Arkansas Tech University MATH 2934: Calculus III Dr. Marcel B Finan
11.2 Vectors in Three Dimensions: Space Vectors In this section, we learn the aspects of the three-dimensional coordinate system. We define the three dimensional coordinate system as shown in Figure 11.2.1(a).
Figure 11.2.1 There are three coordinate axes: The x−axis, the y−axis, and the z−axis. They are oriented as shown in Figure 11.2.1(a). The point O is called the origin of the system. As shown in Figure 11.2.1(b), the coordinate system determine three coordinate planes and these planes divide the space into eight parts, each called an octant: Four octants above the xy−plane and four octants below the xy−plane. The octants are numbered as shown in Figure 11.2.2.
Figure 11.2.2 The first octant is determined by the positive x−, y−, and z−axes. In the two-dimensional system two numbers are required to determine a point.
1 Likewise, in the three-dimensional coordinate system, a point in space P re- quires three numbers: a, b, and c. More precisely, a point P in space is uniquely determined by an ordered triple (a, b, c) as shown in Figure 11.2.3.
Figure 11.2.3 The numbers a, b, and c are the coordinates of P : a is the x−coordinate, b is the y−coordinate and c is the z−coordinate. In terms of sets, the three- dimensional coordinate system will be denoted by the set IR3 = {(x, y, z): x, y, z ∈ IR}. Now, from the point P (a, b, c) we can create a box with each vertex having the coordinates shown in Figure 11.2.4. The point Q(a, b, 0) is called the orthogo- nal projection of P onto the xy−plane. Likewise, R(0, b, c) is the orthogonal projection of P onto the zy−plane and S(a, 0, c) is the orthogonal projection of P onto the xz−plane.
Figure 11.2.4 The Distance Formula The distance between two points P1(x1, y1, z1) and P2(x2, y2, z2) is given by the distance formula
p 2 2 2 d(P1,P2) = (x2 − x1) + (y2 − y1) + (z2 − z1) . This formula is easily derived from Figure 11.2.5. Indeed, using the Pythagorean formula in the highlighted right triangle, we find
2 2 2 2 [d(P1,P2)] = (x2 − x1) + (y2 − y1) + (z2 − z1) .
2 Figure 11.2.5
Example 11.2.1 Find the distance between the two points C(a, b, c) and P (x, y, z).
Solution. Using the distance formula, we find d(C,P ) = p(x − a)2 + (y − b)2 + (z − c)2
The solutions to an equation of the form f(x, y, z) = 0 are represented by a surface in IR3.
Example 11.2.2 Describe and sketch the surface (x − a)2 + (y − b)2 + (z − c)2 = r2, where r > 0.
Solution. The collection of all points in space whose distances from a fixed point C(a, b, c) is equal to a positive number r is called a sphere. We call the point C the center and the number r the radius of the sphere. This definition, implies that d(C,P ) = r or equivalently
(x − a)2 + (y − b)2 + (z − c)2 = r2.
This last equation is called the standard form of the equation of a sphere. The surface is shown in Figure 11.2.6
Figure 11.2.6
3 Example 11.2.3 Find the center and the radius of the sphere: 2x2 +2y2 +2z2 +6x+4y −2z = 1.
Solution. Using the method of completing the square, we find
2x2 + 2y2 + 2z2 + 6x + 4y − 2z =1 9 1 9 1 2 x2 + 3x + + 2(y2 + 2y + 1) + 2 z2 − z + =1 + + 2 + 4 4 2 2 32 12 2 x + + 2(y + 1)2 + 2 z − =8 2 2 32 12 x + + (y + 1)2 + z − =4. 2 2