Lecture AB2.1: Fundamentals of Vector Algebra and Vector Calculus

Transcription

1 Lecture AB2.1: Fundamentals of Vector Algebra and Vector Calculus Two vectors a and b are equal, a = b, if they have the same length and the same direction. If a given vector a has initial point P : (x 1, y 1, z 1 ) and terminal point Q : (x 2, y 2, z 2 ), the three numbers a 1 = x 2 x 1, a 2 = y 2 y 1, a 3 = z 2 z 1 (1) are called the components of vector a with respect to Cartesian coordinate system xyz in space, and we write simply a = [a 1, a 2, a 3 ] Length of a vector in terms of components: a = a a a 2 3 (2) EXAMPLE 1 The vector a with initial point P : (4, 0, 2) and terminal point Q : (6, 1, 2) has the components Hence a = [2, 1, 0] and its length a 1 = 6 4 = 2, a 2 = 1 0 = 1, a 3 = 2 2 = 0 a = ( 1) = 5. If we choose ( 1, 5, 8) as the initial point of a, then, according to (1), the corresponding terminal point is ( 1 + 2, 5 1, 8 + 0) = (1, 4, 8) In the given Cartesian coordinate system, the position vector r of a point A : (x, y, z) is the vector with the origin (0, 0, 0) as the initial point and A as the terminal point. Thus r = [x, y, z] THEOREM A fixed Cartesian coordinate system being given, each vector is uniquely determined by its ordered triple of corresponding components. Conversely, to each ordered triple of real numbers (a 1, a 2, a 3 ), there corresponds precisely one vector a = [a 1, a 2, a 3 ], with (0, 0, 0) corresponding to the zero vector, which has length 0 and no direction. Another representation of a vector is a = [a 1, a 2, a 3 ] = a 1 i + a 2 j + a 3 k (3) i, j, k are unit vectors in the positive directions of the axes of a Cartesian coordinate system. i = [1, 0, 0], j = [0, 1, 0], k = [0, 0, 1]. (4)

2 EXAMPLE 3 a = [4, 0, 1] = 4i + k, b = [2, 5, 1 3 ] = 2i 5j k. Inner Product a b = a b cos γ if a 0, b 0 a b = 0 if a = 0 or b = 0 (1). In components, a = [a 1, a 2, a 3 ], b = [b 1, b 2, b 3 ] and a b = a 1 b 1 + a 2 b 2 + a 3 b 3. The vector a is called orthogonal to vector b if a b = 0. THEOREM The inner product of two nonzero vectors is zero if and only if these vector are perpendicular (cos γ = 0, γ = π/2 (90 o )). Length of a vector and angle between vectors in terms of inner product: a = a a (2) EXAMPLE 1 cos γ = a b a b = a b a a b b (3) Find the inner product and the length of a= [1, 2, 0] and b= [3, 2, 1] as well the angle between these vectors: a b = ( 2) = 1, a = a a = 5, b = b b = 14; γ = arccos a b a b = arccos ( 1) 70 = arccos ( ) = = o Orthonormal Basis The orthonormal basis in the 3-D space is a basis {a, b, c} consisting of orthogonal unit vectors. For a given vector v = l 1 a + l 2 b + l 3 c we have l 1 = a v, l 2 = b v, l 3 = c v The unit vectors i, j, k form an orthonormal basis called the standard basis. EXAMPLE 6 Normal vector to a plane Find a unit vector perpendicular to the plane 4x + 2y + 4z = 7.

3 We may write any plane in the space as a r = a 1 x + a 2 y + a 3 z = c, a = [a 1, a 2, a 3 ] 0, r = [x, y, z]. The unit vector in the direction a is Dividing a r = c by a we obtain n r = p, n = 1 a a. p = c a. n is a unit normal vector to the plane (the other being -n). In our case, a = [4, 2, 4], c = 7, a = = 36 = 6; thus n = (1/6)a = [2/3, 1/3, 2/3], and the plane has the distance p = 7/6 to the origin. Vector Product The vector product a b of two vectors a = [a 1, a 2, a 3 ] and b = [b 1, b 2, b 3 ] is a vector v = a b with the length v = a b sin γ (γ is the angle between a and b) and direction perpendicular to both a and b so that a, b, v form a right-handed triple. or In components, v = [v 1, v 2, v 3 ] = a b = v 1 = a 2 a 3 b 2 b 3, v 2 = a 1 a 2 a 3 b 1 b 2 b 3 a 3 a 1 b 3 b 1 = v 1 i + v 2 j + v 3 k,, v 3 = a 1 a 2 b 1 b 2 EXAMPLE 2 For a = [4, 0, 1] and b = [ 2, 1, 3] a b = = i j k = i 10j + 4k.

