Problem 32 - Thomas Calculus

Question

Cross products of three vectors Show that except in degenerate cases, \((\mathbf{u} \times \mathbf{v}) \times \mathbf{w}\) lies in the plane of \(\mathbf{u}\) and \(\mathbf{v},\) whereas \(\mathbf{u} \times(\mathbf{v} \times \mathbf{w})\) lies in the plane of \(\mathbf{v}\) and \(\mathbf{w} .\) What are the degenerate cases?

Step-by-Step Solution

Answer

\((\mathbf{u} \times \mathbf{v}) \times \mathbf{w}\) is in the plane of \(\mathbf{u}, \mathbf{v}\), and \(\mathbf{u} \times(\mathbf{v} \times \mathbf{w})\) is in the plane of \(\mathbf{v}, \mathbf{w}\). Degenerate cases: vectors are parallel or zero.

Step 1: Understanding Cross Product

The cross product of two vectors, say \( \mathbf{a} \) and \( \mathbf{b} \), results in a new vector \( \mathbf{a} \times \mathbf{b} \) that is orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \). This means the cross product vector lies perpendicular to the plane formed by \( \mathbf{a} \) and \( \mathbf{b} \).

Step 2: Using Cross Product on Vectors

To evaluate \((\mathbf{u} \times \mathbf{v}) \times \mathbf{w}\), observe that \(\mathbf{u} \times \mathbf{v}\) is orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\). Now, using the vector triple product identity: \((\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{b} \cdot \mathbf{c})\mathbf{a}\), we can rewrite the expression.

Step 3: Applying Vector Triple Product Identity

Using the identity: \((\mathbf{u} \times \mathbf{v}) \times \mathbf{w} = (\mathbf{u} \cdot \mathbf{w})\mathbf{v} - (\mathbf{v} \cdot \mathbf{w})\mathbf{u}\), each term clearly shows it is a linear combination of \(\mathbf{u}\) and \(\mathbf{v}\). Thus, \((\mathbf{u} \times \mathbf{v}) \times \mathbf{w}\) remains in the plane spanned by \(\mathbf{u}\) and \(\mathbf{v}\).

Step 4: Analyzing \(\mathbf{u} \times (\mathbf{v} \times \mathbf{w})\)

Now consider \(\mathbf{u} \times (\mathbf{v} \times \mathbf{w})\). Apply the identity: \(\mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \cdot \mathbf{w})\mathbf{v} - (\mathbf{u} \cdot \mathbf{v})\mathbf{w}\). This expression is a linear combination of \(\mathbf{v}\) and \(\mathbf{w}\), showing that \(\mathbf{u} \times (\mathbf{v} \times \mathbf{w})\) lies in the plane of \(\mathbf{v}\) and \(\mathbf{w}\).

Step 5: Identifying Degenerate Cases

The degenerate cases occur when the triple product expressions collapse, that is, when vectors are parallel or zero. Specifically, when \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) are not linearly independent (such as when one or more vectors are zero or they are coplanar). In such cases, the cross product evaluation can result in a zero vector or indeterminacy about the plane.

Key Concepts

Cross Product

The cross product is a fundamental operation in vector calculus, used widely in physics and engineering. When you take the cross product of two vectors, such as \( \mathbf{a} \) and \( \mathbf{b} \), the result is a new vector \( \mathbf{a} \times \mathbf{b} \). This new vector isn't just any vector—it's special because it is perpendicular to both of the original vectors. This means it points straight out of the plane formed by \( \mathbf{a} \) and \( \mathbf{b} \), just like how the Z-axis points out from the X-Y plane.
The length of this cross product vector is related to the area of the parallelogram between the two original vectors. So, if your vectors are pointing in nearly the same direction, the cross product will be small. Conversely, if they are perpendicular, the cross product will be at its maximum.
Cross products are only defined in three-dimensional space, making them unique and specifically useful for analyzing spatial problems. In practical terms, if you are dealing with forces, rotational movements, or need to determine orientations, the cross product is your go-to tool.

Vector Triple Product

The vector triple product involves three vectors and is represented as either \((\mathbf{a} \times \mathbf{b}) \times \mathbf{c}\) or \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c})\). This might sound complicated, but with the vector triple product identity, everything becomes clearer.
This identity simplifies these expressions significantly:

• For \((\mathbf{a} \times \mathbf{b}) \times \mathbf{c}\), the identity states: \[(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{b} \cdot \mathbf{c})\mathbf{a}\]
• For \(\mathbf{a} \times (\mathbf{b} \times \mathbf{c})\): \[\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}\]

What's handy about these identities is they rewrite the triple product into something called a linear combination of two vectors. This means the resulting vector will end up lying in the plane spanned by those two vectors. These kind of calculations are crucial when you want to simplify complex vector expressions without losing the essence of their directional components.

Linear Independence

Linear independence is an important concept in vector spaces. To explain it simply, let's think about what it means for vectors to be linearly independent. Basically, a set of vectors are linearly independent if no vector in the set can be written as a combination of the others. Think of it as each vector contributing its own unique 'direction' to the space.
Imagine you're in a classroom with rows and columns. If all columns look the same, they aren't giving any new direction and hence are not independent. In vector terms, if vectors are linearly independent, they define a unique space: a line, a plane, or an entire 3D space. If they aren't, they just define part of that space.
In the context of the problem, if \( \mathbf{u}, \mathbf{v}, \) and \( \mathbf{w} \) are not linearly independent, such as when one vector is a multiple of another, the planes they try to span collapse into a lower dimension. For example, three vectors in the plane all line up to create a line rather than a genuine 3D space, which we call a degenerate case. This is why understanding and confirming linear independence is necessary when applying vector calculus concepts correctly.