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Chapter 13: Problem 55

Are the statements true or false? Give reasons for your answer. $$\text { The vector } \frac{1}{\sqrt{3}} \vec{i}+\frac{-1}{\sqrt{3}} \vec{j}+\frac{2}{\sqrt{3}} \vec{k} \text { is a unit vector. }$$

Short Answer

Expert verified
False, because the magnitude of the vector is \( \sqrt{2} \), not 1.

Step by step solution

01

Understand what a Unit Vector is

A unit vector is a vector with a magnitude (or length) of 1 unit.
02

Calculate the Magnitude of the Given Vector

To determine if the vector \( \frac{1}{\sqrt{3}} \vec{i}+\frac{-1}{\sqrt{3}} \vec{j}+\frac{2}{\sqrt{3}} \vec{k} \) is a unit vector, calculate its magnitude:\[\|\vec{v}\| = \sqrt{\left( \frac{1}{\sqrt{3}} \right)^2 + \left( \frac{-1}{\sqrt{3}} \right)^2 + \left( \frac{2}{\sqrt{3}} \right)^2 }\]
03

Simplify the Magnitude Calculation

Simplify the expression:\[\|\vec{v}\| = \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{4}{3}}\]
04

Calculate the Sum Inside the Square Root

Combine the terms inside the square root:\[\|\vec{v}\| = \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{4}{3}} = \sqrt{\frac{6}{3}} = \sqrt{2}\]
05

Compare the Magnitude to 1

The magnitude \( \sqrt{2} \) is not equal to 1, therefore the vector is not a unit vector.

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Key Concepts

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

Magnitude of a Vector
The magnitude of a vector tells you about its length or size. You can think of it like finding how long a line is from start to finish. To calculate the magnitude of a vector given as \( \vec{v} = a \vec{i} + b \vec{j} + c \vec{k} \) in three dimensions, use the formula:
  • \( \|\vec{v}\| = \sqrt{a^2 + b^2 + c^2} \)
Using the given vector \( \frac{1}{\sqrt{3}} \vec{i} + \frac{-1}{\sqrt{3}} \vec{j} + \frac{2}{\sqrt{3}} \vec{k} \), its magnitude is found by substituting the coefficients:
  • \( \|\vec{v}\| = \sqrt{\left(\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{-1}{\sqrt{3}}\right)^2 + \left(\frac{2}{\sqrt{3}}\right)^2} \)
  • \( \|\vec{v}\| = \sqrt{\frac{1}{3} + \frac{1}{3} + \frac{4}{3}} = \sqrt{\frac{6}{3}} = \sqrt{2} \)
Since \( \sqrt{2} \) is not equal to 1, the vector is not a unit vector. Understanding magnitude helps us know the size or length of a vector, which is essential in many practical applications, such as physics and engineering.
Vector Algebra
Vector algebra deals with operations that can be performed on vectors such as addition, subtraction, and scalar multiplication. Here are some essential operations:
  • Addition: Combine vectors by adding their corresponding components. For vectors \( \vec{a} = a_1\vec{i} + a_2\vec{j} + a_3\vec{k} \) and \( \vec{b} = b_1\vec{i} + b_2\vec{j} + b_3\vec{k} \), we get \( \vec{a} + \vec{b} = (a_1 + b_1)\vec{i} + (a_2 + b_2)\vec{j} + (a_3 + b_3)\vec{k} \).
  • Subtraction: Similar to addition, but subtract the corresponding components.
  • Scalar Multiplication: Multiply each component of the vector by a scalar (a real number). If \( c \) is a scalar, \( c\vec{v} = c(a\vec{i} + b\vec{j} + c\vec{k}) = ca\vec{i} + cb\vec{j} + cc\vec{k} \).
These operations allow vectors to be manipulated algebraically for solving practical problems, such as determining direction and distance in navigation.
Three-Dimensional Vectors
Three-dimensional vectors provide a way to represent quantities that have both a magnitude and a direction in 3D space. These vectors are written in the form \( \vec{v} = a \vec{i} + b \vec{j} + c \vec{k} \), where \( a, b, \) and \( c \) are components along the x, y, and z axes, respectively.Three-dimensional vectors are crucial for modeling real-world situations where objects move or exist in three-dimensional space. Understanding these vectors is important for:
  • Physics: Analyzing forces and motion in the x, y, and z directions.
  • Engineering: Designing structures and systems that operate in three dimensions.
  • Computer Graphics: Creating and manipulating 3D images.
To fully utilize three-dimensional vectors, it's essential to understand operations like finding their magnitude and performing vector algebra, ensuring precise calculations and interpretations.

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Most popular questions from this chapter

Give reasons for your answer. If \(\|\vec{u}\|=1,\) then the vector \(\vec{v}-(\vec{v} \cdot \vec{u}) \vec{u}\) is perpendicular to \(\vec{u}\)

Given \(\vec{v}=3 \vec{i}+4 \vec{j}\) and force vector \(\vec{F}\) find: (a) The component of \(\vec{F}\) parallel to \(\vec{v}\). (b) The component of \(\vec{F}\) perpendicular to \(\vec{v}\). (c) The work, \(W\), done by force \(\vec{F}\) through displacement \(\vec{v}\). $$\vec{F}=0.2 \vec{i}-0.5 \vec{j}$$

A street vendor sells six items, with prices \(p_{1}\) dollars per unit, \(p_{2}\) dollars per unit, and so on. The vendor's price vector is \(\vec{p}=\left(p_{1}, p_{2}, p_{3}, p_{4}, p_{5}, p_{6}\right)=\) \((1.00,3.50,4.00,2.75,5.00,3.00) .\) The vendor sells \(q_{1}\) units of the first item, \(q_{2}\) units of the second item, and so on. The vendor's quantity vector is \(\vec{q}=\) \(\left(q_{1}, q_{2}, q_{3}, q_{4}, q_{5}, q_{6}\right)=(43,57,12,78,20,35) .\) Find \(\vec{p}\) \(\vec{q},\) give its units, and explain its significance to the vendor.

Compute the angle between the vectors. $$\vec{i}+\vec{j}-\vec{k} \text { and } 2 \vec{i}+3 \vec{j}+\vec{k}$$

Given \(\vec{v}=3 \vec{i}+4 \vec{j}\) and force vector \(\vec{F}\) find: (a) The component of \(\vec{F}\) parallel to \(\vec{v}\). (b) The component of \(\vec{F}\) perpendicular to \(\vec{v}\). (c) The work, \(W\), done by force \(\vec{F}\) through displacement \(\vec{v}\). $$\vec{F}=9 \vec{i}+12 \vec{j}$$
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