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Chapter 11: Problem 55

Find the magnitude and direction angle (to the nearest tenth) for each vector. Give the measure of the direction angle as an angle in \(\left[0,360^{\circ}\right)\). $$\langle 8 \sqrt{2},-8 \sqrt{2}\rangle$$

Short Answer

Expert verified
The magnitude is 16, and the direction angle is 315掳.

Step by step solution

01

Identify the components of the vector

The vector is given as \( \langle 8 \sqrt{2}, -8 \sqrt{2} \rangle \). The \( x \)-component (horizontal) is \( 8 \sqrt{2} \) and the \( y \)-component (vertical) is \( -8 \sqrt{2} \).
02

Calculate the magnitude

The magnitude of a vector \( \langle x, y \rangle \) is calculated using the formula: \[ |\mathbf{v}| = \sqrt{x^2 + y^2} \].\ Plugging in the values, \[ |\mathbf{v}| = \sqrt{(8\sqrt{2})^2 + (-8\sqrt{2})^2} = \sqrt{128 + 128} = \sqrt{256} = 16 \].
03

Calculate the direction angle

The direction angle \( \theta \) can be found using the tangent formula: \( \tan \theta = \frac{y}{x} \). For this vector, \[ \tan \theta = \frac{-8\sqrt{2}}{8\sqrt{2}} = -1 \].\ Thus, \( \theta = \tan^{-1}(-1) \).
04

Determine the correct quadrant for \( \theta \)

Since both components are \( (8 \sqrt{2}, -8 \sqrt{2}) \), the vector lies in the fourth quadrant. The angle with \( \tan \theta = -1 \) in the fourth quadrant is \( 315^\circ \).
05

Check the angle range

Verify the angle is within the range of \( [0, 360^\circ) \). The angle \( 315^\circ \) is already within the correct range.

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Key Concepts

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

Magnitude of a Vector
When working with vectors, one important concept is the magnitude. The magnitude can be thought of simply as the length of the vector. To find it, you'll use a familiar formula similar to the Pythagorean theorem. For a vector defined as \( \langle x, y \rangle \), the magnitude is calculated by the equation:
  • \( |\mathbf{v}| = \sqrt{x^2 + y^2} \)
For the vector \( \langle 8\sqrt{2}, -8\sqrt{2} \rangle \), just square each component, add them, and then take the square root. This calculation yields:
  • \( |\mathbf{v}| = \sqrt{(8\sqrt{2})^2 + (-8\sqrt{2})^2} = 16 \)
Here, 16 represents the distance of the vector from the origin, on the coordinate plane. Knowing how to compute the magnitude helps you to better understand the vector's "size."
Direction Angle
The direction angle of a vector is like the compass direction, telling you which way the vector points, measured from the positive x-axis. To find this angle, you use the tangent function, as it relates the x and y components:
  • \( \tan \theta = \frac{y}{x} \)
Given the components \( y = -8\sqrt{2} \) and \( x = 8\sqrt{2} \), we find:
  • \( \tan \theta = \frac{-8\sqrt{2}}{8\sqrt{2}} = -1 \)
To determine \( \theta \), we take the inverse tangent:
  • \( \theta = \tan^{-1}(-1) \)
As a result of this computation, we locate the angle \( \theta \) in the appropriate quadrant, ensuring it falls within the range \([0, 360^\circ)\).
Vector Components
The components of a vector provide insight into its influence in different directions. They split the vector into its horizontal and vertical contributions. For a vector \( \langle x, y \rangle \):
  • \( x \) is the horizontal (often called the "i" or x-component)
  • \( y \) is the vertical component (often called the "j" or y-component)
Consider the vector \( \langle 8\sqrt{2}, -8\sqrt{2} \rangle \):
  • The x-component is \( 8\sqrt{2} \)
  • The y-component is \( -8\sqrt{2} \)
These components illustrate the push in the positive x direction and corresponding pull in the negative y direction. Understanding vector components is crucial as they are foundational to calculating the vector's magnitude and direction.
Quadrants
The coordinate plane is divided into four regions called quadrants, and each quadrant helps to identify the direction of a vector. The quadrants are labeled counterclockwise starting from the positive x-axis:
  • Quadrant I: Where both x and y are positive
  • Quadrant II: Where x is negative and y is positive
  • Quadrant III: Where both x and y are negative
  • Quadrant IV: Where x is positive and y is negative
For the vector \( \langle 8\sqrt{2}, -8\sqrt{2} \rangle \), since x is positive and y is negative, it falls in Quadrant IV. Knowing which quadrant a vector belongs to helps in interpreting its direction angle and ensuring calculations such as inverse tangent produce the correct angle. This understanding is especially important in ensuring the angle lies in the correct range of \([0, 360^\circ)\).

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Most popular questions from this chapter

A projectile has been launched from the ground with an initial velocity of 88 feet per second. You are given parametric equations that model the path of the projectile. (a) Graph the parametric equations. (b) Approximate \(\theta\), the angle the projectile makes with the horizontal at launch, to the nearest tenth of a degree. (c) On the basis of your answer to part (b), write parametric equations for the projectile, using the cosine and sine functions. $$x=56.56530965 t, y=-16 t^{2}+67.41191099 t$$

Answer each question.How do you graph \((r, \theta)\) by hand if \(r<0 ?\).

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Graph each pair of parametric equations. $$x=\cos ^{3} t, y=\sin ^{3} t ; 0 \leq t \leq 2 \pi$$
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