Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the sum \(\mathbf{u}+\mathbf{v}\), the difference \(\mathbf{u}-\mathbf{v}\), and the magnitudes \(\|\mathbf{u}\|\) and \(\|\mathbf{v}\|\) $$ \mathbf{u}=\langle 12,12\rangle, \mathbf{v}=\langle-2,2\rangle $$

Step-by-Step Solution

Answer

\(\mathbf{u} + \mathbf{v} = \langle 10, 14 \rangle\), \(\mathbf{u} - \mathbf{v} = \langle 14, 10 \rangle\), \(\|\mathbf{u}\| = 12\sqrt{2}\), \(\|\mathbf{v}\| = 2\sqrt{2}\).

Step 1: Calculate the Sum \( \mathbf{u} + \mathbf{v} \)

To find the sum of two vectors \( \mathbf{u} \) and \( \mathbf{v} \), add their corresponding components separately. Given \( \mathbf{u} = \langle 12, 12 \rangle \) and \( \mathbf{v} = \langle -2, 2 \rangle \):\[ \mathbf{u} + \mathbf{v} = \langle 12 + (-2), 12 + 2 \rangle = \langle 10, 14 \rangle \]

Step 2: Calculate the Difference \( \mathbf{u} - \mathbf{v} \)

To find the difference of two vectors \( \mathbf{u} \) and \( \mathbf{v} \), subtract their corresponding components:\[ \mathbf{u} - \mathbf{v} = \langle 12 - (-2), 12 - 2 \rangle = \langle 14, 10 \rangle \]

Step 3: Calculate the Magnitude \( \|\mathbf{u}\| \)

The magnitude of a vector \( \mathbf{u} = \langle x, y \rangle \) is calculated using the formula \( \|\mathbf{u}\| = \sqrt{x^2 + y^2} \):\[ \|\mathbf{u}\| = \sqrt{12^2 + 12^2} = \sqrt{144 + 144} = \sqrt{288} = 12\sqrt{2} \]

Step 4: Calculate the Magnitude \( \|\mathbf{v}\| \)

Similarly, calculate the magnitude of \( \mathbf{v} = \langle -2, 2 \rangle \):\[ \|\mathbf{v}\| = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

Key Concepts

Adding Vectors: Understanding the Basics

Vector addition is a fundamental operation where you combine two vectors to get a resultant vector. Imagine vectors as arrows having both direction and magnitude. To add vectors, you simply align the tail of the second vector to the head of the first and draw a new vector from the tail of the first to the head of the second.
In coordinate form, you add vectors by summing their corresponding components separately. Given vectors \( \mathbf{u} = \langle 12, 12 \rangle \) and \( \mathbf{v} = \langle -2, 2 \rangle \), the sum is:

• First component: \( 12 + (-2) = 10 \)
• Second component: \( 12 + 2 = 14 \)

This results in a new vector \( \langle 10, 14 \rangle \). It’s like navigating a grid; you move right and then up according to your vector components.

Subtraction of Vectors: More Than Just Reversing

Vector subtraction might seem as simple as reversing addition, but it's a bit more engaging. The goal is to find a vector that indicates how one vector differs from another.
Mathematically, it's like tip-to-tail addition, but adjusting the direction of the second vector before adding. Subtracting vectors \( \mathbf{u} = \langle 12, 12 \rangle \) and \( \mathbf{v} = \langle -2, 2 \rangle \) involves:

• Subtracting the first components: \( 12 - (-2) = 14 \)
• Subtracting the second components: \( 12 - 2 = 10 \)

This results in the vector \( \langle 14, 10 \rangle \). So, vector subtraction essentially moves you from the tip of \( \mathbf{v} \) back to the tip of \( \mathbf{u} \).

Understanding the Magnitude of a Vector

The magnitude of a vector is a measure of its length, from its initial to terminal point. It's crucial to understand how long or strong a vector is. The formula for the magnitude of a vector \( \mathbf{a} = \langle x, y \rangle \) is:

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

For \( \mathbf{u} = \langle 12, 12 \rangle \):

• Calculate \( \| \mathbf{u} \| = \sqrt{12^2 + 12^2} = 12\sqrt{2} \)

And for \( \mathbf{v} = \langle -2, 2 \rangle \):

• Calculate \( \| \mathbf{v} \| = \sqrt{(-2)^2 + 2^2} = 2\sqrt{2} \)

The magnitude helps in quantifying how much distance a vector spans in space.

The Role of Coordinate Geometry in Vector Operations

Coordinate geometry gives a spatial backdrop to vector operations, letting us visually represent and manipulate vectors in a plane.
This area of mathematics provides a framework to dissect vectors, clarifying ideas like direction and place in space.
When we talk about vectors \( \mathbf{u} = \langle 12, 12 \rangle \) and \( \mathbf{v} = \langle -2, 2 \rangle \), coordinate geometry helps us:

• Locate these vectors in the 2D plane.
• Visualize their addition or subtraction performances.
• Understand geometric transformations involved.

Essentially, it bridges the gap between numerical calculations and geometric explorations, anchoring our understanding of vectors.