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This document introduces three-dimensional coordinate systems and graphs in three dimensions. It defines the three perpendicular axes (x, y, z) used to locate points in three-dimensional space. Points are represented by ordered triples (a,b,c) indicating their distances from each axis. Equations in three variables determine surfaces in three-dimensional space that can be graphed and analyzed. The document also introduces the distance formula for calculating distances between points in three-dimensional space and defines spheres using this formula. It provides examples of describing, sketching, and finding equations for various three-dimensional surfaces and regions.
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- 1. Section 13.1 Three-Dimensional Coordinate Systems “Living in a 3-dimensional World” 1. Three-Dimensional Rectangular Coordinate System We can sketch the graph of a function of two variables in the plane: the x-coordinate is the “input” value and the y coordinate is the corre- sponding output. To illustrate the graph of a function of two variables, we need two inputs and a single output - this means we need an extra dimension if we want to sketch a graph. We do this as following: • (The right hand rule) With your right hand, point your thumb up, and your next finger and your middle finger outward per- pendicular to each other. Draw three lines in the direction your thumb and fingers are pointing - label the line along your thumb the z-axis, your middle finger the y-axis and the re- maining one the x-axis. The positive axis is in the direction your fingers point. z x y • Any point P in 3 space is completely determined by its distance along the x, y and z-axis. In order to sketch a point P in 3- space, we associate the ordered triple (a, b, c) where P is a directed distance of a units in the x-direction, b units in the y- direction, and c units in the z-direction. We will often denote a point P as P(a, b, c) or (a, b, c). • We call the three axis the coordinate axis. • Any two axis determine a plane which we call a coordinate plane. They are referred to by the axis which determine them - the xy-plane, the yz-plane and the xz-plane. • The three coordinate planes break up three space into eight parts called octants. The first octant is the octant where x, y and z are all positive. • If we drop a perpendicular from any point P(a, b, c) into one of the coordinate planes, we get a point in that plane called the 1
- 2. 2 projection into that plane. We have (0, b, c) as the projection into the yz-plane, we have (a, 0, c) as the projection into the xz-plane, and (a, b, 0) as the projection into the xy-plane. Just as with two dimensions, any equations in 3-dimensions determine a graph in 3-space. Being able to recognize and sketch graphs in 3-space will be very important. Example 1.1. Describe and sketch the surface represented by z = 2. This is all points with a z value of 2, so will be a horizontal plane at z = 2. Describe and sketch the surface represented by y = √ x. At z = 0, the equation y = √ x will simply be the graph of y = √ x in the xy-plane. Since there are no conditions on z, we can extend this graph out vertically, and we the equation y = √ x will still be satisfied. Therefore, this will be the graph of y = √ x in the xy-plane extended vertically (in the z-direction) infinitely. Describe and sketch the surface represented by z = y. Similar to the previous example, at x = 0, the equation z = y will simply be the graph of z = y in the zy-plane. Since there are no conditions on x, we can extend this graph out in the x-direction, and we the equation z = y will still be satisfied. Therefore, this will be the graph of z = y in the zy-plane extended horizontally (in the x- direction) infinitely. In particular, it will be a plane. Example 1.2. If you are stood at (3, 2, 1) and are looking at (1, 2, 3), are you looking up or down? Up since the z value at the point you are looking at is higher than the point you are stood at. If you lift the xy-plane up so it has z-coordinate 1, what will be the equation for this surface? The xy-plane has equation z = 0. If we move it up 1, then it will have equation z = 1. Write down the equation for a surface which when you move in the positive x-direction, z grows exponentially, but z stays fixed in the y-direction. An example of such an equation would be z = ex .
- 3. 3 2. The Distance Formula and the Equation for a Sphere To find the distance between any two points in three space, we use a very similar formula to that in 2-space. The idea is to generalize Pythagoras theorem. Result 2.1. The distance |P1P2| between P1(x1, y1, z1) and P2(x2, y2, z2) is |P1P2| = (x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2. The distance formula is simple to apply to find the distance between points. Example 2.2. (i) Find the shortest distance from point P(3, 8, −2) to the xy-plane. This will just be the distance between the projection of P onto the xy-plane and the point P. (ii) Find the shortest distance from the point (3, 8, −2) to the z- axis. This will just be the absolute value of the z coordinate. A sphere of radius R centered at the point (a, b, c) is by definition the set of all points a distance R from the point (a, b, c) in 3-space. We can use the distance formula to determine a formula for such a sphere. Result 2.3. An equation of a sphere with center (a, b, c) and radius R is (x − a)2 + (y − b)2 + (z − c)2 = r2 . Example 2.4. Show that the graph of the equation x2 +y2 +z2 −6x+ 4y − 2z = 11 is a sphere and find its radius and center. To show it is a sphere, we complete the square in all three variables: x2 +y2 +z2 −6x+4y−2z = (x−3)2 −9+(y+2)2 −4+(z−1)2 −1 = 11 so (x − 3)2 + (y + 2)2 + (z − 1)2 = 25. Thus it is a sphere of radius 5 and center (3, −2, 1). 3. Regions in 3-Space We know how to bound regions in 2-space. Being able to bound regions in 3-space is also important. We illustrate with a couple of examples. Example 3.1. (i) Sketch the region represented by the inequali- ties x2 + y2 + z2 1, x 1/2. This is the “scalp” of a sphere pointing in the x-direction.
- 4. 4 (ii) Find bounds for the region which consists of a hollow ball with outer radius 5 and inner radius 4 centered at (1, 2, 3). Outer ball equation is (x−1)2 +(y −2)2 +(z −3)2 = 25 and the inner ball equation is (x − 1)2 + (y − 2)2 + (z − 3)2 = 16. We want it to be smaller than the outer ball and larger than the inner ball, so we get (x − 1)2 + (y − 2)2 + (z − 3)2 25 and (x − 1)2 + (y − 2)2 + (z − 3)2 16