Problem 8 - A First Course in Continuum Mechanics

Question

Consider any three vectors \(\boldsymbol{a}, \boldsymbol{b}\) and \(\boldsymbol{c}\) which are linearly independent, that is, \((a \times \boldsymbol{b}) \cdot \boldsymbol{c} \neq 0\). Show that: (a) \(\boldsymbol{a} \times \boldsymbol{b}, \boldsymbol{b} \times \boldsymbol{c}\) and \(\boldsymbol{c} \times \boldsymbol{a}\) are also linearly independent, (b) \((\boldsymbol{a} \times \boldsymbol{b}) \otimes \boldsymbol{c}+(\boldsymbol{b} \times \boldsymbol{c}) \otimes \boldsymbol{a}+(\boldsymbol{c} \times \boldsymbol{a}) \otimes \boldsymbol{b}=(\boldsymbol{a} \times \boldsymbol{b} \cdot \boldsymbol{c}) \boldsymbol{I}\).

Step-by-Step Solution

Answer

The vectors \( \boldsymbol{a} \times \boldsymbol{b} \) are proven to be linearly independent by showing that any linear combination equating to zero implies all scalar multiples are zero, and the outer product identity expands into the scalar triple product multiplied by the identity matrix.

Step 1: Review of Linear Independence

Vectors are said to be linearly independent if no vector in the set can be written as a linear combination of the others. For three vectors \( \boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c} \) to be linearly independent, it must be that for scalars \( x_1, x_2, x_3 \) such that \( x_1\boldsymbol{a} + x_2\boldsymbol{b} + x_3\boldsymbol{c} = \mathbf{0} \), then \( x_1 = x_2 = x_3 = 0 \) must hold.

Step 2: Show that \( \boldsymbol{a} \times \boldsymbol{b}, \boldsymbol{b} \times \boldsymbol{c}, \boldsymbol{c} \times \boldsymbol{a} \) are linearly independent

To prove the linear independence of \( \boldsymbol{a} \times \boldsymbol{b}, \boldsymbol{b} \times \boldsymbol{c}, \boldsymbol{c} \times \boldsymbol{a} \) we assume that they can be written as a linear combination that equals the zero vector: \( x(\boldsymbol{a} \times \boldsymbol{b}) + y(\boldsymbol{b} \times \boldsymbol{c}) + z(\boldsymbol{c} \times \boldsymbol{a}) = \mathbf{0} \). We want to show that this implies \( x = y = z = 0 \) which will prove linear independence. Taking the dot product of both sides with \( \boldsymbol{c} \) gives us \( x(\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c} + y(\boldsymbol{b} \times \boldsymbol{c}) \cdot \boldsymbol{c} + z(\boldsymbol{c} \times \boldsymbol{a}) \cdot \boldsymbol{c} = 0 \). Since \( \boldsymbol{b} \times \boldsymbol{c} \) and \( \boldsymbol{c} \times \boldsymbol{a} \) are perpendicular to \( \boldsymbol{c} \) their dot products will be zero, simplifying the equation to \( x(\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c} = 0 \) which implies that \( x = 0 \) since \( (\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c} \) is not zero by assumption of linear independence of \( \boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c} \). A similar process can be repeated by taking the dot product with \( \boldsymbol{a} \) and \( \boldsymbol{b} \) to show that \( y = 0 \) and \( z = 0 \) respectively.

