/*! 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 30 Find the direction angles of the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 30

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

Short Answer

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

Step by step solution

01

Calculation of the Magnitude of the Vector

The magnitude of the vector \(\mathbf{u}=\langle -2, 6, 1 \rangle\) can be obtained using the formula \(\|u\| = \sqrt{X^{2} + Y^{2} + Z^{2}}\). Substituting the values from \(\mathbf{u}=\langle -2, 6, 1 \rangle\) into the formula, gives \(\|u\| = \sqrt{(-2)^{2} + 6^{2} + 1^{2}} = \sqrt{4 + 36 + 1} = \sqrt{41}\)
02

Finding the Cosines of the Direction Angles

To find the cosines of the direction angles, one must use the formulas \(\cos{\alpha}=\frac{X}{\|\mathbf{u}\|}\), \(\cos{\beta}=\frac{Y}{\|\mathbf{u}\|}\), \(\cos{\gamma}=\frac{Z}{\|\mathbf{u}\|}\). Substituting the values from \(\mathbf{u}=\langle -2, 6, 1 \rangle\) and \(\|u\| = \sqrt{41}\), gives \(\cos{\alpha} = \frac{-2}{\sqrt{41}}\), \(\cos{\beta} = \frac{6}{\sqrt{41}}\), \(\cos{\gamma} = \frac{1}{\sqrt{41}}\).
03

Finding the Direction Angles

The direction angles are obtained by finding the inverse cosine of the respective cosines. Therefore, \(\alpha = \cos^{-1}(\frac{-2}{\sqrt{41}})\), \(\beta = \cos^{-1}(\frac{6}{\sqrt{41}})\), \(\gamma = \cos^{-1}(\frac{1}{\sqrt{41}})\)

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!

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

The vertices of a triangle are given. Determine whether the triangle is an acute triangle, an obtuse triangle, or a right triangle. Explain your reasoning. $$ (-3,0,0),(0,0,0),(1,2,3) $$

The initial and terminal points of a vector \(v\) are given. (a) Sketch the directed line segment, (b) find the component form of the vector, and (c) sketch the vector with its initial point at the origin. Initial point: (2,-1,-2) Terminal point: (-4,3,7)

Find the magnitude of \(v\). Initial point of \(\mathbf{v}:(1,-3,4)\) Terminal point of \(\mathbf{v}:(1,0,-1)\)

Determine whether \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal parallel, or neither. $$ \begin{array}{l} \mathbf{u}=-2 \mathbf{i}+3 \mathbf{j}-\mathbf{k} \\ \mathbf{v}=2 \mathbf{i}+\mathbf{j}-\mathbf{k} \end{array} $$

Find the vector \(z,\) given that \(\mathbf{u}=\langle 1,2,3\rangle\) \(\mathbf{v}=\langle 2,2,-1\rangle,\) and \(\mathbf{w}=\langle 4,0,-4\rangle\) \(\mathbf{z}=5 \mathbf{u}-3 \mathbf{v}-\frac{1}{2} \mathbf{w}\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Geometry

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.