Problem 2 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(\mathbf{a}=\langle 3,-1\rangle, \mathbf{b}=\langle 1,-1\rangle\), and \(\mathbf{c}=\langle 0,5\rangle\). Find each of the following: (a) \(-4 a+3 b\) (b) \(\mathbf{b} \cdot \mathbf{c}\) (c) \((\mathbf{a}+\mathbf{b}) \cdot \mathbf{c}\) (d) \(2 \mathbf{c} \cdot(3 \mathbf{a}+4 \mathbf{b})\) (e) \(\|\mathbf{b}\| \mathbf{b} \cdot \mathbf{a}\) (f) \(\|\mathbf{c}\|^{2}-\mathbf{c} \cdot \mathbf{c}\)

Step-by-Step Solution

Answer

(a) \(\langle -9, 1 \rangle\); (b) \(-5\); (c) \(-10\); (d) \(-70\); (e) \(4\sqrt{2}\); (f) \(0\).

Step 1: Compute -4a + 3b

First, multiply vector \( \mathbf{a} = \langle 3, -1 \rangle \) by \(-4\): \(-4 \mathbf{a} = \langle -4 \times 3, -4 \times (-1) \rangle = \langle -12, 4 \rangle\). Next, multiply vector \( \mathbf{b} = \langle 1, -1 \rangle \) by \(3\): \(3 \mathbf{b} = \langle 3 \times 1, 3 \times (-1) \rangle = \langle 3, -3 \rangle\). Finally, add these two results: \(-4 \mathbf{a} + 3 \mathbf{b} = \langle -12, 4 \rangle + \langle 3, -3 \rangle = \langle -9, 1 \rangle\).

Step 2: Calculate b · c

Use the dot product formula: If \( \mathbf{b} = \langle x_1, y_1 \rangle \) and \( \mathbf{c} = \langle x_2, y_2 \rangle \), then \( \mathbf{b} \cdot \mathbf{c} = x_1 \times x_2 + y_1 \times y_2 \). For \( \mathbf{b} = \langle 1, -1 \rangle \) and \( \mathbf{c} = \langle 0, 5 \rangle \), this becomes: \( 1 \times 0 + (-1) \times 5 = 0 - 5 = -5 \).

Step 3: Compute (a + b) · c

First, find \( \mathbf{a} + \mathbf{b} \): \( \mathbf{a} = \langle 3, -1 \rangle \) and \( \mathbf{b} = \langle 1, -1 \rangle \) gives \( \mathbf{a} + \mathbf{b} = \langle 3+1, -1+(-1) \rangle = \langle 4, -2 \rangle \). Then, use the dot product: \( \langle 4, -2 \rangle \cdot \langle 0, 5 \rangle = 4 \times 0 + (-2) \times 5 = 0 - 10 = -10 \).

Step 4: Evaluate 2c · (3a + 4b)

First, calculate \( 3 \mathbf{a} + 4 \mathbf{b} \): \( 3 \mathbf{a} = \langle 9, -3 \rangle \) and \( 4 \mathbf{b} = \langle 4, -4 \rangle \), thus \( 3 \mathbf{a} + 4 \mathbf{b} = \langle 9+4, -3+(-4) \rangle = \langle 13, -7 \rangle \). Next, multiply \( \mathbf{c} = \langle 0, 5 \rangle \) by 2: \( 2 \mathbf{c} = \langle 0, 10 \rangle \). Finally, find the dot product: \( \langle 0, 10 \rangle \cdot \langle 13, -7 \rangle = 0 \times 13 + 10 \times (-7) = -70 \).

Step 5: Compute ||b|| b · a

First, compute \( \| \mathbf{b} \| \) using the magnitude formula \( \| \mathbf{b} \| = \sqrt{x^2 + y^2} \). For \( \mathbf{b} = \langle 1, -1 \rangle \), \( \| \mathbf{b} \| = \sqrt{1^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2} \). Then, calculate \( \mathbf{b} \cdot \mathbf{a} \) as \( \langle 1, -1 \rangle \cdot \langle 3, -1 \rangle = 1 \times 3 + (-1) \times (-1) = 3 + 1 = 4 \). Thus, \( \| \mathbf{b} \| \mathbf{b} \cdot \mathbf{a} = \sqrt{2} \times 4 = 4 \sqrt{2} \).

Step 6: Evaluate ||c||² - c · c

First, find \( \| \mathbf{c} \|^2 \) using the magnitude squared formula: \( \| \mathbf{c} \|^2 = (0^2 + 5^2) = 0 + 25 = 25 \). Since \( \mathbf{c} \cdot \mathbf{c} = \| \mathbf{c} \|^2 \), the value is \( 25 \). Hence, \( \| \mathbf{c} \|^2 - \mathbf{c} \cdot \mathbf{c} = 25 - 25 = 0 \).

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector calculus. It combines two vectors to produce a single scalar, making it a crucial tool for various applications in mathematics and physics. The dot product for two vectors, \( \mathbf{b} = \langle x_1, y_1 \rangle \) and \( \mathbf{c} = \langle x_2, y_2 \rangle \), is calculated using the formula:

• \( \mathbf{b} \cdot \mathbf{c} = x_1 \times x_2 + y_1 \times y_2 \)

In this calculation, each component of the vectors is multiplied together and then summed. This operation results in a scalar, which measures how much one vector projects onto another. The dot product is especially useful in determining the angle between two vectors or in checking if they are orthogonal (perpendicular). When vectors are orthogonal, their dot product is zero.

Vector Operations

Vector operations involve various calculations that you can perform with vectors, similar to operations with numbers. Some common vector operations include addition, subtraction, and scalar multiplication. Here are the key points for these operations:

• **Addition**: To add two vectors, such as \( \mathbf{a} = \langle x_1, y_1 \rangle \) and \( \mathbf{b} = \langle x_2, y_2 \rangle \), simply add corresponding components: \( \mathbf{a} + \mathbf{b} = \langle x_1 + x_2, y_1 + y_2 \rangle \).
• **Subtraction**: Similarly, subtracting vectors involves subtracting the components of one vector from another: \( \mathbf{a} - \mathbf{b} = \langle x_1 - x_2, y_1 - y_2 \rangle \).
• **Scalar Multiplication**: This operation involves multiplying each component of a vector by a scalar, \( k \): \( k \cdot \mathbf{a} = \langle kx_1, ky_1 \rangle \).

Vector operations are essential for solving many problems in physics and engineering, providing a way to manipulate directions and magnitudes in a multidimensional space.

Magnitude of Vectors

The magnitude of a vector, also referred to as its length or norm, gives the distance of the vector from the origin in the coordinate space. It is calculated using the Pythagorean theorem applied in multiple dimensions. For a vector \( \mathbf{b} = \langle x, y \rangle \), the magnitude is defined as:

• \( \| \mathbf{b} \| = \sqrt{x^2 + y^2} \)

The magnitude is an essential concept in vector calculus since it provides a measure of how long a vector is, independent of its direction. It is particularly useful when normalizing a vector, which involves scaling it to have a magnitude of 1 while maintaining its direction. This process is vital in applications where direction is necessary, but magnitude is irrelevant, such as in unit vector calculations.