/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Find the direction angles of the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 38

Find the direction angles of the vector. $$ \mathbf{u}=\langle-2,6,1\rangle $$

Short Answer

Expert verified
The direction angles \(\alpha, \beta\), and \(\gamma\) of the vector \(\mathbf{u} = \langle-2,6,1\rangle\) are \(\alpha = \cos^{-1}\left(\frac{-2}{\sqrt{41}}\right)\), \(\beta = \cos^{-1}\left(\frac{6}{\sqrt{41}}\right)\), and \(\gamma = \cos^{-1}\left(\frac{1}{\sqrt{41}}\right)\).

Step by step solution

01

Compute the Magnitude of the Vector

In the first step, calculate the magnitude of the vector \(\mathbf{u}\) which is given by the formula \(||\mathbf{u}|| = \sqrt{u_1^2 + u_2^2 + u_3^2}\). So, \(||\mathbf{u}|| = \sqrt{(-2)^2 + 6^2 + 1^2} = \sqrt{41}\).
02

Determine the Cosines of the Direction Angles

Next, use the formula \(\cos(\alpha) = \frac{u_1}{||\mathbf{u}||}\), \(\cos(\beta) = \frac{u_2}{||\mathbf{u}||}\), and \(\cos(\gamma) = \frac{u_3}{||\mathbf{u}||}\) to determine the cosines of the direction angles for vector \(\mathbf{u}\). Thus, \(\cos(\alpha) = \frac{-2}{\sqrt{41}}\), \(\cos(\beta) = \frac{6}{\sqrt{41}}\), and \(\cos(\gamma) = \frac{1}{\sqrt{41}}\).
03

Determine the Direction Angles

Finally, compute the direction angles themselves, which are the inverse cosines of the values determined in the previous step. Using a calculator or a software tool, we can determine that \(\alpha = \cos^{-1}\left(\frac{-2}{\sqrt{41}}\right)\), \(\beta = \cos^{-1}\left(\frac{6}{\sqrt{41}}\right)\), and \(\gamma = \cos^{-1}\left(\frac{1}{\sqrt{41}}\right)\). These are the direction angles of the vector \(\mathbf{u}\). However, the answer will typically be represented in degrees.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Vector Magnitude
In vector mathematics, the magnitude of a vector represents its length and is a crucial part of understanding the vector's properties.
The magnitude is computed using the formula \[||\mathbf{u}|| = \sqrt{u_1^2 + u_2^2 + u_3^2}\].
This formula essentially uses the Pythagorean theorem in three dimensions.
If you take the vector \(\mathbf{u} = \langle-2,6,1\rangle\), calculate the magnitude like this:
  • Square each component: \((-2)^2 = 4\), \(6^2 = 36\), and \(1^2 = 1\).
  • Add these squares: \(4 + 36 + 1 = 41\).
  • Take the square root to get the magnitude: \(\sqrt{41}\).
This tells us how far the vector reaches from the origin in a 3D space.
Direction Cosines
Direction cosines help us determine the angle a vector makes with each axis in a coordinate system.
These are the cosine values of the direction angles and give us a sense of how the vector is oriented.To find the direction cosines, you use:
  • \(\cos(\alpha) = \frac{u_1}{||\mathbf{u}||}\)
  • \(\cos(\beta) = \frac{u_2}{||\mathbf{u}||}\)
  • \(\cos(\gamma) = \frac{u_3}{||\mathbf{u}||}\)
Applying this to our vector \(\mathbf{u} = \langle-2,6,1\rangle\):
  • \(\cos(\alpha) = \frac{-2}{\sqrt{41}}\)
  • \(\cos(\beta) = \frac{6}{\sqrt{41}}\)
  • \(\cos(\gamma) = \frac{1}{\sqrt{41}}\)
These fractional values are the cosines of the angles the vector makes with the x, y, and z axes.
Trigonometry in Vectors
Trigonometry plays a crucial role in understanding the relationship between vectors and their direction angles.
By using inverse trigonometric functions, we can convert the direction cosines into actual angles.To find the direction angles, we calculate the inverse cosine:
  • \(\alpha = \cos^{-1}\left(\frac{-2}{\sqrt{41}}\right)\)
  • \(\beta = \cos^{-1}\left(\frac{6}{\sqrt{41}}\right)\)
  • \(\gamma = \cos^{-1}\left(\frac{1}{\sqrt{41}}\right)\)
These angles are typically expressed in degrees, offering a tangible measure of direction.
The use of trigonometric functions allows us to interpret how vectors interact in space and make sense of their orientation and distance effects.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find two vectors in opposite directions that are orthogonal to the vector \(\mathbf{u}\). (The answers are not unique.) $$ \mathbf{u}=\langle 0,-3,6\rangle $$

(a) find the unit tangent vectors to each curve at their points of intersection and (b) find the angles \(\left(0 \leq \theta \leq 90^{\circ}\right)\) between the curves at their points of intersection. $$ (y+1)^{2}=x, \quad y=x^{3}-1 $$

\(\begin{array}{ll}\text { } & \mathbf{}, & \text { describe } & \text { the family of planes }\end{array}\) represented by the equation, where \(c\) is any real number.\(c y+z=0\)

Sketch the solid that has the given description in cylindrical coordinates. \(-\pi / 2 \leq \theta \leq \pi / 2,0 \leq r \leq 3,0 \leq z \leq r \cos \theta\)

\mathrm{\\{} P r o g r a m m i n g ~ Given vectors \(\mathbf{u}\) and \(\mathbf{v}\) in component form, write a program for a graphing utility in which the output is the component form of the projection of \(\mathbf{u}\) onto \(\mathbf{v}\).
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.