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Chapter 7: Problem 41

Vector \(\mathbf{V}\) is in standard position, and makes an angle of \(40^{\circ}\) with the positive \(x\)-axis. Its magnitude is 18 . Write \(\mathbf{V}\) in component form \(\langle a, b\rangle\) and in vector component form \(a \mathbf{i}+b \mathbf{j}\).

Short Answer

Expert verified
\( \mathbf{V} = \langle 13.788, 11.574 \rangle \) or \( \mathbf{V} = 13.788 \mathbf{i} + 11.574 \mathbf{j} \).

Step by step solution

01

Understanding Vector Components

To express a vector in component form, we need to find the horizontal and vertical components of the vector. These are determined using its angle with the positive x-axis and its magnitude. The components can be expressed as \( \langle a, b \rangle \), where \( a \) is the horizontal component and \( b \) is the vertical component.
02

Calculating the Horizontal Component

The horizontal component \( a \) is calculated using cosine of the angle. The formula is: \( a = \| \mathbf{V} \| \cos(\theta) \). Here, \( \| \mathbf{V} \| = 18 \) and \( \theta = 40^{\circ} \). Calculate: \( a = 18 \cos(40^{\circ}) \).
03

Calculating the Vertical Component

The vertical component \( b \) is calculated using sine of the angle. The formula is: \( b = \| \mathbf{V} \| \sin(\theta) \). Using \( \| \mathbf{V} \| = 18 \) and \( \theta = 40^{\circ} \), calculate \( b = 18 \sin(40^{\circ}) \).
04

Calculating the Cosine and Sine Values

Use a calculator to find \( \cos(40^{\circ}) \) which is approximately 0.766, and \( \sin(40^{\circ}) \) which is approximately 0.643.
05

Finalize Components

Substitute the cosine and sine values into the component formulas: \( a = 18 \times 0.766 = 13.788 \) and \( b = 18 \times 0.643 = 11.574 \). Thus, the vector in component form is \( \langle 13.788, 11.574 \rangle \).
06

Writing in Vector Component Form

Using the components \( a \) and \( b \), express the vector in the form \( a \mathbf{i} + b \mathbf{j} \). Therefore, \( \mathbf{V} = 13.788 \mathbf{i} + 11.574 \mathbf{j} \).

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Key Concepts

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

Magnitude
Magnitude describes the overall size or length of a vector. It is a measurement of how far the vector extends in space, regardless of its direction. For a vector \(\mathbf{V}\), the magnitude is often denoted by \(|\mathbf{V}|\) and is typically found using the Pythagorean theorem or other formulas involving the vector鈥檚 components if they are known.
  • When given as a scalar, the magnitude does not affect direction, but simply indicates the size.
  • In this exercise, the vector \(\mathbf{V}\) has a magnitude of 18, which tells us that the vector is 18 units long.
Knowing the magnitude helps in calculating the horizontal and vertical components, essential for describing the vector in component form.
Angle with x-axis
The angle a vector makes with the x-axis is crucial in determining its direction in a plane. This angle, measured in degrees or radians, tells how slanted the vector is relative to the positive x-axis.
  • In this exercise, the vector \(\mathbf{V}\) makes a 40掳 angle with the positive x-axis.
  • This angle helps in using trigonometric functions to find the vector's components.
Understanding this angle allows us to utilize the sine and cosine functions effectively to resolve the vector into its x (horizontal) and y (vertical) components.
Component Form
Component form of a vector breaks it down into its horizontal and vertical parts. Expressing a vector in component form makes it easy to perform calculations and visualize in two-dimensional space.
  • It is denoted as \(\langle a, b \rangle\), where \(a\) and \(b\) are the components along the x and y axes, respectively.
  • This form allows us to see how much the vector moves along each axis separately.
A better grasp on component form aids in applying basic vector operations like addition, subtraction, and scaling through simple arithmetic on these components.
Horizontal Component
The horizontal component of a vector, represented as \(a\), shows how much the vector extends along the x-axis.
  • Calculated using the cosine trigonometric function: \(a = |\mathbf{V}| \cos(\theta)\).
  • In the example given, we use \(\cos(40^{\circ})\) to find that \(a = 18 \times 0.766\) which gives approximately 13.788.
The horizontal component is key in determining how much of the vector's magnitude is directed along the horizontal plane.
Vertical Component
The vertical component, denoted as \(b\), indicates the amount of the vector's magnitude projected along the y-axis.
  • It utilizes the sine function: \(b = |\mathbf{V}| \sin(\theta)\).
  • For our problem, we calculate this by \(b = 18 \times 0.643\), yielding approximately 11.574.
Understanding this component allows you to measure how much the vector moves upwards or downwards in a planar context.
Trigonometric Functions
Trigonometric functions, specifically sine and cosine, play a vital role in vector mathematics. They help resolve vectors into their respective components.
  • Cosine corresponds to the horizontal direction: \(\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\).
  • Sine correlates with the vertical component: \(\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}\).
These functions allow us to take a vector's angle and magnitude and break it down into a more manageable form, facilitating further calculations or applications.

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Most popular questions from this chapter

Find the work performed when the given force \(\mathbf{F}\) is applied to an object, whose resulting motion is represented by the displacement vector d. Assume the force is in pounds and the displacement is measured in feet. \(\mathbf{F}=85 i, \mathbf{d}=6 i\)

A 155 -foot antenna is on top of a tall building. From a point on the ground, the angle of elevation to the top of the antenna is \(28.5^{\circ}\), while the angle of elevation to the bottom of the antenna from the same point is \(23.5^{\circ}\). How tall is the building?

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Each problem that follows refers to triangle \(A B C\).If \(B=120^{\circ}, C=20^{\circ}\), and \(c=28\) inches, find \(b\).
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