/*! 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 138 For the following exercises, fin... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 138

For the following exercises, find the measure of the angle between the three- dimensional vectors a and \(\mathbf{b}\) . Express the answer in radians rounded to two decimal places, if it is not possible to express it exactly. $$ \mathbf{a}=\mathbf{i}-2 \mathbf{j}+\mathbf{k}, \quad \mathbf{b}=\mathbf{i}+\mathbf{j}-2 \mathbf{k} $$

Short Answer

Expert verified
The angle between the vectors is approximately 2.09 radians.

Step by step solution

01

Understand the Formula

To find the angle \( \theta \) between two vectors \( \mathbf{a} \) and \( \mathbf{b} \), we can use the dot product formula: \[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \] Where \( \mathbf{a} \cdot \mathbf{b} \) is the dot product of \( \mathbf{a} \) and \( \mathbf{b} \), and \( \|\mathbf{a}\| \) and \( \|\mathbf{b}\| \) are the magnitudes of vectors \( \mathbf{a} \) and \( \mathbf{b} \), respectively.
02

Compute the Dot Product

The dot product of \( \mathbf{a} = \mathbf{i} - 2 \mathbf{j} + \mathbf{k} \) and \( \mathbf{b} = \mathbf{i} + \mathbf{j} - 2 \mathbf{k} \) is calculated as follows: \[ \mathbf{a} \cdot \mathbf{b} = (1)(1) + (-2)(1) + (1)(-2) = 1 - 2 - 2 = -3 \]
03

Find the Magnitude of the Vectors

The magnitude of vector \( \mathbf{a} \) is: \[ \| \mathbf{a} \| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \] The magnitude of vector \( \mathbf{b} \) is: \[ \| \mathbf{b} \| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]
04

Compute the Cosine of the Angle

Substitute the dot product and magnitudes back into the formula: \[ \cos \theta = \frac{-3}{\sqrt{6} \cdot \sqrt{6}} = \frac{-3}{6} = -0.5 \]
05

Calculate the Angle in Radians

To find \( \theta \), use the inverse cosine function: \[ \theta = \cos^{-1}(-0.5) \] This gives: \( \theta = \frac{2\pi}{3} \) radians, which is approximately 2.09 radians when rounded to two decimal places.

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.

Dot Product
The dot product is an important concept in vector mathematics. It finds the product of two vectors by multiplying corresponding elements and then summing the results.
This operation is essential because it helps determine the angle between vectors.
To find the dot product of two vectors \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \) and \( \mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k} \), use the formula:
\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]
This generates a scalar quantity, not a vector.
  • If the dot product is zero, the vectors are perpendicular (orthogonal).
  • A positive dot product means the vectors point in a similar direction.
  • A negative product indicates opposite general directions.
This makes dot products extremely useful for understanding vector relationships in various fields like physics and computer graphics.
Vector Magnitude
The magnitude of a vector represents its length or size.
In mathematical terms, it is the Euclidean distance from the vector's initial point to its terminal point.
If a vector \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \), its magnitude is calculated as:
\[ \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]
This measurement is important because it provides the scale needed in vector operations.
  • The larger the magnitude, the longer the vector is.
  • Magnitude of a vector is never negative.
When finding the angle between vectors, magnitudes help scale the formula appropriately. The magnitude tells us about the strength or the extent of the vector's impact.
Cosine Formula
The cosine formula is key in finding the angle between vectors.
The angle \( \theta \) between vectors \( \mathbf{a} \) and \( \mathbf{b} \) is determined using:
\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \]
This relation shows us how two vectors are aligned relative to each other. It uses the dot product and magnitudes of the vectors.
  • If \( \cos \theta = 1 \), the vectors are parallel.
  • If \( \cos \theta = 0 \), the vectors are perpendicular.
  • If \( \cos \theta = -1 \), the vectors are diametrically opposite.
To actually find \( \theta \), you would need to apply the inverse cosine function, \( \cos^{-1} \).
This method is popular in disciplines like physics and computer simulations to define directional relationships.

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

Determine a unit vector perpendicular to the plane passing through the \(z\) -axis and point \(A(3,1,-2)\) .

For the following exercises, find the vector and parametric equations of the line with the given properties. The line that passes through point \((2,-3,7)\) that is parallel to vector \(\langle 1,3,-2\rangle\)

Show there is no plane perpendicular to \(\mathbf{n}=\mathbf{i}+\mathbf{j}\) that passes through points \(P(1,2,3)\) and \(Q(2,3,4)\)

show that line \(\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}\) is parallel to plane \(x-2 y+z=6\)

[T] A solar panel is mounted on the roof of a house. The panel may be regarded as positioned at the points of coordinates (in meters) \(A(8,0,0)\), \(B(8,18,0)\), \(C(0,18,8)\) and \(D(0,0,8)\) (see the following figure). a. Find the general form of the equation of the plane that contains the solar panel by using points A, B, and C, and show that its normal vector is equivalent to \(\overrightarrow{A B} \times \overrightarrow{A D}\) b. Find parametric equations of line \(L_{1}\) that passes through the center of the solar panel and has direction vector \(\mathbf{s}=\frac{1}{\sqrt{3}} \mathbf{i}+\frac{1}{\sqrt{3}} \mathbf{j}+\frac{1}{\sqrt{3}} \mathbf{k}, \quad\) which points toward the position of the Sun at a particular time of day. c. Find symmetric equations of line \(L_{2}\) that passes through the center of the solar panel and is perpendicular to it. d. Determine the angle of elevation of the Sun above the solar panel by using the angle between lines \(L_{1}\) and \(L_{2}\)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

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.