Section 2.1: Three-Dimensional Coordinate Systems

Transcription

1 Section 2.1: Three-Dimensional Coordinate Systems ² Review on 2D Cartesian Coordinate Systems on Planes Let s rst take a look at 2D Cartesian coordinate system: it consists of x axis (horizontal, pointing to the right) and y axis (vertical, upward) with appropriate units in both axes. Any point P on the plane is associated with a pair of real numbers (x, y) in a unique way as follows: x = intersection of the vertical line passing through P and x axis (called x coordinate), y = intersection of the horizontal line passing through P and y axis (called y coordinate). For instance, Point P (1, 2) is formed as the intersection of a vertical line across x axis at x = 1 and a horizontal line across y axis at y = 2. The plane is also called xy plane. Distance & Midpoint Formula: Given two point P (x 1, y 1 ) and Q (x 2, y 2 ).The distance and midpoint between P and Q are, respectively q dist (P, Q) = (x 1 x 2 ) 2 + (y 1 y 2 ) 2 µ x1 + x 2 M (P, Q) =, y 1 + y

2 Graph of any equation F (x, y) = 0 is a curve. In particular, x = a is a vertical line and y = b is a horizontal line. Both are called coordinate lines, and they form rectangular grid on xy-plane that can be used to de ne coordinates. The graph of any linear function y = mx + b is a straight line with slope m and y intercept y = b. Any circle centered at (a, b) with radius R has the standard equation (x a) 2 + (y b) 2 = R 2 3D Cartesian Coordinate Systems This idea of identifying points with numbers extends to three dimensional space by introducing a third axis called z axis that is perpendicular to the xy plane at the origin and is pointing towards "you", forming the right-hand rule: 2

3 The plane formed by y axis and z axis is called yz plane, and the plane formed by x axis and z axis is called zx plane. We usually use the following cycle (like "Recycle Symbol") for naming in order to keep "right-hand rule: x! y! z! x. All these three planes are called coordinate planes. In hand drawing, we often make yz plane on the paper and x axis pointing towards us: Any point P is assigned to a unique triple of three ordered numbers (x, y, z) using planes parallel to three coordinate planes as follows. x ( in the following gure, x = 1) is the intersection of x axis and the plane across P and parallel to yz plane 3

4 z O y x y (y = 2 in the gure) is the intersection of y axis and the plane across P and parallel to zx plane, 4

5 and z ( z = 3 in the gure) is the intersection of z axis and the plane across P and parallel to xy plane The combined the graph looks like 5

6 or, after a rotation, The process of identifying a point P with a triple of ordered numbers (x, y, z) is reversible, and x = 1 is called x coordinate of P (1, 2, 3) 6

7 y = 2 is called y coordinate of P (1, 2, 3) z = 3 is called z coordinate of P (1, 2, 3). ² Method of dimension reduction Another easy way to assign 3D coordinate to a point P is to use the idea of dimension reduction: reducing a 3D problem to a 2D problem. This method will be applied later again. Given any point P. We rst draw a line passing through P and is perpendicular to xy plane. This line should meets xy plane at a point Q on xy plane. We call Q the orthogonal projection of P onto xy plane. Z P(x,y,z) z O Y X Q(x,y,0) If P is above xy plane, z, the z coordinate of P, is the distance between P and Q. If P is below xy plane, then z = (distance between P and Q). Now the projection point Q is a point on xy plane. Finding the coordinate of Q on xy plane becomes a 2D problem. Let (x, y) be the 2D coordinate of Q. Then (x, y, z) is 3D coordinate of P. ² Graph of an equation F (x, y, z) = 0 7

8 consists of all points satisfying the equation; that is, f(x, y, x) j F (x, y, z) = 0g. For convenience, we call any plane parallel to xy plane, or yz plane, or zx plane a coordinate plane. Coordinate planes are the graphs of either x = a or y = b or z = c: x = 1 and x = 2 8

9 y = 1, y = 1 z = 1, z = 1 Distance Formula: The distance, dist (P, Q) = jp Qj, between two points P (x 1, y 1, z 1 ) and 9

10 z Q P E O C y D x Q (x 2, y 2, z 2 ) is dist (P, Q) = q (x 1 x 2 ) 2 + (y 1 y 2 ) 2 + (z 1 z 2 ) 2. We shall use the dimension reduction to verify this formula. In the above gure, we project point P and Q vertically (parallel to z axiz) onto xy plane. Let C & D, respectively, their projection points. The line segment CD lies on xy plane, and is called orthogonal PROJECTION of P Q onto xy plane. E is the point on the vertical line QD such that P E // xy plane (i.e., P E is parallel to xy plane). Note that if we assume that P (x 1, y 1, z 1 ) and Q (x 2, y 2, z 2 ), then and E (x 2, y 2, z 1 ), C (x 1, y 1, 0), D (x 2, y 2, 0), dist (P, Q) 2 = jp Qj 2 = jp Ej 2 + jqej 2 = jcdj 2 + (z 2 z 1 ) 2 = (x 1 x 2 ) 2 + (y 1 y 2 ) 2 + (z 2 z 1 ) 2, 10

