Problem 1 - Essential Calculus Early Transcendentals

Question

Find the cross product a \(\times\) b and verify that it is orthogonal to both a and b. $$\mathbf{a}=\langle 6,0,-2\rangle, \quad \mathbf{b}=\langle 0,8,0\rangle$$

Step-by-Step Solution

Answer

The cross product \( \mathbf{a} \times \mathbf{b} = \langle 16, 0, 48 \rangle \) and it is orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \).

Step 1: Compute the Cross Product

To find the cross product \( \mathbf{a} \times \mathbf{b} \), use the determinant of a matrix with unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) in the first row, and vectors \( \mathbf{a} = \langle 6,0,-2 \rangle \) and \( \mathbf{b} = \langle 0,8,0 \rangle \) in the second and third rows respectively:\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \6 & 0 & -2 \0 & 8 & 0 \\end{vmatrix}\]Evaluate the determinant:\[\mathbf{a} \times \mathbf{b} = \mathbf{i}(0 \cdot 0 - (-2) \cdot 8) - \mathbf{j}(6 \cdot 0 - (-2) \cdot 0) + \mathbf{k}(6 \cdot 8 - 0 \cdot 0)\]Simplify:\[\mathbf{a} \times \mathbf{b} = \mathbf{i}(16) - \mathbf{j}(0) + \mathbf{k}(48) = \langle 16, 0, 48 \rangle\]

Step 2: Verify Perpendicularity to Vector a

To check if \( \mathbf{a} \times \mathbf{b} \) is orthogonal to \( \mathbf{a} \), compute the dot product \( \mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) \):\[\mathbf{a} = \langle 6, 0, -2 \rangle, \quad \mathbf{a} \times \mathbf{b} = \langle 16, 0, 48 \rangle\]\[\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 6 \cdot 16 + 0 \cdot 0 + (-2) \cdot 48\]Calculate:\[96 + 0 - 96 = 0\]Since the dot product is zero, \( \mathbf{a} \times \mathbf{b} \) is orthogonal to \( \mathbf{a} \).

Step 3: Verify Perpendicularity to Vector b

To check if \( \mathbf{a} \times \mathbf{b} \) is orthogonal to \( \mathbf{b} \), compute the dot product \( \mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) \):\[\mathbf{b} = \langle 0, 8, 0 \rangle, \quad \mathbf{a} \times \mathbf{b} = \langle 16, 0, 48 \rangle\]\[\mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) = 0 \cdot 16 + 8 \cdot 0 + 0 \cdot 48\]Calculate:\[0 + 0 + 0 = 0\]Since the dot product is zero, \( \mathbf{a} \times \mathbf{b} \) is orthogonal to \( \mathbf{b} \).

Key Concepts

Orthogonality

In the world of vectors, orthogonality refers to the perpendicular relationship between two vectors. Two vectors are said to be orthogonal if their dot product is zero. Imagine standing upright and pointing one arm to the left while the other points forward. In this position, your arms form a 90-degree angle, which is exactly what it means to be orthogonal.A real-world application of orthogonality is often observed in computer graphics, where calculating normal vectors that are orthogonal to a surface is essential for lighting and rendering effects. If we consider vectors \( \mathbf{a} \) and \( \mathbf{b} \) given in the exercise, the task was to find the cross product \( \mathbf{a} \times \mathbf{b} \) and verify that this new vector is orthogonal to both originals. This is because the cross product of two vectors is always orthogonal to the plane formed by them, simplifying many calculations in three-dimensional space.

Dot Product

The dot product is a fundamental concept in vector analysis that provides a way to multiply vectors and obtain a scalar. It helps determine the angle between two vectors and is deeply tied to the idea of orthogonality. The dot product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is calculated as:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]If the result is zero, this means the vectors are orthogonal. For the exercise at hand, the dot product was used to verify that the cross product vector is indeed orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \). Why is it important? Knowing when two vectors are orthogonal can greatly simplify calculations and is crucial for understanding geometric relations in vector spaces. It indicates that the vectors are independent in direction within that space, which is particularly handy in fields like physics and engineering.

Vector Analysis

Vector analysis is the study and manipulation of vectors, which are mathematical entities with both magnitude and direction. This branch of mathematics plays a critical role in various fields like physics, engineering, and computer science. It helps describe quantities such as force, velocity, and acceleration.In vector analysis, operations like the dot product and cross product are fundamental. While the dot product gives a scalar (or a number) that tells about the angle relationship between two vectors, the cross product results in a new vector that is orthogonal to the original vectors.The exercise demonstrated these concepts by finding the cross product of vectors \( \mathbf{a} \) and \( \mathbf{b} \) to verify orthogonality. Highlights of vector analysis include:

• Understanding how vectors transform under different coordinate systems.
• Using operations to solve real-world problems, such as finding areas and volumes using vector products.
• Providing the tools necessary for analyzing forces in physics.

Vector analysis serves as a universal language for solving complex spatial problems, enabling practitioners to model and predict real-world phenomena effectively.