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In Exercises \(26-31,\) approximate the component form of the vector \(\vec{v}\) using the information given about its magnitude and direction. Round your approximations to two decimal places. \|\vec{v}\|=63.92 \text { ; when drawn in standard position } \vec{v} \text { makes a } 78.3^{\circ} \text { angle with the positive } x \text { -axis }

Short Answer

Expert verified
The approximate component form of the vector is (13.06, 62.55).

Step by step solution

01

Understand the Vector Problem

We are given the magnitude and direction angle of a vector \( \vec{v} \). The task is to find the component form of this vector. The vector is in standard position, which means it starts at the origin.
02

Use Trigonometric Functions

To find the components \((v_x, v_y)\) of the vector \(\vec{v}\), use the equations: \( v_x = \|\vec{v}\| \cdot \cos(\theta) \) and \( v_y = \|\vec{v}\| \cdot \sin(\theta) \), where \(\theta\) is the angle with the positive x-axis.
03

Substitute Given Values

Use the given values: \(\|\vec{v}\| = 63.92\) and \(\theta = 78.3^\circ\). First, calculate \(v_x = 63.92 \cdot \cos(78.3^\circ)\) and \(v_y = 63.92 \cdot \sin(78.3^\circ)\). Ensure your calculator is in degree mode when finding the cosine and sine.
04

Perform Calculations for \(v_x\)

Calculate \(v_x = 63.92 \cdot \cos(78.3^\circ)\). Use a calculator to find \(\cos(78.3^\circ)\), which is approximately 0.2043. Then, \(v_x \approx 63.92 \cdot 0.2043 = 13.06\).
05

Perform Calculations for \(v_y\)

Calculate \(v_y = 63.92 \cdot \sin(78.3^\circ)\). Use a calculator to find \(\sin(78.3^\circ)\), which is approximately 0.9789. Then, \(v_y \approx 63.92 \cdot 0.9789 = 62.55\).
06

Present the Component Form

Combine the calculated \(v_x\) and \(v_y\) to express the vector in component form: \(\vec{v} \approx (13.06, 62.55)\) after rounding to two decimal places.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnitude and Direction
When we talk about vectors, understanding their magnitude and direction is crucial. Magnitude refers to the length or size of the vector. Imagine it as how far or powerful the vector is in its effect. In this exercise, the magnitude of the vector \(\vec{v}\) is given as 63.92.

Direction, on the other hand, is about where the vector is pointing. It's often described by an angle, and in our exercise, this vector forms a 78.3° angle with the positive x-axis. This tells us how the vector is oriented in space.

Together, magnitude and direction completely define a vector. To solve problems, having both helps us figure out exactly how the vector behaves.
Trigonometric Functions
Trigonometry plays a big role when working with vectors, especially when converting them into component form. The two most relevant trigonometric functions in this context are cosine and sine.

- **Cosine (\( \cos \):** Measures the adjacent side of the angle over the hypotenuse in a right triangle.- **Sine (\( \sin \):** Measures the opposite side over the hypotenuse in a right triangle.By knowing the magnitude of the vector and its direction angle, we use these trigonometric functions to find the vector's components. Using the formulas:
  • \( v_x = \|\vec{v}\| \cdot \cos(\theta) \)
  • \( v_y = \|\vec{v}\| \cdot \sin(\theta) \)
The calculations provide us with the horizontal and vertical components of the vector, respectively. This mathematical bridge translates the vector's properties from a geometric perspective to numeric data.
Standard Position
A vector in standard position means that it originates from the origin of the coordinate plane, which is point (0,0).

This sets a common ground for describing any vector's position and direction because we always measure the vector's angle from the positive x-axis.

By starting at the origin, we simplify calculations and ensure consistency in how vectors are described and compared. In standard position, any vector can be easily described by its direction angle and magnitude without the complexity of translating it from different starting points.
Angle with x-axis
The angle a vector makes with the x-axis is vital for understanding its direction. This angle, often denoted as \( \theta \), helps in determining how the vector spreads across the coordinate plane. In the given exercise, the angle \( \theta = 78.3^\circ \) helps to describe precisely how the vector is oriented relative to the horizontal axis.

This angle is measured counter-clockwise from the positive x-axis, which is the conventional direction in mathematics. Understanding this angle gives us the direction part of the vector's description, while magnitude gives us size. This duo of magnitude and angle thoroughfully positions the vector in the coordinate space, enabling precise calculations and geometric interpretations.

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Most popular questions from this chapter

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