/*! 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 48 Find the angle between the diago... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 48

Find the angle between the diagonal of a cube and the diagonal of one of its sides.

Short Answer

Expert verified
The angle between the diagonal of a cube and the diagonal of one of its sides is approximately 35.26 degrees.

Step by step solution

01

Define the parameters and draw the cube

First, let's denote the edge of the cube as 'a'. Now, visualize a cube and draw it with its edges and diagonals.
02

Find the Face Diagonal

Use the Pythagorean theorem across one face of the cube to calculate the length of the face diagonal. In a right triangle formed by one edge of the cube, the face diagonal, and another edge, the hypotenuse would be the face diagonal. By Pythagorean theorem, the face diagonal \(d_f\) is given by \(\sqrt{a^2 + a^2} = a\sqrt{2}\)
03

Find the Cube Diagonal

Now, to find the diagonal of the cube, we need to use the Pythagorean theorem again. This time, in a right triangle formed by an edge, the face diagonal (already calculated in step 2), and the hypotenuse would be the cube diagonal. Hence, cube diagonal \(d_c\) is given by \(\sqrt{a^2 + (a\sqrt{2})^2} = a\sqrt{3}\)
04

Find the Required Angle

In a right triangle formed by a side of the cube, the diagonal of the cube \(d_c\) and the diagonal of the face \(d_f\), the required angle is the angle between \(d_c\) and \(d_f\). This angle \(酶\) can be found using the cosine formula, which in our case would be: \(Cos 酶 = \frac{d_f}{d_c} = \frac{a\sqrt{2}}{a\sqrt{3}} = \sqrt{\frac{2}{3}}\). The angle \(酶\) is therefore given by the arccos of this value, which is approximately 35.26 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!

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

In Exercises 45 and \(46,\) 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: (-1,2,3) Terminal point: (3,3,4)

Find the component form and magnitude of the vector \(u\) with the given initial and terminal points. Then find a unit vector in the direction of \(\mathbf{u}\). \(\frac{\text { Initial Point }}{(-4,3,1)}\) \(\frac{\text { Terminal Point }}{(-5,3,0)}\)

Determine whether \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal parallel, or neither. $$ \begin{array}{l} \mathbf{u}=-\frac{1}{3}(\mathbf{i}-2 \mathbf{j}) \\ \mathbf{v}=2 \mathbf{i}-4 \mathbf{j} \end{array} $$

In Exercises 33-36, (a) find the projection of \(\mathbf{u}\) onto \(\mathbf{v}\), and (b) find the vector component of u orthogonal to v. $$ \mathbf{u}=\langle 2,3\rangle, \quad \mathbf{v}=\langle 5,1\rangle $$

Find the component of \(u\) that is orthogonal to \(\mathbf{v},\) given \(\mathbf{w}_{\mathbf{1}}=\operatorname{proj}_{\mathbf{v}} \mathbf{u}\). $$ \mathbf{u}=\langle 8,2,0\rangle, \quad \mathbf{v}=\langle 2,1,-1\rangle, \quad \operatorname{proj}_{\mathbf{v}} \mathbf{u}=\langle 6,3,-3\rangle $$
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure 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.