/*! 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 Consider the following points ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider the following points \(P\) and \(Q\) a. Find PQ and state your answer in two forms: \(\langle a, b, c\rangle\) and \(a \mathbf{i}+b \mathbf{j}+c \mathbf{k}\) b. Find the magnitude of \(\overrightarrow{P Q}\). c. Find two unit vectors parallel to \(\overline{P Q}\) $$P(0,0,2), Q(-2,4,0)$$

Short Answer

Expert verified
Question: Calculate the vector from point P(0,0,2) to point Q(-2,4,0). Represent the vector in the form of \(\langle a, b, c \rangle\) and \(\mathbf{i}+\mathbf{j}+\mathbf{k}\). Then find the magnitude of the vector and two unit vectors parallel to the vector. Answer: The vector PQ is given as \(\langle -2, 4, -2 \rangle\) and \(-2\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\). The magnitude of the vector is \(\sqrt{24}\). The two unit vectors parallel to the vector are \(\left\langle -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}} \right\rangle\) and \(\left\langle \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right\rangle\).

Step by step solution

01

Find the Vector PQ

To find the vector PQ, subtract the coordinates of point P from the coordinates of point Q: \(\overrightarrow{PQ} = Q - P\). Using the given coordinates: $$\overrightarrow{PQ} = (-2 - 0, 4 - 0, 0 - 2) = (-2, 4, -2)$$
02

Represent PQ in Two Forms

We can represent the vector PQ in the two requested forms: 1. \(\langle a, b, c \rangle\) form: \(\langle -2, 4, -2 \rangle\) 2. \(\mathbf{i}+\mathbf{j}+\mathbf{k}\) form: \(-2\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\)
03

Find the Magnitude of PQ

To find the magnitude of \(\overrightarrow{PQ}\), use the formula: $$\lVert \overrightarrow{PQ} \rVert = \sqrt{(-2)^{2} + 4^{2} + (-2)^{2}}$$ $$\lVert \overrightarrow{PQ} \rVert = \sqrt{4 + 16 + 4} = \sqrt{24}$$
04

Normalize PQ to Find Unit Vectors

To find two unit vectors parallel to PQ, normalize the vector PQ by dividing it by its magnitude. This results in one unit vector, and for the second unit vector, simply negate the first unit vector: $$\text{Unit vector 1} = \frac{\overrightarrow{PQ}}{\lVert \overrightarrow{PQ} \rVert} = \frac{\langle -2, 4, -2 \rangle}{\sqrt{24}} = \left\langle -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, -\frac{1}{\sqrt{6}} \right\rangle$$ $$\text{Unit vector 2} = -\text{Unit vector 1} = \left\langle \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right\rangle$$

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 91Ó°ÊÓ!

Key Concepts

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

Vector Subtraction
Vector subtraction is a fundamental operation in vector algebra that allows us to determine the directional difference between two points or the change in position. Imagine you are walking from one point in a space to another; the displacement can be represented as a vector subtraction, indicating the start and end of your journey.

To perform vector subtraction, we simply subtract the corresponding components of one vector from another. If we have vectors \(\mathbf{A}\) and \(\mathbf{B}\), their subtraction \(\mathbf{A} - \mathbf{B}\) is carried out by subtracting each component of \(\mathbf{B}\) from each corresponding component of \(\mathbf{A}\). For instance, if \(\mathbf{A} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{B} = \langle b_1, b_2, b_3 \rangle\), then the resulting vector \(\mathbf{A} - \mathbf{B} = \langle a_1 - b_1, a_2 - b_2, a_3 - b_3 \rangle\).

This operation helps in diverse fields such as physics, engineering, and computer graphics, where it's essential for understanding relative movements and forces.
Magnitude of a Vector
The magnitude of a vector, often referred to as its 'length' or 'norm', is a measure of how long the vector is. This scalar quantity represents the distance from the vector's initial point to its terminal point in a coordinate space. To find the magnitude of a vector with components \(a, b, c\), we use the Pythagorean theorem extended into three dimensions, resulting in the formula \(\sqrt{a^2 + b^2 + c^2}\).

