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Introduction to Vectors Part 3

by: Saadiq Shaik

Introduction to Vectors Part 3 MATH 241

Marketplace > University of Maryland - College Park > Math > MATH 241 > Introduction to Vectors Part 3
Saadiq Shaik


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About this Document

Covers 11.5 and 11.6.
Calculus III
Dr. Roohollah Ebrahimian
Class Notes
Math, Calculus, Multivariable, planes, Lines, space, curves, intercepts
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This 4 page Class Notes was uploaded by Saadiq Shaik on Saturday September 24, 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 4 views. For similar materials see Calculus III in Math at University of Maryland - College Park.


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Date Created: 09/24/16
Introduction to Vectors Part 3 Saadiq Shaik September 2016 1 Lines in Space Vectors and lines are highly related. A line in space can be written as a vector equation r = r0+ tL where r represents a vector and 0 represents a vector with a point on the line l. t represents a coe▯cient to L, which is a line parallel to l. Example 1 Find a vector equation of the line that contains (-1, 4, 0) and is parallel to 2i - 3j - 2k Solution Putting the vectors in unit vector form, we get r = ▯i + 4j 0 L = 2i ▯ 3j ▯ 2k and end up with r = (▯i + 4j) + t(2i ▯ 3j ▯ 2k) t is also known as a parameter, and the vector equation can be rewritten as three parametric equations of l x = x0+ at;y = y0+ bt;z = z0+ ct 1 and solving for t leads to the three symmetric equations of l x ▯ x0 y ▯ y0 z ▯ z0 = = a b c The distance from a point 1 to a line is given by the formula jjL ▯ P P jj D = 0 1 jjLjj where L is the same parallel line and P is a point on the line. 0 Example 2 Find the distance between the point (1, 2, 3) and the line from example 1. Solution P1is (1;2;3), 0 is (▯1;4;0) and L is (2;▯3;▯2). So D = jj(2;▯3;▯2) ▯ (2;▯2;3)jj jj(2;▯3;▯2)jj s 273 D = 17 Remark Keep in mind that you should always include an interval when you parametrize a line. 2 Planes in Space Planes are surfaces important in the study of vectors. Every plane is de▯ned by a vector perpendicular to the plane and a point on that plane and any three points not collinear lie on the same plane. Given a normal vector N = (a;b;c) and a point 0 = (x0;0 ;0 ) on the plane, we arrive at the formula a(x ▯ x0) + b(y ▯ 0 ) + c(z 0 z ) = 0 2 which we arrive from N ▯ P P = 0 0 1 Example 3 Find the normal vector of a plane given by the formula 2(x▯3)+3(y▯4) = 0. Solution With the de▯nition in mind, we see that the normal vector N here is (2;3;0). We can ▯nd the distance from a point to a plane as well. This is given by the formula jN ▯ 0 1 j D = jjNjj where 1 is a point not on the plane. Example 4 Find the distance from the point (0;9;0) to the plane de▯ned by the points P = (3;2;1), Q = (5;0;0) and R = (7;1;1). Solution We need a point on the plane and a normal vector to ▯gure out the distance. ▯! ▯! We can create two vectors PQ = (2;▯2;▯1) and QR = (2;1;1) and cross them to ▯nd a vector perpendicular to the plane. This works because the vectors lie on the plane and so a vector normal to them is normal to the plane. N = PQ ▯ QR = (▯3;0;6) ▯▯! P0P1= (5;▯9;0) using point Q D = j(▯3;0;6) ▯ (5;▯9;0)j jj(▯3;0;6)jj 15 D = p 45 and simplifying p D = 5 3 Remarks Two planes are parallel whenever their normal vectors are scalar multiples. For plane sketching, ▯nd the x, y, and z intercepts of the plane. 4


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