Problem 13 - Essential Calculus Early Transcendentals

Question

(a) Show that \(\mathbf{i} \cdot \mathbf{j}=\mathbf{j} \cdot \mathbf{k}=\mathbf{k} \cdot \mathbf{i}=0\) (b) Show that \(\mathbf{i} \cdot \mathbf{i}=\mathbf{j} \cdot \mathbf{j}=\mathbf{k} \cdot \mathbf{k}=1\)

Step-by-Step Solution

Answer

Vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are orthogonal and unit vectors, hence \( \mathbf{i} \cdot \mathbf{j} = \mathbf{j} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{i} = 0 \) and \( \mathbf{i} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{j} = \mathbf{k} \cdot \mathbf{k} = 1 \).

Step 1: Understand the Dot Product

The dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is defined as \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos\theta \), where \( \theta \) is the angle between the vectors. The dot product is zero if the vectors are perpendicular (\( \theta = 90^\circ \)).

Step 2: Verify Orthogonality for i, j, k

Vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are unit vectors along the x, y, and z axes, respectively. They are orthogonal to each other. Thus, \( \mathbf{i} \cdot \mathbf{j} = \mathbf{j} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{i} = 0 \).

Step 3: Use Definition of Dot Product of Same Vector

The dot product of a vector with itself is its magnitude squared: \( \mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2 \). For a unit vector, its magnitude is 1.

Step 4: Calculate Dot Product for i, j, k with Themselves

Since \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are unit vectors, \( \mathbf{i} \cdot \mathbf{i} = |\mathbf{i}|^2 = 1 \), \( \mathbf{j} \cdot \mathbf{j} = |\mathbf{j}|^2 = 1 \), and \( \mathbf{k} \cdot \mathbf{k} = |\mathbf{k}|^2 = 1 \).

Key Concepts

Orthogonal Vectors

Orthogonal vectors are vectors that meet at a right angle, or 90 degrees. In simple terms, they are perpendicular to each other. When two vectors are orthogonal, their dot product is zero. This is because the cosine of the angle between them, 90 degrees, is zero. The dot product formula is: \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\theta) \).

• If \( \theta = 90^{\circ} \), then \( \cos(\theta) = 0 \), leading to \( \mathbf{a} \cdot \mathbf{b} = 0 \).
• Common examples of orthogonal vectors in three-dimensional space include the unit vectors \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \).

These vectors represent the positive directions along the x, y, and z axes, respectively. So, since they are orthogonal to each other, their dot products are \( \mathbf{i} \cdot \mathbf{j} = \mathbf{j} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{i} = 0 \). This orthogonal relationship is essential in understanding vector geometry and physics, as it simplifies calculations and clarifies vector relationships.

Unit Vectors

Unit vectors are vectors with a magnitude of exactly one unit. They are used to indicate direction. In physics and engineering, they help describe directions in space without altering the vector’s magnitude.

• The standard unit vectors in three-dimensional space are \( \mathbf{i}, \mathbf{j}, \mathbf{k} \).
• They correspond to the x-axis, y-axis, and z-axis directions, respectively.

The importance of unit vectors can be seen when working with the dot product. Typically, when calculating the dot product of a vector with itself, we take its magnitude squared. For a unit vector \( \mathbf{a} \), \( \mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2 = 1^2 = 1 \). Therefore, for the unit vectors \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \), their dot products with themselves are:

• \( \mathbf{i} \cdot \mathbf{i} = 1 \)
• \( \mathbf{j} \cdot \mathbf{j} = 1 \)
• \( \mathbf{k} \cdot \mathbf{k} = 1 \)

Understanding unit vectors is crucial for defining other vectors in terms of directional components and simplifying complex vector equations.

Vector Magnitude

Vector magnitude refers to the length or size of a vector. It is a scalar value that represents how long the vector is, irrespective of direction. The magnitude of a vector \( \mathbf{a} = (a_1, a_2, a_3) \) in a 3-dimensional space is calculated using the formula: \[| \mathbf{a} | = \sqrt{a_1^2 + a_2^2 + a_3^2}\] This formula comes from extending the Pythagorean theorem to three dimensions.

• For a unit vector, the magnitude is always 1.
• For example, \( | \mathbf{i} | = 1\), \( | \mathbf{j} | = 1\), and \( | \mathbf{k} | = 1\).

Understanding vector magnitude is important when analyzing vectors since it allows us to separate the vector's direction from its strength or extent. Whether in physics under vector forces or in computer graphics dealing with new coordinates, the vector magnitude offers a measure for the length of the vector, and thus helps describe the vector’s full properties.