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Introduction to Vectors Part 1

by: Saadiq Shaik

Introduction to Vectors Part 1 MATH 241

Marketplace > University of Maryland - College Park > Math > MATH 241 > Introduction to Vectors Part 1
Saadiq Shaik


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About this Document

This covers the basics of the three dimensional coordinate system and vectors while also providing examples.
Calculus III
Dr. Roohollah Ebrahimian
Class Notes
Math, Calculus, Multivariable, vectors, coordinates, magnitude
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This 4 page Class Notes was uploaded by Saadiq Shaik on Saturday September 10, 2016. The Class Notes belongs to MATH 241 at University of Maryland - College Park taught by Dr. Roohollah Ebrahimian in Fall 2016. Since its upload, it has received 39 views. For similar materials see Calculus III in Math at University of Maryland - College Park.


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Date Created: 09/10/16
Introduction to Vectors Part 1 Saadiq Shaik September 2016 1 Cartesian Coordinates in Space A vector is de▯ned as a quantity with both a magnitude and direction. We will be dealing with vectors in three dimensional space, so familiarize yourself with the three dimensional coordinate system. The coordinate planes are the xy-plane, yz-plane, and xz-plane. There are eight octants, but the one we will be dealing with most is the ▯rst octant, which contains positive x, y, and z values. Imagine a point P with the coordinates (a ;b ;c ) and a point Q with 1 1 1 the coordinates 2a2;b3;c ) in three dimensional space. The distance between them is given by the distance formula, q jPQj = (1 ▯ a2) + (1 ▯ b2) + (1 ▯ 2 ) Example 1 p p p Find the distance between the points P = (▯1;0;2 and Q = (1; 2; 3). Solution Simple substitution gives us the answer. q p p p jPQj = (▯1 ▯ 1) + (0 ▯ 2) + (2 3 ▯ 3) p jPQj = 4 + 2 + 3 p jPQj = 9 = 3 1 Likewise, the midpoint of P and Q is given by the midpoint formula, ▯a1+ a 2 b1+ b 2 c1+ c 2▯ M = ; ; 2 2 2 Example 2 1 1 Find the midpoint of points Q = ( ;2;0) and R = ( ;322): Solution Substituting the values in gives us ▯ 1+ 1 1 + 3 0 + 2 ▯ M = 2 2; ; 2 2 2 ▯ 1 4 2 ▯ M = ; ; 2 2 2 1 M = ( 22;1) A sphere of radius r centered at (x0;y0;z0) is de▯ned by the equation of a sphere in space where (x;y;z) is the set of all points on the sphere r = (x ▯ x )0+ (y ▯ y ) 0 (z ▯ z ) 0 2 Example 3 Show that x + y + z = 4x + 4y ▯ 6z is an equation of a sphere. Find its center and radius. Solution By moving the terms to the right and completing the square we get 2 2 2 (x ▯ 4x + 4) + (y ▯ 4y + 4) + (z + 6z + 9) = 4 + 4 + 9 or (x ▯ 2) + (y ▯ 2) + (z + 3) = 17 p This is an equation of the sphere with center (2, 2, -3) and 17. 2 2 Vectors in Space Given the points P = (a ;b ;c ) and Q = (a ;b ;c ), the vector PQ is de▯ned 0 0 0 1 1 1 as (a1▯ a 0b 1 b ;0 ▯1c ) 0 and can be rewritten with the unit vectors i = (1;0;0), j = (0;1;0), and k = (0;0;1) as (a1▯ a 0i + (b 1 b 0j + (c 1 c )0 or ai + bj + ck where a, b, and c are the di▯erences of the x, y, and z coordinates respec- tively. They are called the vector components. Example 4 ▯! Given P = (1;2;3) and Q = (0;2;4), write the vector PQ in unit vector notation. Solution We start by ▯nding the coe▯cients a, b and c, a = 0 ▯ 1 = ▯1;b = 2 ▯ 2 = 0;c = 4 ▯ 3 = 1 and then plug them into the formula to obtain ▯1i + 0j + 1k or ▯i + k The magnitude of a vector a is given by the magnitude formula q jjajj = a1+ a 2 a 3 This can also be called the length or norm of the vector. 3 Example 5 Find the magnitude of the vector a = 2i + 3j +3k. Solution We simply ▯nd the vector components and substitute them to arrive at q 2 2 p 2 jjajj = 2 + 3 + 3 or p jjajj = 4 + 9 + 3 = 16 = 4 A vector in a plane, where3a = 0, can be rewritten as a = jjajj(cos(▯i) + sin(▯j)) where ▯ is the angle between the x-axis and a. Here is an important point regarding vectors: jjcajj = jcjjjajj where c is some constant. If b = jcjjjajj, then the two vectors are parallel. The unit vector in the direction of a vector a is given by a jjajj Vectors can added by adding their components and subtracted by subtracting their components, as well. 4


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