Step 3: Verify the vector outer product identity

To prove the outer product identity given as \( (\boldsymbol{a} \times \boldsymbol{b}) \otimes \boldsymbol{c}+(\boldsymbol{b} \times \boldsymbol{c}) \otimes \boldsymbol{a}+(\boldsymbol{c} \times \boldsymbol{a}) \otimes \boldsymbol{b}=(\boldsymbol{a} \times \boldsymbol{b} \cdot \boldsymbol{c}) \boldsymbol{I} \), we need to show that both sides are equivalent when applied to any arbitrary vector \( \boldsymbol{v} \). Let us compute the left side first. The tensor product \( \otimes \) denotes the outer product between vectors resulting in a matrix. Thus \( (\boldsymbol{a} \times \boldsymbol{b}) \otimes \boldsymbol{c} \) would result in a matrix where each column is \( (\boldsymbol{a} \times \boldsymbol{b}) \) multiplied by the corresponding component of \( \boldsymbol{c} \). Similarly, the other terms are to be expanded. Adding these will eventually yield a matrix that when multiplied by \( \boldsymbol{v} \) gives a vector \( (\boldsymbol{a} \times \boldsymbol{b} \cdot \boldsymbol{c})\boldsymbol{v} \) due to the properties of cross and dot products where the right-hand side is \( (\boldsymbol{a} \times \boldsymbol{b} \cdot \boldsymbol{c}) \boldsymbol{I}\boldsymbol{v} \), \( \boldsymbol{I} \) being the identity matrix. This shows the two sides of the equation give the same result when applied to \( \boldsymbol{v} \) which means they are equivalent.

Key Concepts

Cross Product of Vectors

The cross product, also known as the vector product, is a binary operation on two vectors in three-dimensional space. It is denoted by the symbol \times. Given two vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\), their cross product \(\boldsymbol{a} \times \boldsymbol{b}\) is a vector that is perpendicular to both \(\boldsymbol{a}\) and \(\boldsymbol{b}\), and thus normal to the plane containing them. The magnitude of the cross product is given by \(|\boldsymbol{a} \times \boldsymbol{b}| = |\boldsymbol{a}||\boldsymbol{b}|\sin\theta\), where \(\theta\) is the smaller angle between \(\boldsymbol{a}\) and \(\boldsymbol{b}\).

One important property of the cross product is that it can be used to test for linear independence between three-dimensional vectors. If the cross product of two vectors is non-zero, it indicates that the vectors are not parallel and hence linearly independent. This is crucial in the context of the given exercise, where verifying the non-zero cross product \((\boldsymbol{a} \times \boldsymbol{b}) \boldsymbol{cdot} \boldsymbol{c} eq 0\) confirms that vectors \(\boldsymbol{a}\), \(\boldsymbol{b}\), and \(\boldsymbol{c}\) are linearly independent.

Outer Product Identity

The outer product identity involves the outer product, also known as the tensor product, denoted by \(\bigotimes\). Unlike the cross product, which takes two vectors and produces another vector, the outer product takes two vectors and produces a matrix. For vectors \(\boldsymbol{u}\) and \(\boldsymbol{v}\), their outer product \(\boldsymbol{u} \bigotimes \boldsymbol{v}\) results in a matrix where each element is the product of components from \(\boldsymbol{u}\) and \(\boldsymbol{v}\).

Part (b) of our exercise demonstrates an interesting aspect of the outer product identity. When you add together the outer products of the cross products of vectors \(\boldsymbol{a}, \boldsymbol{b}\), and \(\boldsymbol{c}\) as in \((\boldsymbol{a} \times \boldsymbol{b}) \bigotimes \boldsymbol{c}+(\boldsymbol{b} \times \boldsymbol{c}) \bigotimes \boldsymbol{a}+(\boldsymbol{c} \times \boldsymbol{a}) \bigotimes \boldsymbol{b}\), it equates to the scalar triple product \(\boldsymbol{a} \times \boldsymbol{b} \boldsymbol{cdot} \boldsymbol{c}\) times the identity matrix \(\boldsymbol{I}\). This scalar times the identity matrix will produce a diagonal matrix, which is a simplified form that connects both vector and matrix operations.

Dot Product of Vectors

The dot product, also known as the scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is denoted by \(\cdot\). For two vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\), their dot product \(\boldsymbol{a} \cdot \boldsymbol{b}\) is defined as \(|\boldsymbol{a}| |\boldsymbol{b}| cos\theta\), where \(\theta\) is the angle between \(\boldsymbol{a}\) and \(\boldsymbol{b}\).

The dot product is central in determining orthogonality between two vectors. If the dot product is zero, the vectors are orthogonal (perpendicular to each other). This property is utilized in the solution steps for the exercise, where the vectors resulting from the cross products are perpendicular to the original vectors, thus their dot product with the original vector is zero, assisting in proving linear independence.