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Chapter 10: Problem 44

\(41-46\) . Find the horizontal and vertical components of the vector with given length and direction, and write the vector in terms of the vectors \(\mathbf{i}\) and \(\mathbf{j}\) . $$ |\mathbf{v}|=800, \quad \theta=125^{\circ} $$

Short Answer

Expert verified
The vector is \( \mathbf{v} = -458.88 \mathbf{i} + 655.36 \mathbf{j} \).

Step by step solution

01

Identify Vector Components

The problem provides the length (magnitude) of the vector and its direction. The horizontal component can be found using \[ v_x = |\mathbf{v}| \cos(\theta) \]and the vertical component using \[ v_y = |\mathbf{v}| \sin(\theta) \]where \(|\mathbf{v}|\) is the magnitude and \(\theta\) is the angle from the positive x-axis.
02

Calculate Horizontal Component

Use the formula for the horizontal component:\[ v_x = 800 \cos(125^{\circ}) \]Calculate \(\cos(125^{\circ})\) which is approximately \(-0.5736\). Then:\[ v_x = 800 \times (-0.5736) = -458.88 \]
03

Calculate Vertical Component

Use the formula for the vertical component:\[ v_y = 800 \sin(125^{\circ}) \]Calculate \(\sin(125^{\circ})\) which is approximately \(0.8192\). Then:\[ v_y = 800 \times 0.8192 = 655.36 \]
04

Write Vector in Terms of \(\mathbf{i}\) and \(\mathbf{j}\)

Combine the horizontal and vertical components to express the vector in terms of the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\):\[ \mathbf{v} = -458.88 \mathbf{i} + 655.36 \mathbf{j} \].

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Key Concepts

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

Horizontal Component
When we talk about the horizontal component of a vector, we're identifying the part of the vector that stretches along the horizontal axis, typically the x-axis. This component tells us how much of the vector runs left or right.

To find the horizontal component, use the formula:
  • \( v_x = ||\mathbf{v}| \cos(\theta) \)
Here, \(|\mathbf{v}|\) is the magnitude of the vector, and \(\theta\) is the angle the vector makes with the positive x-axis. It is like projecting the vector onto the horizontal plane.

In the example given with a magnitude of 800 and an angle of 125掳, the calculation becomes:
  • \( v_x = 800 \cos(125掳) \)
  • Using \(\cos(125掳) \approx -0.5736\), the horizontal component \( v_x \) equates to approximately -458.88.
The negative sign indicates that the vector extends in the negative x-direction.
Vertical Component
The vertical component of a vector represents how much of the vector extends up or down along the vertical axis, usually the y-axis. It shows the vector's influence in the upward or downward direction.

To compute the vertical component, you can use the formula:
  • \( v_y = |\mathbf{v}| \sin(\theta) \)
This signifies projecting the vector onto the vertical plane.

In the example provided, where \(|\mathbf{v}| = 800\) and \(\theta = 125掳\), calculate:
  • \( v_y = 800 \sin(125掳) \)
  • With \(\sin(125掳) \approx 0.8192\), the vertical component \( v_y \) is about 655.36.
This result indicates the vector has a strong upward component in the positive y-direction.
Unit Vectors
Unit vectors are fundamental in vector calculus and physics. They provide directions without scaling any length. A unit vector has a magnitude of exactly 1, and they are immensely useful for expressing vectors in a standardized way.

The most common unit vectors are:
  • \(\mathbf{i}\) - Represents the unit vector in the horizontal direction.
  • \(\mathbf{j}\) - Represents the unit vector in the vertical direction.
When expressing any vector \(\mathbf{v}\) in terms of \(\mathbf{i}\) and \(\mathbf{j}\), you're essentially breaking it down into its components along these unit vectors. In the solution, the vector is expressed as:
  • \(\mathbf{v} = -458.88 \mathbf{i} + 655.36 \mathbf{j}\)
This representation literally describes how far and in which direction the vector moves along the x-axis and y-axis.
Magnitude of a Vector
The magnitude of a vector is a measure of its length or size. Think of it as the distance from the origin of the vector to its endpoint in a standard coordinate system. It provides no direction, only size.

In our example, the magnitude was given as:\[ |\mathbf{v}| = 800 \].

Magnitude is calculated by combining both the horizontal and vertical components using the Pythagorean theorem:
  • \(|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}\)
This equation confirms the length of the vector remains consistent with the calculated components. It is important, especially in physics, to understand not just the direction through its components but also how much these components `weigh` in terms of magnitude.

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Most popular questions from this chapter

Find the indicated quantity, assuming \(\mathbf{u}=2 \mathbf{i}+\mathbf{j}, \mathbf{v}=\mathbf{i}-3 \mathbf{j},\) and \(\mathbf{w}=3 \mathbf{i}+4 \mathbf{j}\) $$ (\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}-\mathbf{v}) $$

Find the component of \(\mathbf{u}\) along \(\mathbf{v}\) $$ \mathbf{u}=7 \mathbf{i}-24 \mathbf{j}, \quad \mathbf{v}=\mathbf{j} $$

A plane has normal vector \(\mathbf{n}\) and passes through the point \(P\) (a) Find an equation for the plane. (b) Find the intercepts and sketch a graph of the plane. $$ \mathbf{n}=\langle 3,2,0\rangle, \quad P(1,2,7) $$

Find the work done by the force \(\mathbf{F}\) in moving an object from \(P\) to \(Q .\) $$ \mathbf{F}=400 \mathbf{i}+50 \mathbf{j} ; \quad P(-1,1), Q(200,1) $$

Determine whether or not the given vectors are perpendicular. $$ \langle 4,-2,-4\rangle,\langle 1,-2,2\rangle $$
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