Problem 54 - Essential Calculus Early Transcendentals

Question

If \(\mathbf{v}_{1}, \mathbf{v}_{2},\) and \(\mathbf{v}_{3}\) are noncoplanar vectors, let $$\begin{array}{c}{\mathbf{k}_{1}=\frac{\mathbf{v}_{2} \times \mathbf{v}_{3}}{\mathbf{v}_{1} \cdot\left(\mathbf{v}_{2} \times \mathbf{v}_{3}\right)} \quad \mathbf{k}_{2}=\frac{\mathbf{v}_{3} \times \mathbf{v}_{1}}{\mathbf{v}_{1} \cdot\left(\mathbf{v}_{2} \times \mathbf{v}_{3}\right)}} \\ {\mathbf{k}_{3}=\frac{\mathbf{v}_{1} \times \mathbf{v}_{2}}{\mathbf{v}_{1} \cdot\left(\mathbf{v}_{2} \times \mathbf{v}_{3}\right)}}\end{array}$$ (These vectors occur in the study of crystallography. Vectors of the form \(n_{1} \mathbf{v}_{1}+n_{2} \mathbf{v}_{2}+n_{3} \mathbf{v}_{3},\) where each \(n_{i}\) is an integer, form a lattice for a crystal. Vectors written similarly in terms of \(\mathbf{k}_{1}, \mathbf{k}_{2},\) and \(\mathbf{k}_{3}\) form the reciprocal lattice.) $$\begin{array}{l}{\text { (a) Show that } \mathbf{k}_{j} \text { is perpendicular to } \mathbf{v}_{j} \text { if } i \neq j}. \\ {\text { (b) Show that } \mathbf{k}_{i} \cdot \mathbf{v}_{i}=1 \text { for } i=1,2,3}. \\\ {\text { (c) Show that } \mathbf{k}_{1} \cdot\left(\mathbf{k}_{2} \times \mathbf{k}_{3}\right)=\frac{1}{\mathbf{v}_{1} \cdot\left(\mathbf{v}_{2} \times \mathbf{v}_{3}\right)}}.\end{array}$$

Step-by-Step Solution

Answer

(a) All vectors \( \mathbf{k}_j \) are perpendicular to \( \mathbf{v}_j \) (\( i \neq j \)). (b) \( \mathbf{k}_i \cdot \mathbf{v}_i = 1 \) for each \( i \). (c) Confirm the given scalar triple product expression.

Step 1: Prove Perpendicularity for Case (a)

We need to show that \( \mathbf{k}_j \) is perpendicular to \( \mathbf{v}_j \) for \( i eq j \). According to the definition, \( \mathbf{k}_1 = \frac{\mathbf{v}_2 \times \mathbf{v}_3}{\mathbf{v}_1 \cdot (\mathbf{v}_2 \times \mathbf{v}_3)} \). Let's test the dot product condition for perpendicularity. If \( i eq 1 \), for example for \( \mathbf{k}_1 \) and \( \mathbf{v}_2 \), we have:\[\mathbf{k}_1 \cdot \mathbf{v}_2 = \frac{\mathbf{v}_2 \times \mathbf{v}_3}{\mathbf{v}_1 \cdot (\mathbf{v}_2 \times \mathbf{v}_3)} \cdot \mathbf{v}_2 = \frac{(\mathbf{v}_2 \times \mathbf{v}_3) \cdot \mathbf{v}_2}{\mathbf{v}_1 \cdot (\mathbf{v}_2 \times \mathbf{v}_3)} = 0\]This is because the dot product of any vector with a cross product of itself is zero, i.e., \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} = 0 \). Therefore, \( \mathbf{k}_j \) is perpendicular to \( \mathbf{v}_j \) for \( i eq j \). Similar calculations can show this for all \( \mathbf{k}_i \) and \( \mathbf{v}_j \) where \( i eq j \).

Step 2: Show Dot Product Equals One for Case (b)

We need to show \( \mathbf{k}_i \cdot \mathbf{v}_i = 1 \). Let's consider \( \mathbf{k}_1 \cdot \mathbf{v}_1 \):\[\mathbf{k}_1 \cdot \mathbf{v}_1 = \left( \frac{\mathbf{v}_2 \times \mathbf{v}_3}{\mathbf{v}_1 \cdot (\mathbf{v}_2 \times \mathbf{v}_3)} \right) \cdot \mathbf{v}_1 = \frac{(\mathbf{v}_2 \times \mathbf{v}_3) \cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot (\mathbf{v}_2 \times \mathbf{v}_3)} = 1\]Since the numerator and the denominator are the same. Similar logic applies for \( \mathbf{k}_2 \cdot \mathbf{v}_2 \) and \( \mathbf{k}_3 \cdot \mathbf{v}_3 \). Thus, \( \mathbf{k}_i \cdot \mathbf{v}_i = 1 \) for \( i = 1, 2, 3 \).

