Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(\mathbf{a}\) and \(\mathbf{b}\) be nonparallel vectors, and let \(\mathbf{c}\) be any nonzero vector. Show that \((\mathbf{a} \times \mathbf{b}) \times \mathbf{c}\) is a vector in the plane of \(\mathbf{a}\) and b.

Step-by-Step Solution

Answer

The vector \((\mathbf{a} \times \mathbf{b}) \times \mathbf{c}\) lies in the plane of \(\mathbf{a}\) and \(\mathbf{b}\) due to the vector triple product identity.

Step 1: Understanding the Cross Product

The cross product \(\mathbf{a} \times \mathbf{b}\) results in a vector orthogonal to both \(\mathbf{a}\) and \(\mathbf{b}\). This vector does not lie in the plane formed by \(\mathbf{a}\) and \(\mathbf{b}\), as it is perpendicular to both.

Step 2: Apply Cross Product Property

Using the vector triple product identity: \(\mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \cdot \mathbf{w})\mathbf{v} - (\mathbf{u} \cdot \mathbf{v})\mathbf{w}\), set \(\mathbf{u} = \mathbf{c}\), \(\mathbf{v} = \mathbf{a}\), and \(\mathbf{w} = \mathbf{b}\).

Step 3: Substitute into Triple Product

Substituting the vectors into the identity gives: \((\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{c} \cdot \mathbf{b})\mathbf{a} - (\mathbf{c} \cdot \mathbf{a})\mathbf{b}\).

Step 4: Expansion Result

The result \((\mathbf{c} \cdot \mathbf{b})\mathbf{a} - (\mathbf{c} \cdot \mathbf{a})\mathbf{b}\) is a linear combination of \(\mathbf{a}\) and \(\mathbf{b}\). This means the resulting vector lies in the plane formed by \(\mathbf{a}\) and \(\mathbf{b}\).

Key Concepts

Cross Product

In vector calculus, the cross product is a binary operation on two vectors in three-dimensional space. It results in a third vector that is orthogonal to the original two. This property distinguishes it from the dot product, which results in a scalar. To calculate the cross product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\), you use the determinant of a matrix formed by these vectors and the unit vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\). The formula is:

• \(\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \)

This operation gives us a vector perpendicular to the plane of \(\mathbf{a}\) and \(\mathbf{b}\). Because it is orthogonal, it does not lie within the plane.
The magnitude of the cross product vector is given by \(|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin\theta\), where \(\theta\) is the angle between \(\mathbf{a}\) and \(\mathbf{b}\). This is a key property, as it equals the area of the parallelogram formed by \(\mathbf{a}\) and \(\mathbf{b}\).

Vector Triple Product

The vector triple product involves taking the cross product twice, essentially operating it on three vectors. This is useful for simplifying expressions involving multiple vectors and understanding geometrical relationships in three-dimensional space. The vector triple product identity is crucial:

• \(\mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \cdot \mathbf{w}) \mathbf{v} - (\mathbf{u} \cdot \mathbf{v}) \mathbf{w} \)

This identity simplifies the triple cross product and expresses it as a combination of dot products and scaled vectors. When solving \((\mathbf{a} \times \mathbf{b}) \times \mathbf{c}\), you substitute \(\mathbf{u} = \mathbf{c}\), \(\mathbf{v} = \mathbf{a}\), and \(\mathbf{w} = \mathbf{b}\), resulting in:

• \((\mathbf{c} \cdot \mathbf{b})\mathbf{a} - (\mathbf{c} \cdot \mathbf{a}) \mathbf{b}\)

This shows it is a linear combination of \(\mathbf{a}\) and \(\mathbf{b}\), meaning this vector lies in their plane.
The beauty of this identity is how it allows us to analyze vector relationships and simplifies complex vector expressions into more manageable forms.

Linear Combination

A linear combination is a fundamental concept in vector spaces where we express a vector as a combination of other vectors, each multiplied by a corresponding scalar coefficient.
For example, consider the expression \((\mathbf{c} \cdot \mathbf{b})\mathbf{a} - (\mathbf{c} \cdot \mathbf{a}) \mathbf{b}\). Here, we have:\

• The vector \(\mathbf{a}\) scaled by \((\mathbf{c} \cdot \mathbf{b})\)
• The vector \(\mathbf{b}\) scaled by \(-(\mathbf{c} \cdot \mathbf{a})\)

These scaled vectors combine to form a new vector inside the plane spanned by \(\mathbf{a}\) and \(\mathbf{b}\).
Linear combinations are foundational in understanding vector spaces because they can represent any vector as a sum of scaled basis vectors. This idea is crucial when you wish to decompose vector components along specific directions or in solving systems of linear equations. By analyzing such combinations, one can predict vector behavior from the algebraic relationships of the coefficients and vectors.