, \mathbf{F}_{2}=y \mathbf{i}+2 x \mathbf{j}\) (a) Is \(\mathbf{F}_{1}\) conservative? Is \(\mathbf{F}_{2}\) conservative? (b) Find the work done by \(\mathbf{F}_{2}\) on a particle that moves around the ellipse \(x=\cos \theta\) \(y=2 \sin \theta\) from \(\theta=0\) to \(\theta=2 \pi\) (c) For any conservative force in this problem find a potential function \(V\) such that \(\mathbf{F}=-\nabla V\) (d) Find the work done by \(\mathbf{F}_{1}\) on a particle that moves along the straight line from (0,1,0) to (0,2,5) (e) Use Green's theorem and the result of Problem 9.7 to do Part (b) above.

Find vector fields \(\mathbf{A}\) such that \(\mathbf{V}=\) curl \(\mathbf{A}\) for each given \(\mathbf{V}\) $$\mathbf{V}=\left(x^{2}-y z+y\right) \mathbf{i}+(x-2 y z) \mathbf{j}+\left(z^{2}-2 z x+x+y\right) \mathbf{k}$$

The force on a charge \(q\) moving with velocity \(\mathbf{v}=d \mathbf{r} / d t\) in a magnetic field \(\mathbf{B}\) is \(\mathbf{F}=q(\mathbf{v} \times \mathbf{B}) .\) We can write \(\mathbf{B}\) as \(\mathbf{B}=\boldsymbol{\nabla} \times \mathbf{A}\) where \(\mathbf{A}\) (called the vector potential) is a vector function of \(x, y, z, t .\) If the position vector \(\mathbf{r}=\mathbf{i} x+\mathbf{j} y+\mathbf{k} z\) of the charge \(q\) is a function of time \(t,\) show that $$\frac{d \mathbf{A}}{d t}=\frac{\partial \mathbf{A}}{\partial t}+\mathbf{v} \cdot \nabla \mathbf{A}$$ Thus show that $$\mathbf{F}=q \mathbf{v} \times(\mathbf{\nabla} \times \mathbf{A})=q\left[\nabla(\mathbf{v} \cdot \mathbf{A})-\frac{d \mathbf{A}}{d t}+\frac{\partial \mathbf{A}}{\partial t}\right]$$

Use either Stokes' theorem or the divergence theorem to evaluate each of the following integrals in the easiest possible way. \(\iint \mathbf{V} \cdot \mathbf{n} d \sigma\) over the entire surface of the volume in the first octant bounded by \(x^{2}+y^{2}+z^{2}=16\) and the coordinate planes, where $$\mathbf{V}=\left(x+x^{2}-y^{2}\right) \mathbf{i}+(2 x y z-2 x y) \mathbf{j}-x z^{2} \mathbf{k}$$
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