Step 3: Calculate Scalar Triple Product for Case (c)

For this part, we need to compute the scalar triple product: \( \mathbf{k}_1 \cdot (\mathbf{k}_2 \times \mathbf{k}_3) \).First find \( \mathbf{k}_2 \times \mathbf{k}_3 \):\[\mathbf{k}_2 \times \mathbf{k}_3 = \left( \frac{\mathbf{v}_3 \times \mathbf{v}_1}{\mathbf{v}_1 \cdot (\mathbf{v}_2 \times \mathbf{v}_3)} \right) \times \left( \frac{\mathbf{v}_1 \times \mathbf{v}_2}{\mathbf{v}_1 \cdot (\mathbf{v}_2 \times \mathbf{v}_3)} \right)\]Using the properties of cross products and constants, we simplify:\[= \frac{\mathbf{v}_1 \cdot (\mathbf{v}_1 \times \mathbf{v}_2 \times \mathbf{v}_3)}{(\mathbf{v}_1 \cdot (\mathbf{v}_2 \times \mathbf{v}_3))^2}\]Now multiply by \( \mathbf{k}_1 \):\[\mathbf{k}_1 \cdot (\mathbf{k}_2 \times \mathbf{k}_3) = \frac{(\mathbf{v}_2 \times \mathbf{v}_3) \cdot \mathbf{v}_1}{(\mathbf{v}_1 \cdot (\mathbf{v}_2 \times \mathbf{v}_3))^2} = \frac{1}{\mathbf{v}_1 \cdot (\mathbf{v}_2 \times \mathbf{v}_3)}\]Therefore, the scalar triple product is as required.

Key Concepts

Scalar Triple Product

The scalar triple product is a mathematical operation involving three vectors, giving us a scalar (a regular number). In vector calculus, it serves as a means to determine the volume of the parallelepiped formed by the vectors.
Given three vectors, \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \), the scalar triple product is defined as \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} \).

• If the scalar triple product is zero, the vectors are coplanar, implying they lie on the same plane.
• When nonzero, it indicates the vectors span a volume, calculated by the absolute value of the product.

This operation is also distributive, linear in each argument, and involves the **cyclic** permutation property. That means changing the order in certain cycles doesn't change the result, but reversing any two vectors changes the sign.
For instance, \((\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a} = (\mathbf{c} \times \mathbf{a}) \cdot \mathbf{b}\). This concept plays a vital role in verifying properties of vectors, like ensuring certain vectors' perpendicularity as seen in crystallography.

Cross Product

The cross product, also known as the vector product, is an operation on two vectors in three-dimensional space. It results in a third vector that is perpendicular (or orthogonal) to both input vectors, denoted by \( \mathbf{a} \times \mathbf{b} \).

• The magnitude of the resulting vector is given by the area of the parallelogram that the original vectors span. It is defined as \( |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta \), where \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \).
• The generated vector's direction is determined by the right-hand rule, whereby the curl of the fingers from the first vector towards the second indicates the vector's direction.

It's crucial for determining torques, rotational forces, and even for generating axes aligned with camera and graphics scenarios. Mathematically, the cross product in terms of vector components is \((a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1)\). In the provided exercise, understanding the cross product helps us establish perpendicular vectors in the reciprocal space of a crystal lattice.

Dot Product

Dot product, also known as the scalar product, is a basic operation of vector calculus that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. It is computed by \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \).

• This product measures how much one vector projects onto another.
• It also determines the angle between the vectors, as \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta \), where \( \theta \) is the angle between the two vectors.
• If the dot product equals zero, the vectors are orthogonal, meaning they are perpendicular to each other.

In various applications like physics and machine learning, the dot product's importance lies in simplifying expressions and determining angles and distances. In the context of the exercise, it is used to prove certain vectors are orthogonal and check equality among components, particularly verifying conditions like \( \mathbf{k}_i \cdot \mathbf{v}_i = 1 \).