11 z M B Q P E x O C N D y where in the very last equality, we used 2D distance formula for jcdj. Midpoint Formula: The midpoint, M (P, Q), the point in the middle of the line segment connecting two points P (x 1, y 1, z 1 ) and Q (x 2, y 2, z 2 ) is µ x1 + x 2 M (P, Q) =, y 1 + y 2, z 1 + z In this gure, N is the vertical projection of M onto xy plane, and N happens to the midpoint of CD.Thus, by 2D midpoint formula, if we view C and D as 2D points, i.e., then µ x1 + x 2 N, y 1 + y C (x 1, y 1 ), D (x 2, y 2 ), µ x1 + x 2 =) M 2 To calculate z coordinate z, observe that Triangle MQB is similar to P QE =) B is the midpoint of QE.., y 1 + y 2, z. 2 11

12 We may project Q, B and E onto z axis and view them as one dimensional point on z axis with coordinates µ z1 + z 2 Q (z 2 ), E (z 1 ) =) B (by one dimensional midpoint formula). 2 We thus veri ed 3D midpoint formula. Example 2.1. Given P (2, 1, 7) and Q (1, 3, 5). Find (a) their distance dist (P, Q) and (b) their midpoint M (P, Q). Solution: (a) q dist (P, Q) = (x 1 x 2 ) 2 + (y 1 y 2 ) 2 + (z 1 z 2 ) 2 q = (2 1) 2 + ( 1 + 3) 2 + (7 + 5) 2 = p 149 = (b) µ x1 + x 2 M (P, Q) =, y 1 + y 2, z 1 + z µ = 2, 1 3, µ 3 = 2, 2, 1. ² Equation of Sphere A sphere centered at C (h, k, l) with radius R consists of all points P (x, y, z) whose distance to the center is R; that is q dist (C, P ) = (x h) 2 + (y k) 2 + (z l) 2 = R or after squaring both sides (x h) 2 + (y k) 2 + (z l) 2 = R 2. We call it the standard form of the sphere with center (h, k, l) and radius R. Example 2.2. Find the radius and center of the sphere x 2 + y 2 + z 2 + 4x 6y + 2z + 6 = 0. 12

13 Solution: We use the technique called "completing square". First we re-group the equation as x 2 + 4x + y 2 6y + (z 2 + 2z) = 6, x 2 + 4x y 2 6y (z 2 + 2z ) = , Thus (x + 2) 2 + (y 3) 2 + (z + 1) 2 = 8, (x + 2) 2 + (y 3) 2 + (z + 1) 2 = ³p 8 2. Center = ( 2, 3, 1), Radius = p 8. Example 2.3. Explain and sketch the region described by (a) 1 < x 2, and (b) 1 x 2 + y 2 + z 2 < 4. Solution: (a) To graph, we always start with graphing equations. The graphs for x = 1 and x = 2 are two coordinate planes 13

14 So the graph of the inequalities consists of the region between these two planes, including the plane x = 2 on the left but excluding the plane x = 1on the right. (b) The graphs of equations x 2 + y 2 + z 2 = 1 and x 2 + y 2 + z 2 = 4 14

15 are two spheres centered at the origin with radius r = 1 and r = 2,respectively. Therefore, the region 1 x 2 + y 2 + z 2 < 4 consists of all points such that (1) inside the larger sphere with radius R = 2 ( x 2 + y 2 + z 2 < 4 ), (2) outside the smaller sphere with radius R = 1 ( x 2 + y 2 + z 2 1 ) and (3) including the smaller sphere ( x 2 + y 2 + z 2 = 1 ). So The region = between two spheres, including the smaller one, but excluding the larger one. Homework: 1. Sketch the poins (2, 3, 1) and ( 1, 2, 2). You must describe in words how you do that. 2. What are the orthogonal projection of (2, 4, 3) on xy, yz, and zx planes? How to sketch this point? 3. Find distance and midpoint between P (1, 2, 3) and Q (3, 4, 1). 4. Using distance formula, determine whether three points lie on straight line. (a) A (2, 4, 2), B (3, 7, 2), C (1, 3, 3) (b) A (1, 4, 4), B (2, 1, 5), C (4, 5, 3) 5. Find the equation of the sphere if one of its diameters has endpoints (2, 1, 3) and (4, 3, 9). 6. Find the center and the radius of the sphere x 2 + y 2 + z 2 + 4x 2y + 6z 2 = 0 7. Describe in words the region three-dimensional space. (a) 1 < y 2 (b) 1 z < 4 (c) x 2 + y 2 + z

16 8. Consider the point P such that the distance from P to A ( 1, 5, 3) is twice the distance from P to B (6, 2, 2). Show that the set of all such points is a sphere. Find the center and radius of the sphere. 16