Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(\mathbf{a}=\langle\sqrt{2}, \sqrt{2}, 0\rangle, \mathbf{b}=\langle 1,-1,1\rangle\), and \(\mathbf{c}=\langle-2,2,1\rangle\). Find each of the following: (a) \(a^{\cdot} \mathbf{c}\) (b) \((a-c) \cdot b\) (c) \(\mathbf{a} /\|\mathbf{a}\|\) (d) \((\mathbf{b}-\mathbf{c}) \cdot \mathbf{a}\) (e) \(\frac{\mathbf{b} \cdot \mathbf{c}}{\|\mathbf{b}\| \mathbf{c} \|}\) (f) \(\mathbf{a} \cdot \mathbf{a}-\|\mathbf{a}\|^{2}\)

Step-by-Step Solution

Answer

(a) 0, (b) 3, (c) \(\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 0\rangle\), (d) 0, (e) \(-\frac{1}{\sqrt{3}}\), (f) 0.

Step 1: Calculate \( \mathbf{a} \cdot \mathbf{c} \)

To find the dot product \( \mathbf{a} \cdot \mathbf{c} \), use the formula for dot product: \( \mathbf{a} \cdot \mathbf{c} = a_1c_1 + a_2c_2 + a_3c_3 \). Thus, substituting the values gives: \( \sqrt{2} \times (-2) + \sqrt{2} \times 2 + 0 \times 1 = -2\sqrt{2} + 2\sqrt{2} + 0 = 0 \).

Step 2: Calculate \((\mathbf{a} - \mathbf{c}) \cdot \mathbf{b} \)

First, find \( \mathbf{a} - \mathbf{c} = \langle \sqrt{2} - (-2), \sqrt{2} - 2, 0 - 1 \rangle = \langle \sqrt{2}+2, \sqrt{2}-2, -1 \rangle \). Now compute the dot product with \( \mathbf{b} \): \( (\sqrt{2}+2) \times 1 + (\sqrt{2}-2) \times (-1) + (-1) \times 1 = \sqrt{2}+2 - \sqrt{2} + 2 - 1 = 3 \).

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

First, find the magnitude \( \|\mathbf{a}\| = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2 + 0^2} = \sqrt{2+2} = 2 \). Then, \( \mathbf{a} / \|\mathbf{a}\| = \langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 0 \rangle \).

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

First compute \( \mathbf{b} - \mathbf{c} = \langle 1 - (-2), -1 - 2, 1 - 1 \rangle = \langle 3, -3, 0 \rangle \). Then find the dot product with \( \mathbf{a} \): \( 3 \times \sqrt{2} + (-3) \times \sqrt{2} + 0 \times 0 = 3\sqrt{2} - 3\sqrt{2} = 0 \).

Step 5: Calculate \( \frac{\mathbf{b} \cdot \mathbf{c}}{\|\mathbf{b}\| \mathbf{c} \|} \)

First find the dot product \( \mathbf{b} \cdot \mathbf{c} = 1 \times (-2) + (-1) \times 2 + 1 \times 1 = -2 - 2 + 1 = -3 \). Then calculate magnitudes: \( \|\mathbf{b}\| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \) and \( \|\mathbf{c}\| = \sqrt{(-2)^2 + 2^2 + 1^2} = \sqrt{9} = 3 \). Therefore, \( \frac{\mathbf{b} \cdot \mathbf{c}}{\|\mathbf{b}\| \|\mathbf{c}\|} = \frac{-3}{3\sqrt{3}} = \frac{-1}{\sqrt{3}} \).

Step 6: Calculate \( \mathbf{a} \cdot \mathbf{a} - \|\mathbf{a}\|^2 \)

Compute \( \mathbf{a} \cdot \mathbf{a} = (\sqrt{2})^2 + (\sqrt{2})^2 + 0^2 = 2 + 2 + 0 = 4 \) and \( \|\mathbf{a}\|^2 = (2)^2 = 4 \). Thus \( \mathbf{a} \cdot \mathbf{a} - \|\mathbf{a}\|^2 = 4 - 4 = 0 \).

Key Concepts

Dot Product

In vector calculus, the dot product is a fundamental operation that combines two vectors to produce a scalar. It is computed by multiplying the corresponding components of two vectors and then summing these products. For vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{c} = \langle c_1, c_2, c_3 \rangle \), the dot product is calculated as:

• \( \mathbf{a} \cdot \mathbf{c} = a_1c_1 + a_2c_2 + a_3c_3 \)

The dot product results in a scalar and reveals a lot about the relationship between the two vectors:

• If the dot product is zero, it implies that the vectors are orthogonal, meaning they are at right angles to each other. This was the case in Step 1 and Step 4.

Dot products are widely used in physics and engineering to project vectors onto one another, calculate work done by a force, or determine the angle between two vectors.

Vector Magnitude

Vector magnitude represents the "length" of the vector in space. For a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \), the magnitude \( \|\mathbf{a}\| \) is calculated by:

• \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \)

This is akin to finding the distance from the origin of a coordinate system to the point described by the vector. In Step 3, the magnitude of vector \( \mathbf{a} = \langle \sqrt{2}, \sqrt{2}, 0 \rangle \) was calculated to be 2, revealing that it forms a right triangle with its components under the Pythagorean theorem.
Understanding magnitude is crucial for normalizing a vector, a process that involves dividing each component by the vector's magnitude to yield a unit vector, which has a magnitude of 1. This was shown in Step 3 of the solution.

Vector Operations

Vector operations include addition, subtraction, multiplication (dot and cross products), and scalar multiplication. Working with vectors often involves these operations to solve problems in physics, engineering, and computer graphics.

• **Addition and Subtraction:** Adding or subtracting corresponding components of vectors. For example, \( \mathbf{a} - \mathbf{c} \) was calculated to determine a new vector.
• **Scalar Multiplication:** Involves multiplying each component of a vector by a scalar, which scales the vector but does not change its direction.
• **Dot Product:** Already discussed but foundational; it provides a way to multiply vectors to find useful scalar values like work or projections.

These operations expand our ability to interact with multidimensional data in numerous disciplines, making them foundational in linear algebra and calculus applications.

Orthogonal Vectors

Orthogonal vectors are vectors that meet at a right angle (90 degrees). When two vectors are orthogonal, their dot product equals zero, as this signifies no projection from one vector onto the other.

• In vector notation, if \( \mathbf{a} \cdot \mathbf{b} = 0 \), then \( \mathbf{a} \) and \( \mathbf{b} \) are orthogonal.

Orthogonality is a critical concept in vector calculus and linear algebra, as it leads to a simplified understanding of vector projections and cross products, and is frequently used in applications like computer graphics to calculate perpendicularities.
In Steps 1 and 4 of the solution, demonstrating that certain computed dot products equal zero confirmed orthogonality, showing that these vectors stand at right angles in vector space.