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Chapter 12 - Vectors and the Geometry of Space

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Chapter 12 - Vectors and the Geometry of Space Calculus III Chapter 12 - Vectors and the Geometry of Space 1. Three-Dimensional Coordinate System Third-semester calculus is the study of functions of more than one variable. Much of what we do will be to generalize concepts from your previous two semesters. However, multivariable calculus will soon take on a life of its own and there are many instances where the picture is significantly more complicated than in single variable calculus. Nevertheless, you will often be able to rely on intuition you have built up in previous courses. Initially, we will focus on building up the geometry of three-dimensional space. Much of what we do will extend beyond this, but pictures are harder to come by. The three-dimensional coordinate system is formed by three coordinate axes labeled the x- axis, y-axis, and z-axis, all pairwise perpendicular. The origin, denoted O, is the intersection of all three axes. In order to standardize direction, we use the right-hand rule. Right-hand rule: Extend your right hand, with fingers curled and thumb pointed up. Your fingers align with the positive direction of the x-axis, your arm with the positive direction of the y-axis, and your thumb with the positive direction of the z-axis. These axes form three planes, labeled the xy-plane, xz-plane, and yz-plane. These planes in turn divide space into eight octants. We will not concern ourselves with numbering the eight octants, except for the first octant which is bounded by the positive part of each plane. A point P in three-dimensional space can be represented by an ordered triple (a; b; c), in- dicating the distance from the origin O in the x, y, and z direction, respectively. Hence, O is represented by (0; 0; 0). This gives a one-to-one bijection between points in space and ordered triples in the Cartesian product R × R × R = f(x; y; z): x; y; z 2 Rg. Given a point P (a; b; c), we can drop a perpendicular to the xy-plane1. This perpendicular intersects the xy-plane at the point Q(a; b; 0). We call this point the projection of P onto the xy-plane. One can project similarly onto the xz- and yz-planes. 1What this means precisely will be defined later. Roughly, it means that we draw a line from P through the xy-plane such that it is perpendicular to every line that lies on that plane. 1 Given a function f : Rn ! R (i.e., a function of n-variables), the graph of f is the set of points P such that f(P ) = 0. When n = 3, we call this graph a surface. Example 1. (1) The surface represented by z − 3 is a plane parallel to the xy-plane through the point (0; 0; 3). (2) The surface represented by x2 + y2 − 1 is a cylinder of radius 1, perpendicular to the xy-plane. The distance between two points P1(x1; y1; z1) and P2(x2; y2; z2), denoted jP1P2j, is the length of the line segment whose endpoints are P1 and P2. To find this distance, we form a box whose opposite endpoints are P1 and P2. Set two other points on this box, A(x2; y1; z1) and B(x2; y2; z1). Applying the Pythagorean Theorem (twice) gives 2 2 2 jP1P2j = jP1Bj + jBP2j (PT) 2 2 2 = jP1Aj + jABj + jBP2j (PT) 2 2 2 = jx2 − x1j + jy2 − y1j + jz2 − z1j (distance formula on real line): Hence, we have now derived the following. Distance Formula in Three Dimensions: The distance jP1P2j between the points P1(x1; y1; z1) and P2(x2; y2; z2) is p 2 2 2 jP1P2j = (x2 − x1) + (y2 − y1) + (z2 − z1) : A sphere of radius r > 0 centered at C(h; k; `) is the set of points whose distance from (h; k; `) is r. Thus, it follows from the distance formula that the equation of a sphere is (x − h)2 + (y − k)2 + (z − `)2 = r2: 2 2. Vectors A vector is, informally (for now), a quantity that has magnitude and direction. We will formalize this eventually, for now we think of a vector v as a ray, having an initial point A −! (the tail) and a terminal point B (the tip), denoted v = AB. The length of v is the distance jABj. We say that two vectors u and v are equivalent if they have the same magnitude and direction, even if they are in different positions, and write u = v. The zero vector, denoted by 0, is the unique vector