/*! 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 49 For the following exercises, fin... [FREE SOLUTION] | 91Ó°ÊÓ

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For the following exercises, find the unit vector in the direction of the given vector a and express it using standard unit vectors.Determine whether \(\overrightarrow{A B}\) and \(\overrightarrow{P Q}\) are equivalent vectors, where \(A(1,1,1), B(3,3,3), P(1,4,5)\), and \(Q(3,6,7)\).

Short Answer

Expert verified
The unit vector is \(\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\); \(\overrightarrow{AB}\) and \(\overrightarrow{PQ}\) are equivalent.

Step by step solution

01

Understand the Problem

We need to find the unit vector in the direction of a vector and check whether two vectors are equivalent. The vectors are determined by their endpoint coordinates.
02

Determine Vectors \(\overrightarrow{AB}\) and \(\overrightarrow{PQ}\)

The vectors can be found using the formula: \(\overrightarrow{XY} = (x_2 - x_1, y_2 - y_1, z_2 - z_1)\). So, \(\overrightarrow{AB} = (3-1, 3-1, 3-1) = (2, 2, 2)\) and \(\overrightarrow{PQ} = (3-1, 6-4, 7-5) = (2, 2, 2)\).
03

Check Equivalence of Vectors

Two vectors are equivalent if they have the same magnitude and direction. Since both \(\overrightarrow{AB}\) and \(\overrightarrow{PQ}\) are \((2, 2, 2)\), they are equivalent.
04

Unit Vector Calculation

A unit vector \(\mathbf{u}\) in the direction of \(\boldsymbol{v} = (a, b, c)\) is given by \(\mathbf{u} = \frac{1}{\|\boldsymbol{v}\|}(a, b, c)\), where \(\|\boldsymbol{v}\| = \sqrt{a^2 + b^2 + c^2}\).
05

Magnitude of \(\overrightarrow{AB}\) or \(\overrightarrow{PQ}\)

The magnitude \(\|\overrightarrow{AB}\| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{12} = 2\sqrt{3}\).
06

Calculate the Unit Vector

The unit vector in the direction of \(\overrightarrow{AB}\) is \(\frac{1}{2\sqrt{3}}(2, 2, 2) = \left(\frac{2}{2\sqrt{3}}, \frac{2}{2\sqrt{3}}, \frac{2}{2\sqrt{3}}\right) = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\).

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Key Concepts

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

Unit Vector
A unit vector is a vector that has a magnitude of exactly 1. Unit vectors are particularly useful in mathematics and physics as they are primarily used to describe directions, independent of the vector's magnitude. To obtain a unit vector in the direction of any given vector, you need to divide each component of the vector by its magnitude.

In mathematical terms, if you have a vector \( \boldsymbol{v} = (a, b, c) \), the unit vector \( \mathbf{u} \) is calculated as:\[\mathbf{u} = \frac{1}{\|\boldsymbol{v}\|}(a, b, c)\]where \( \|\boldsymbol{v}\| \) represents the magnitude of the vector \( \boldsymbol{v} \). By doing this, you scale the original vector down to a length of 1, while retaining the original direction.

To illustrate, if a vector \( \overrightarrow{AB} \) is \((2, 2, 2)\), its magnitude is \( 2\sqrt{3} \). Thus, the unit vector in the direction of \( \overrightarrow{AB} \) becomes \( \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \).
Vector Equivalence
Vector equivalence is a simple but crucial concept in understanding vector operations. Two vectors are considered equivalent if they have both identical magnitude and direction, which means they are essentially the same vector even if they originate from different points.

To determine if two vectors are equivalent, compare their respective components. If every component matches exactly, the vectors are equivalent. For example, consider vectors \( \overrightarrow{AB} = (2, 2, 2) \) and \( \overrightarrow{PQ} = (2, 2, 2) \). Since both vectors have the same components, they are equal in both direction and magnitude.

Equivalent vectors essentially represent the same movement or force in space, regardless of their initial starting points, such as vector \( \overrightarrow{A B} \) beginning at \( A \) and \( \overrightarrow{P Q} \) starting at \( P \).
Magnitude of a Vector
The magnitude of a vector, sometimes known as the length or norm, measures how long a vector is when represented in space. The magnitude is computed using the Pythagorean theorem extended to three dimensions. If the vector is \( \boldsymbol{v} = (a, b, c) \), the formula for its magnitude \( \|\boldsymbol{v}\| \) is:

\[\|\boldsymbol{v}\| = \sqrt{a^2 + b^2 + c^2}\]This calculation shows the straight-line distance from the tail to the head of the vector.

For instance, if we take the vector \( \overrightarrow{AB} = (2, 2, 2) \), its magnitude would be calculated as \( \sqrt{2^2 + 2^2 + 2^2} = \sqrt{12} = 2\sqrt{3} \). Here, the components indicate displacement in the X, Y, and Z directions, and the magnitude tells us about the total effect of these displacements.

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Most popular questions from this chapter

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 \(\mathbf{s}=\frac{1}{\sqrt{3}} \mathbf{i}+\frac{1}{\sqrt{3}} \mathbf{j}+\frac{1}{\sqrt{3}} \mathbf{k}\), 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 \(L_{2}\).

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