4 Fundamentals of vector calculus We will consider two kinds of functions, vector functions, whose values are vectors, v = v(p ) = [v 1 (P ), v 2 (P ), v 3 (P ), ] and scalar functions, whose values are scalars, f = f(p ) depending on the point P in space. A vector function defines a vector field and a scalar function defines a scalar field. If we introduce Cartesian coordinates x, y, z then we can also write v = [v 1 (x, y, z), v 2 (x, y, z), v 3 (x, y, z)] f = f(x, y, z). EXAMPLE 1 Scalar function (Euclidean distance in space) In a Cartesian coordinate system, the distance f(p ) of any point P : [x, y, z] from a fixed point P 0 : [x 0, y 0, z 0 ] in space is a scalar function f = f(p ) = f(x, y, z) = (x x 0 ) 2 + (y y 0 ) 2 + (z z 0 ) 2. EXAMPLE 3 Vector field (field of force) The magnitude of the gravitational force p directed from a point P to a point P 0 in space is a scalar function p = c, c = const. (2) r2 In a Cartesian coordinate system, P : [x, y, z], P 0 : [x 0, y 0, z 0 ], and r = (x x 0 ) 2 + (y y 0 ) 2 + (z z 0 ) 2. Introducing the vector r = [x x 0, y y 0, z z 0 ] = (x x 0 )i + (y y 0 )j + (z z 0 )k we have r = r, 1 r is a unit vector in the direction of p, and r p = p ( 1 ) r r = c r r = cx x 0 i c y y 0 j c z z 0 k (3) 3 r 3 r 3 r 3 This vector function decribes the gravitational force acting on a particle at the point P attracted by a particle fixed at the point P 0. Let, for example, P 0 : [0, 0, 0] be the origin of a Cartesian coordinate system; then for points P situated on the unit circle x 2 + y 2 = 1 on the plane z = 0 we have r = 1 and the vector function (3) has the form p = p ( r) = cr, (3 ) so that at every point on the circle the magnitude of p is constant and the direction of p is opposite to that of the position vector r. This statement is valid for any other circle of the radius a (x 2 + y 2 = a 2 ) and for any sphere of the radius a (x 2 + y 2 + z 2 = a 2 ).

5 PROBLEM Initial point P : (1, 1, 0), terminal point Q : (4, 5, 0). P Q = v = [4 1, 5 1, 0 0] = [3, 4, 0] = 3i + 4j. v = = 25 = 5. PROBLEM Initial point P : (1, 2, 3), terminal point Q : (2, 4, 6). P Q = v = [2 1, 4 2, 6 3] = [1, 2, 3] = i + 2j + 3k. v = = 14. PROBLEM Initial point P : (1, 0, 0), P Q = v = [2, 3, 0] = 2i + 3j. Find terminal point Q : (x 2, y 2, z 2 ). P Q = v = [x 2 1, y 2 0, z 2 0] = [2, 3, 0]. Thus, x 2 = = 3, y 2 = = 3, z 2 = = 0] and Q : (3, 3, 0) v = = 13. PROBLEM a = [3, 2, 1] = 3i 2j + k, b = [0, 3, 0] = 3j. Find a + b, a + b. a + b = [3 + 0, 2 + 3, 1 + 0] = [3, 1, 1] = = 11. a = = 14. b = 3 2 = 3. a + b = a + b < a + b. PROBLEM a = [1, 3, 2] = i + 3j + 2k, b = [2, 0, 5] = 2i 5k, c = [4, 2, 1] = 4i 2j + k. We calculate the inner product a b = ( 5) = = 8 = b a. PROBLEM b + 3c = [2 2, 0, 2 ( 5)] + [3 4, 3 ( 2), 3 1] = [4 + 12, 0 6, ] = [16, 6, 7]. a (2b + 3c) = ( 6) + 2 ( 7) = = 16 = 2a b + 3a c due to linearity of the inner product.

6 PROBLEM Find the angle between planes x + y + z = 1 and x + 2y + 3z = 6. We may write any plane in the space as a r = a 1 x + a 2 y + a 3 z = c, a = [a 1, a 2, a 3 ] 0, r = [x, y, z]. In our case, for the first plane and a r = x + y + z = 1, a = [1, 1, 1] b r = x + 2y + 3z = 6, b = [1, 2, 3], r = [x, y, z] for the second plane. c 1 = 1, a = = 3; c 2 = 6, b = = 14. The unit normal vector to the first plane is The unit normal vector to the second plane is n 1 = 1 a a = 1 3 a. n 2 = 1 b b = 1 14 b. The angle between planes is equal to the angle between their normal vectors which coincides with the angle γ between vectors a and b that have the same directions. The latter angle is determined in terms of the inner product: cos γ = a b a b = = 6 42 = , so that γ = arccos = o. PROBLEM Find the vector and inner products given a = [1, 2, 0], b = [ 3, 2, 0], and c = [2, 3, 4]. a b = = k b a = a b = 8k. = 8k; a b = b a = 1 ( 3) = = 1.

7 PROBLEM a c = = i j k = 8i 4j k = [8, 4, 1]; PROBLEM a c = c a = = = 81 = 9. a c = = = 8. The scalar field (pressure) is given by f(x, y) = 9x 2 + 4y 2. Find the pressure at the points (2, 4), (0.5, 3.25), ( 17, 1/ 6). f(2, 4) = = 4 (9 + 16) = 100. f(0.5, 3.25) = (3.25) 2 = = f( 17, 1/ 6) = (1/6) = / PROBLEM The isobars (curves of constant pressure) are ellipses 9x 2 +4y 2 = c, c > 0, e.g., 9x 2 +4y 2 = 1: PROBLEM The isotherms (curves of constant temperature) are determined from ln x 2 + y 2 = c; they are circles x 2 + y 2 = C = e c > 0. PROBLEM We have arctan y/x = c; therefore, tan arctan y/x = y/x = C = tan c and the isotherms are straight lines y = Cx. PROBLEM We have x 2 y 2 = c, and the isotherms are parabolas y = ± x 2 c, c 0 or straight lines y = x, c = 0. PROBLEM The level surfaces are planes 4x + 3y z = c. PROBLEM The level surfaces are elliptic cylinders x 2 + 3y 2 = c, c > 0.