with length zero and no direction. Definition of Vector Addition: If u and v are vectors positioned so that the initial point of v is the terminal point of u, then the sum of u and v is the vector from the initial point of u to the terminal point of v. −! −−! −! Hence, if u = AB and v = BC, then u+v = AC. Note that vector addition is commutative, as can be seen by the parallelogram law: Definition of Scalar Multiplication: If c is a scalar and v is a vector, then the scalar multiple cv is the vector whose length is jcj times the length of v and whose direction is the same as v if c > 0 and opposite if c < 0. If c = 0, then cv = 0. Two vectors are said to be parallel if they are scalar multiples of one another. We call −v = (−1)v the negative of v. By the difference of vectors u and v we mean u − v = u + (−1)v: If we fix a coordinate system, we can denote a vector by its terminal point P , fixing its initial point to be the origin (since position doesn't matter). Of course, a vector may have many representations, i.e., different initial and terminal points. We call this one the position vector of the point P . The coordinates for this point are called the components of the vector. In two-dimensions, we write a = ha1; a2i, and in three-dimensions, we write a = 3 ha1; a2; a3i. We denote by Vn the set of all n-dimensional vectors. So if a 2 Vn, then −! a = ha1; a2; : : : ; ani. Given any representation AB of the vector a, where A(x1; y1; z1) and B(x2; y2; z2), the position vector has coordinates a = hx2 − x1; y2 − y1; z2 − z1i: The length (or magnitude) of the vector v is the length of any representation and is denoted p 2 2 by jvj. Hence, the length of the two-dimensional vector a = ha1; a2i is jaj = a1 + a2. The p 2 2 2 length of the three-dimensional vector a = ha1; a2; a3i is jaj = a1 + a2 + a3. Given components, it is straightforward to find the sum, difference, and scalar multiple of vectors. Let a = ha1; a2i, v = hb1; b2i, and c a scalar. Then a + b = ha1 + b1; a2 + b2i; a − b = ha1 − b1; a2 − b2i; ca = hca1; ca2i: This extends in the obvious way to three or more dimensions. Note that, jcaj = jcjjaj. This can be verified directly, q p 2 2 2 2 2 jcaj = (ca1) + (ca2) = c (a1 + a2) = jcjjaj: Caution: It is not true that ja + bj = jaj + jbj. Properties of Vectors For all a; b; c 2 Vn and all scalars c; d, (i) a + b = b + a (v) c(a + b) = ca + cb (ii) (a + b) + c = a + (b + c) (vi) (c + d)a = ca + da (iii) a + 0 = a (vii) c(da) = (cd)a (iv) a + (−a) = a + (−1)a = 0 (viii) 1a = a Another way to represent vectors is as sums (linear combinations) of the standard basis vectors. In V2, these are the vectors i = h1; 0i and j = h0; 1i. In V3, these are the vectors i = h1; 0; 0i; j = h0; 1; 0i; k = h0; 0; 1i: If a = ha1; a2; a3i, then a = a1i + a2j + a3k. Of course, it is important to know whether you are working in V2 or V3. A unit vector is a vector whose length is 1. Hence, the standard basis vectors are, by definition, unit vectors. In general, if a 6= 0, the the unit vector in the same direction as a is 1 u = a: jaj We can check this easily. Let c = 1=jaj. Then juj = jcaj = jcjjaj = jaj. 4 Application. A 100-lb weight hangs from two wires as shown in Figure 19 on page 804. Find the tensions (forces) T1 and T2 in both wires and the magnitudes of the tensions. This is left mostly as a reading exercise. We will walk through some of the initial steps. We regard T1 and T2 as vectors in V2. Draw a line through the common initial point parallel to the \top" of the triangle and drop perpendiculars from the terminal points of both vectors to this line. This gives two right triangles with hypotenuses jT1j and jT2j, respectively. Hence, using right triangle trigonometry, we have ◦ ◦ T1 = −|T1j cos 50 i + jT1j sin 50 j ◦ ◦ T2 = jT2j cos 32 i + jT2j sin 32 j: The resultant force experience by an object is the vector sum of the forces acting on it. In this case, the resultant T1 + T2 of the tensions counterbalances the weight w = −100j and so we have T1 + T2 = −w = 100j. This equation must hold in each component. Hence we have the system of equations, ◦ ◦ −|T1j cos 50 + jT2j cos 32 = 0 ◦ ◦ jT1j sin 50 + jT2j sin 32 = 100: We can solve this system using elementary means2.
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