Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

For the three-dimensional vectors \(\mathbf{u}\) and \(\mathbf{v}\) in Problems 13-16, find the sum \(\mathbf{u}+\mathbf{v}\), the difference \(\mathbf{u}-\mathbf{v}\), and the magnitudes \(\|\mathbf{u}\|\) and \(\|\mathbf{v}\|\). $$ \mathbf{u}=\langle 0,0,0\rangle, \mathbf{v}=\langle-3,3,1\rangle $$

Step-by-Step Solution

Answer

The vector sum is \( \langle -3, 3, 1 \rangle \), the difference is \( \langle 3, -3, -1 \rangle \), \( \|\mathbf{u}\| = 0 \), and \( \|\mathbf{v}\| = \sqrt{19} \).

Step 1: Understanding Vector Addition

The sum of two vectors \( \mathbf{u} + \mathbf{v} \) is found by adding their corresponding components. Here, \( \mathbf{u} = \langle 0, 0, 0 \rangle \) and \( \mathbf{v} = \langle -3, 3, 1 \rangle \).

Step 2: Calculating Vector Sum

Add the corresponding components: \( 0 + (-3), 0 + 3, 0 + 1 \). Thus, \( \mathbf{u} + \mathbf{v} = \langle -3, 3, 1 \rangle \).

Step 3: Understanding Vector Subtraction

The difference of two vectors \( \mathbf{u} - \mathbf{v} \) is found by subtracting their corresponding components. This involves taking the components of \( \mathbf{v} \) and subtracting them from \( \mathbf{u} \).

Step 4: Calculating Vector Difference

Subtract the corresponding components: \( 0 - (-3), 0 - 3, 0 - 1 \). Thus, \( \mathbf{u} - \mathbf{v} = \langle 3, -3, -1 \rangle \).

Step 5: Calculating Magnitude of \(\mathbf{u}\)

The magnitude of a vector \( \mathbf{u} = \langle x, y, z \rangle \) is calculated using the formula \( \| \mathbf{u} \| = \sqrt{x^2 + y^2 + z^2} \). For \( \mathbf{u} = \langle 0, 0, 0 \rangle \), the magnitude is \( \| \mathbf{u} \| = \sqrt{0^2 + 0^2 + 0^2} = 0 \).

Step 6: Calculating Magnitude of \(\mathbf{v}\)

Similarly, the magnitude of \( \mathbf{v} = \langle -3, 3, 1 \rangle \) is \( \| \mathbf{v} \| = \sqrt{(-3)^2 + 3^2 + 1^2} = \sqrt{9 + 9 + 1} = \sqrt{19} \).

Key Concepts

Understanding Vector Addition

Vector addition is a fundamental concept in vector algebra. It involves combining two or more vectors to produce a resultant vector. This is helpful in physics and engineering, where multitudes of forces can act at once. Given two vectors, vector addition is performed by adding their corresponding components together. For example, consider the vectors \( \mathbf{u} = \langle 0, 0, 0 \rangle \) and \( \mathbf{v} = \langle -3, 3, 1 \rangle \). The sum \( \mathbf{u} + \mathbf{v} \) is found by taking each component of \( \mathbf{u} \) and \( \mathbf{v} \) and adding them together:

• First component: \( 0 + (-3) = -3 \)
• Second component: \( 0 + 3 = 3 \)
• Third component: \( 0 + 1 = 1 \)

Thus, the result of the vector addition is \( \mathbf{u} + \mathbf{v} = \langle -3, 3, 1 \rangle \). This new vector indicates the cumulative effect of the two vectors combined.

Demystifying Vector Subtraction

Vector subtraction is similar to addition, but instead of adding components, you subtract them. This operation is useful when you need to find the relative change between vectors or analyze forces' opposite effects. To subtract one vector from another, you take the components of the second vector and subtract them from the first vector's components. For the vectors \( \mathbf{u} = \langle 0, 0, 0 \rangle \) and \( \mathbf{v} = \langle -3, 3, 1 \rangle \), the subtraction \( \mathbf{u} - \mathbf{v} \) is calculated as follows:

• First component: \( 0 - (-3) = 3 \)
• Second component: \( 0 - 3 = -3 \)
• Third component: \( 0 - 1 = -1 \)

Therefore, the difference between these vectors is \( \mathbf{u} - \mathbf{v} = \langle 3, -3, -1 \rangle \). This resulting vector demonstrates how much \( \mathbf{u} \) must change to become \( \mathbf{v} \).

Understanding the Magnitude of a Vector

The magnitude of a vector is akin to its length or size, which is crucial for understanding the vector's impact or extent. Calculating the magnitude helps in determining how strong a force represented by the vector is or how far its effect can be felt. The formula for finding the magnitude of a vector \( \mathbf{v} = \langle x, y, z \rangle \) is given by:\[\| \mathbf{v} \| = \sqrt{x^2 + y^2 + z^2}\]For example, let's calculate the magnitudes for

• \( \| \mathbf{u} \| \) where \( \mathbf{u} = \langle 0, 0, 0 \rangle \) is 0. Since all components are zero, the magnitude of \( \mathbf{u} \) is \( \sqrt{0^2 + 0^2 + 0^2} = 0 \).
• \( \| \mathbf{v} \| \) where \( \mathbf{v} = \langle -3, 3, 1 \rangle \), gives a magnitude of \( \sqrt{(-3)^2 + 3^2 + 1^2} = \sqrt{9 + 9 + 1} = \sqrt{19} \).

Knowing the magnitudes allows us to understand and compare vectors beyond their direction, facilitating comprehensive vector analysis.