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 = f(x; y; z): x; y; z 2 IRg: 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 3 1 Hence, the center is C − 2 ; −1; 2 and the radius is 2 Example 11.2.4 Describe the region in IR3 represented by the compound inequality 1 ≤ x2 + y2 + z2 ≤ 4; z ≤ 0: Solution. An equivalent form of the given inequality is 1 ≤ px2 + y2 + z2 ≤ 2: This inequality represents the points in IR3 whose distance from the origin is at least 1 and at most 2. Since z ≤ 0; we consider only those points that lie on or below the xy−plane. Thus, the inequality represents those points that are between (or on) the spheres x2 + y2 + z2 = 1 and x2 + y2 + z2 = 4 and beneath (or on) the xy−plane as shown in Figure 11.2.7 Figure 11.2.7 As for plane vectors, a space vector ~v can be uniquely represented by an ordered triple ~v =< v1; v2; v3 > : Many of the definitions and formulas established for the plane vectors can be extended to space vectors. For example, the sum and difference of the vectors ~u =< u1; u2; u3 > and ~v =< v1; v2; v3 > are ~u + ~v =< u1 + v1; u2 + v2; u3 + v3 > 4 and ~u − ~v =< u1 − v1; u2 − v2; u3 − v3 > : The multiplication of the vector ~u =< u1; u2; u3 > by a scalar c is the vector c~u =< cu1; cu2; cu3 > : Also, a space vector ~v =< v1; v2; v3 > can be represented as follows ~ ~v = v1~i + v2~j + v3k where ~i =< 1; 0; 0 >; ~j =< 0; 1; 0 >; and ~k =< 0; 0; 1 > are the standard unit vectors. We define the zero vector to be the vector ~0 =< 0; 0; 0 > : p 2 2 2 The length of a vector ~v =< v1; v2; v3 > is the scalar jj~vjj = v1 + v2 + v3: If P (p1; p2; p3) is the tail of a vector ~v and Q(q1; q2; q3) is the tip then the components of ~v are ~v =< q1 − p1; q2 − p2; q3 − p3 > : We say that two vectors ~u =< u1; u2; u3 > and ~v =< v1; v2; v3 > are equal or equivalent if and only if u1 = v1; u2 = v2; u3 = v3: The unit vector ~u in the direction of ~v =< v1; v2; v3 > is the vector v v v ~u = 1 ; 2 ; 3 : jj~vjj jj~vjj jj~vjj The properties of the plane vector space discussed in the previous section are also valid for space vectors. Example 11.2.5 Given the tail P (−2; 3; 1) and the tip Q(0; −4; 4) of a vector ~v; find the compo- nents of ~v; its length and the unit vector ~u in the direction of ~v: Solution. We have ~v =< 0 − (−2); −4 − 3; 4 − 1 >=< 2; −7; 3 > : p Thus, jj~vjj = p22 + (−7)2 + 32 = 62: The unit vector in the direction of ~v is the vector 2 7 3 ~u = p ; −p ; p 62 62 62 Example 11.2.6 Given two points P (a; b; c) and Q(a0; b0; c0): Find the coordinates of the midpoint M(xM ; yM ; zM ): Solution. −−! 1 −−! By the definition of midpoint, we have PM = 2 P Q: That is, < xM − a; yM − 1 0 0 0 D a0−a b0−b c0−c E b; zM − c >= 2 < a − a; b − b; c − c >= 2 ; 2 ; 2 : By the defini- a0−a b0−b tion of equality of two vectors, we have xM − a = 2 ; yM − b = 2 ; and 5 c0−c a+a0 b+b0 zM − c = 2 : Solving each equation, we find xM = 2 ; yM = 2 ; and c+c0 zM = 2 Parallel Vectors The operation of multiplying a vector by a scalar leads to the following defini- tion: We say that two vectors ~u and ~v are parallel if and only if ~u = c~v for some non-zero scalar c: If c > 0 the two vectors point in the same direction. If c < 0; the two vectors point in opposite directions. Example 11.2.7 Determine whether the vector ~u with initial point A(2; 4; 3) and terminal point B(−4; 7; 5) is parallel to the vector ~v =< 12; −6; 4 > : Solution. The components of the vector ~u are ~u =< −4 − 2; 7 − 4; 5 − 3 >=< −6; 3; 2 > : If ~u is parallel to ~v then there is a scalar c such that ~v = c~u: This equation is equivalent to < 12; −6; 4 >= c < −6; 3; 2 >=< −6c; 3c; 2c > : Thus, we have −6c = 12; 3c = −6; 2c = 4: The first equation yields c = −2: The second equation yields c = −2 but the third equation yields c = 2: We conclude that c does not exist and hence the two vectors are not parallel The concept of parallel vectors can be used to show that three points are collinear, i.e., belong to the same straight line. Example 11.2.8 Show that the points P (1; −2; 3);Q(2; 1; 0); and R(4; 7; −6) are collinear. Solution. We have −−! PQ = < 2 − 1; 1 − (−2); 0 − 3 >=< 1; 3; −3 > −! PR = < 4 − 1; 7 − (−2); −6 − 3 >=< 3; 9; −9 > : −! −−! These two vectors have the same tail P: Since PR = 3P Q; the two vectors are parallel. Hence, the three points are collinear Applications of Vectors Vectors, whether in plane or space, have a wide range of applications. We describe one in the following example. Example 11.2.9 The guy wire supporting a 100-foot tower has tension of 550 pounds. Using Figure 11.2.8, write the component form of the vector F~ representing the ten- sion in the wire.