Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(\mathbf{a}=\mathbf{i}+2 \mathbf{j}-\mathbf{k}, \mathbf{b}=\mathbf{j}+\mathbf{k}\), and \(\mathbf{c}=-\mathbf{i}+\mathbf{j}+2 \mathbf{k}\). Find each of the following: (a) \(\mathbf{a} \cdot \mathbf{b}\) (b) \((\mathbf{a}+\mathbf{c}) \cdot \mathbf{b}\) (c) \(\mathbf{a} / \| \mathbf{a}\) (d) \((\mathbf{b}-\mathbf{c}) \cdot \mathbf{a}\) (e) \(\frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|}\) (f) \(\mathbf{b} \cdot \mathbf{b}-\|\mathbf{b}\|^{2}\)

Step-by-Step Solution

Answer

(a) 1, (b) 4, (c) \( \frac{1}{\sqrt{6}} \mathbf{i} + \frac{2}{\sqrt{6}} \mathbf{j} - \frac{1}{\sqrt{6}} \mathbf{k} \), (d) 2, (e) \( \frac{1}{2\sqrt{3}} \), (f) 0.

Step 1: Calculate Dot Product \( \mathbf{a} \cdot \mathbf{b} \)

The dot product is computed using the formula \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \). Given \( \mathbf{a} = \mathbf{i} + 2 \mathbf{j} - \mathbf{k} \) and \( \mathbf{b} = \mathbf{j} + \mathbf{k} \), we find:\[ \mathbf{a} \cdot \mathbf{b} = (1)(0) + (2)(1) + (-1)(1) = 0 + 2 - 1 = 1 \]

Step 2: Compute \( (\mathbf{a} + \mathbf{c}) \cdot \mathbf{b} \)

First, calculate \( \mathbf{a} + \mathbf{c} \):\[ \mathbf{a} + \mathbf{c} = (\mathbf{i} + 2\mathbf{j} - \mathbf{k}) + (-\mathbf{i} + \mathbf{j} + 2\mathbf{k}) = 3\mathbf{j} + \mathbf{k} \]Now find the dot product with \( \mathbf{b} \):\[ (\mathbf{a} + \mathbf{c}) \cdot \mathbf{b} = (0)(0) + (3)(1) + (1)(1) = 3 + 1 = 4 \]

Step 3: Calculate \( \mathbf{a} / \| \mathbf{a} \| \)

First, find the magnitude of \( \mathbf{a} \):\[ \| \mathbf{a} \| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]Normalize \( \mathbf{a} \):\[ \mathbf{a} / \| \mathbf{a} \| = \frac{1}{\sqrt{6}} \mathbf{i} + \frac{2}{\sqrt{6}} \mathbf{j} - \frac{1}{\sqrt{6}} \mathbf{k} \]

Step 4: Find \( (\mathbf{b} - \mathbf{c}) \cdot \mathbf{a} \)

Calculate \( \mathbf{b} - \mathbf{c} \):\[ \mathbf{b} - \mathbf{c} = (\mathbf{j} + \mathbf{k}) - (-\mathbf{i} + \mathbf{j} + 2\mathbf{k}) = \mathbf{i} - \mathbf{k} \]Dot product with \( \mathbf{a} \):\[ (\mathbf{b} - \mathbf{c}) \cdot \mathbf{a} = (1)(1) + (0)(2) + (-1)(-1) = 1 + 1 = 2 \]

Step 5: Compute \( \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|} \)

Already found \( \mathbf{a} \cdot \mathbf{b} = 1 \) and \( \| \mathbf{a} \| = \sqrt{6} \). Now find \( \| \mathbf{b} \| \):\[ \| \mathbf{b} \| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2} \]Plug into the formula:\[ \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|} = \frac{1}{\sqrt{6} \cdot \sqrt{2}} = \frac{1}{\sqrt{12}} = \frac{1}{2\sqrt{3}} \]

Step 6: Confirm \( \mathbf{b} \cdot \mathbf{b} - \|\mathbf{b}\|^2 \)

Since \( \mathbf{b} \cdot \mathbf{b} = \|\mathbf{b}\|^2 \), we have:\[ \mathbf{b} \cdot \mathbf{b} = (0^2 + 1^2 + 1^2) = 2 \]\[ \| \mathbf{b} \|^2 = 2 \] Thus the expression evaluates to zero:\[ \mathbf{b} \cdot \mathbf{b} - \|\mathbf{b}\|^2 = 2 - 2 = 0 \]

Key Concepts

Dot Product

The dot product, also known as the scalar product, is an essential operation in vector calculus. It combines two vectors to produce a scalar value. This operation is particularly useful in understanding angles and projections between vectors.
To compute the dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\), we use the formula:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)

Here, \(a_1\), \(a_2\), and \(a_3\) are the components of vector \(\mathbf{a}\), while \(b_1\), \(b_2\), and \(b_3\) are components of vector \(\mathbf{b}\).
An important property of the dot product is that it is positive if the vectors point roughly in the same direction, negative if they are directing oppositely, and zero when the vectors are perpendicular. This operation is fundamental in calculations involving work and energy in physics. By understanding these principles, you can compute dot products effectively in a variety of contexts.

Vector Magnitude

The magnitude of a vector is a measure of its length and is a crucial concept in vector analysis. It provides insight into the size or extent of a vector.
For a three-dimensional vector \( \mathbf{v} = v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k} \), the magnitude is calculated using the formula:

• \( \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2 + v_3^2} \)

This computation essentially involves finding the square root of the sum of the squares of its components. It mirrors the Pythagorean theorem in three dimensions.
Knowing a vector's magnitude is pivotal, especially when normalizing vectors. Normalization involves dividing each component of a vector by its magnitude, turning the vector into a unit vector, which retains the direction but has a magnitude of one. This process is especially useful in applications of physics and engineering, where consistent vector sizes simplify calculations and modeling.

Vector Operations

Vector operations form the backbone of many vector calculus problems. They include addition, subtraction, and scalar multiplication of vectors, and each of these operations follows specific rules.

• **Addition:** For vectors \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\), addition is performed component-wise:\[ \mathbf{a} + \mathbf{b} = (a_1 + b_1)\mathbf{i} + (a_2 + b_2)\mathbf{j} + (a_3 + b_3)\mathbf{k} \]


• **Subtraction:** Similar to addition, subtraction is also done component-wise:\[ \mathbf{a} - \mathbf{b} = (a_1 - b_1)\mathbf{i} + (a_2 - b_2)\mathbf{j} + (a_3 - b_3)\mathbf{k} \]


• **Scalar Multiplication:** Multiplying a vector by a scalar \(c\) results in each component being multiplied by \(c\): \[ c\mathbf{a} = (ca_1)\mathbf{i} + (ca_2)\mathbf{j} + (ca_3)\mathbf{k} \]

Understanding these operations is crucial, as they are frequently encountered in mathematics and physics problems. Mastering them is essential to progress in topics like vector calculus, linear algebra, and applied mathematics.