This concept is crucial when calculating movements or forces in any given direction. For example, in the case of the vector \(\overrightarrow{PQ}\) with components \(\langle -2, 4, -2 \rangle\), its magnitude can be calculated as \(\sqrt{(-2)^2 + 4^2 + (-2)^2}\), which simplifies to \(\sqrt{24}\).

Understanding how to find vector magnitude is essential not only for solving physical problems but also in various mathematical and engineering applications.
Unit Vector
A unit vector is a vector that has a magnitude of one. Its primary function is to indicate direction, as its length is fixed. To convert any non-zero vector into a unit vector, we divide the vector by its own magnitude.

This process, known as normalization, gives us a vector that maintains the same direction but scales down to unit length. In mathematical terms, if vector \(\mathbf{A}\) has components \(\langle a_1, a_2, a_3 \rangle\), its unit vector \(\mathbf{u}\) is given by \(\mathbf{u} = \frac{\mathbf{A}}{\lVert \mathbf{A} \rVert} = \langle \frac{a_1}{\lVert \mathbf{A} \rVert}, \frac{a_2}{\lVert \mathbf{A} \rVert}, \frac{a_3}{\lVert \mathbf{A} \rVert} \rangle\).

Unit vectors are pivotal in defining vector directions in various vector spaces, making them invaluable in physics for representing force directions, for example.
Vector Components
Vector components are the building blocks of a vector; they represent the vector in terms of its influence along axes in a coordinate system. These components can be thought of as projections of the vector onto the coordinate axes, typically denoted as 'x', 'y', and 'z'.

For a vector represented in three dimensions as \(\mathbf{V} = \langle a, b, c \rangle\) or in unit vector notation as \(a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\), each letter (a, b, c) represents its component along the respective axis. This notation allows us to easily visualize and compute operations involving vectors since each dimension's contribution is clearly stated.

Furthermore, understanding vector components is useful in breaking down complex motion into simpler, one-dimensional movements, which can then be analyzed separately.

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

Linear combinations A sum of scalar multiples of two or more vectors (such as \(c_{1} \mathbf{u}+c_{2} \mathbf{v}+c_{3} \mathbf{w},\) where \(c_{i}\) are scalars) is called a linear combination of the vectors. Let \(\mathbf{i}=\langle 1,0\rangle, \mathbf{j}=\langle 0,1\rangle\) \(\mathbf{u}=\langle 1,1\rangle,\) and \(\mathbf{v}=\langle-1,1\rangle\) Express \langle 4,-8\rangle as a linear combination of \(\mathbf{u}\) and \(\mathbf{v}\)

Determine whether the following statements are true using a proof or counterexample. Assume \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are nonzero vectors in \(\mathbb{R}^{3}\). $$\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w})=\mathbf{w} \cdot(\mathbf{u} \times \mathbf{v})$$

Consider the 12 vectors that have their tails at the center of a (circular) clock and their heads at the numbers on the edge of the clock. a. What is the sum of these 12 vectors? b. If the 12: 00 vector is removed, what is the sum of the remaining 11 vectors? c. By removing one or more of these 12 clock vectors, explain how to make the sum of the remaining vectors as large as possible in magnitude. d. Consider the 11 vectors that originate at the number 12 at the top of the clock and point to the other 11 numbers. What is the

Parallel vectors of varying lengths Find vectors parallel to \(\mathbf{v}\) of the given length. $$\mathbf{v}=\langle 3,-2,6\rangle ; \text { length }=10$$

Equations of planes Find an equation of the following planes. The plane passing through the point \(P_{0}(1,0,4)\) that is parallel to the plane \(-x+2 y-4 z=1\)

See all solutions

Recommended explanations on Math Textbooks

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.