**Problem Statement:**Let $ \alpha $ be the angle between $ \vec{v} $ and $ \vec{w} $, and $ \beta $ be the angle between $ \vec{w} $ and $ \vec{z} $. Is $ \alpha > \beta $? Support your answer mathematically.---To determine whether $ \alpha $ is greater than $ \beta $, we can use the concept of dot products in vector mathematics, which relates to the cosine of the angle between two vectors.For any two vectors $ \vec{a} $ and $ \vec{b} $, the dot product is defined as:\[\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta)\]where $ \theta $ is the angle between $ \vec{a} $ and $ \vec{b} $.1. **Calculate $ \alpha $:** - Use $ \vec{v} \cdot \vec{w} = |\vec{v}| |\vec{w}| \cos(\alpha) $2. **Calculate $ \beta $:** - Use $ \vec{w} \cdot \vec{z} = |\vec{w}| |\vec{z}| \cos(\beta) $By comparing these cosine values, we can determine the relationship between $ \alpha $ and $ \beta $.- If $\cos(\alpha) < \cos(\beta)$, then $\alpha > \beta$.- If $\cos(\alpha) > \cos(\beta)$, then $\alpha < \beta$.Based on the given problem, calculate or assume values for the dot products and magnitudes, then analyze the cosine values to conclude if $ \alpha > \beta $. The image displays mathematical expressions representing vectors in two-dimensional space. These vectors are expressed in both component form and in terms of the unit vectors $ \hat{i} $ and $ \hat{j} $.1. $ \vec{v} = \langle 6, -3 \rangle $ This vector $\vec{v}$ is written in component form, indicating that it has an x-component of 6 and a y-component of -3.2. $ \vec{w} = -2\hat{i} + 5\hat{j} $ Vector $\vec{w}$ is expressed in terms of the unit vectors $\hat{i}$ and $\hat{j}$. It has an x-component of -2 and a y-component of 5.3. $ \vec{z} = -5\hat{j} $ Vector $\vec{z}$ is fully expressed as $-5\hat{j}$, indicating that it is oriented entirely along the y-axis with a magnitude of 5 in the negative direction.4. $ \vec{a} = \langle x, y \rangle $ This vector $\vec{a}$ is a general representation in component form, where $x$ and $y$ are variables representing the components of the vector along the x and y axes, respectively.5. $ \vec{b} = -y\hat{i} + x\hat{j} $ Vector $\vec{b}$ is expressed using $\hat{i}$ and $\hat{j}$, indicating an x-component of $-y$ and a y-component of $x$.There are no graphs or diagrams in this image. The focus is on different ways of expressing vectors mathematically.

Given: v→=6,-3w→=-2i→+5j→z→=-5j→ To Find: Let α be the angle between v→ and w→and β be the angle…

Answered: Let a be the angle between ở and w and…

Math

Trigonometry

Let a be the angle between ở and w and ß be the angle between w and ž. Is a > B? Support your answer mathematically.

Trigonometry (11th Edition)

11th Edition

ISBN:9780134217437

Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Chapter1: Trigonometric Functions

Section: Chapter Questions

Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.

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Question

**Problem Statement:**

Let $ \alpha $ be the angle between $ \vec{v} $ and $ \vec{w} $, and $ \beta $ be the angle between $ \vec{w} $ and $ \vec{z} $. Is $ \alpha > \beta $? Support your answer mathematically.

---

To determine whether $ \alpha $ is greater than $ \beta $, we can use the concept of dot products in vector mathematics, which relates to the cosine of the angle between two vectors.

For any two vectors $ \vec{a} $ and $ \vec{b} $, the dot product is defined as:

\[
\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta)
\]

where $ \theta $ is the angle between $ \vec{a} $ and $ \vec{b} $.

1. **Calculate $ \alpha $:**
- Use $ \vec{v} \cdot \vec{w} = |\vec{v}| |\vec{w}| \cos(\alpha) $

2. **Calculate $ \beta $:**
- Use $ \vec{w} \cdot \vec{z} = |\vec{w}| |\vec{z}| \cos(\beta) $

By comparing these cosine values, we can determine the relationship between $ \alpha $ and $ \beta $.

- If $\cos(\alpha) < \cos(\beta)$, then $\alpha > \beta$.
- If $\cos(\alpha) > \cos(\beta)$, then $\alpha < \beta$.

Based on the given problem, calculate or assume values for the dot products and magnitudes, then analyze the cosine values to conclude if $ \alpha > \beta $.

The image displays mathematical expressions representing vectors in two-dimensional space. These vectors are expressed in both component form and in terms of the unit vectors $ \hat{i} $ and $ \hat{j} $.

1. $ \vec{v} = \langle 6, -3 \rangle $

This vector $\vec{v}$ is written in component form, indicating that it has an x-component of 6 and a y-component of -3.

2. $ \vec{w} = -2\hat{i} + 5\hat{j} $

Vector $\vec{w}$ is expressed in terms of the unit vectors $\hat{i}$ and $\hat{j}$. It has an x-component of -2 and a y-component of 5.

3. $ \vec{z} = -5\hat{j} $

Vector $\vec{z}$ is fully expressed as $-5\hat{j}$, indicating that it is oriented entirely along the y-axis with a magnitude of 5 in the negative direction.

4. $ \vec{a} = \langle x, y \rangle $

This vector $\vec{a}$ is a general representation in component form, where $x$ and $y$ are variables representing the components of the vector along the x and y axes, respectively.

5. $ \vec{b} = -y\hat{i} + x\hat{j} $

Vector $\vec{b}$ is expressed using $\hat{i}$ and $\hat{j}$, indicating an x-component of $-y$ and a y-component of $x$.

There are no graphs or diagrams in this image. The focus is on different ways of expressing vectors mathematically.

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