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Spring 2004 Math 253/501–503 11 Three Dimensional Analytic Geometry and Vectors 11.2 Vectors and the Dot Product Tue, 20/Jan C

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Spring 2004 Math 253/501–503 11 Three Dimensional Analytic Geometry and Vectors 11.2 Vectors and the Dot Product Tue, 20/Jan C Spring 2004 Math 253/501–503 • vector subtraction / vector difference: Formally, − = + .− / 11 Three Dimensional Analytic a b a 1 b. Just subtract components slotwise. − = ;:::; − ;:::; = − ;:::; − Geometry and Vectors a b [a1 an] [b1 bn] [a1 b1 an bn] 11.2 Vectors and the Dot Product • standard basis vectors:InVn (the set of all n-D vectors), th Tue, 20/Jan c 2004, Art Belmonte these are the vectors e1;:::en,whereek hasa1inthek slot and n − 1 zeros in its other slots. In V3,wehavethe { ; ; } Summary following aliases for e1 e2 e3 ,anorthonormal basis. i = e1 = [1; 0; 0] VECTORS: Terms and concepts j = e2 = [0; 1; 0] • vector: A quantity having magnitude (length) and direction. k = e3 = [0; 0; 1] : – Geometrically, an equivalence class of directed Note that we may write a given vector in terms of standard (hyper)line segments in n-D space having the same basis vectors. For example, magnitude and direction (analogy: 1 and 2 are 2 4 = ; ; equivalent fractions). A particular member of this a [a1 a2 a3] equivalence class has a point of application—the “tail” = [a1; 0; 0] + [0; a2; 0] + [0; 0; a3] of the vector. = a1 [1; 0; 0] + a2 [0; 1; 0] + a3 [0; 0; 1] – Analytically, an ordered n-tuple of (real) numbers, = a1i + a2j + a3k: called components or elements: a = [a1; a2;:::;an]. While Stewart uses angle brackets h:::ito delimit • unit vector: A vector whose length is 1. Given a nonzero vectors, most folks (other authors, MATLAB, TI-89) 6= −→ vector a 0, the unit vector in the direction of a is a-“hat.” use square brackets; so shall we! We also write a as a . 1 a aˆ = a = : • position vector: The distinguished member of the kak kak equivalence class that starts at the origin of n-D space. −→ • resultant vector: The vector sum of several vectors. For • AB: Vector from A .a ;:::;a /to B .b ;:::;b /, realized 1 n 1 n example, the resultant force is the vector sum of several as [b − a ;:::;b −a ]; i.e., end−start in each slot. 1 1 n n forces. Again, the resultant velocity is the vector sum of • magnitude: The length of a (real) vector a = [a1;:::;an]is several velocities. n kak=v a2 VECTORS: Properties u k ukX=1 t In the following, c and d are scalars; a and b are vectors. via repeated application of the Pythagorean Theorem. Thus 1. a + b = b + a (commutativity) −→ n = . − /2: AB v bk ak 2. a + .b + c/ = .a + b/ + c (associativity) u ukX=1 t 3. a + 0 = a (additive identity: the zero vector) • zero vector: This is the n-D vector all of whose components + .− / = − are zero: 0 = [0; 0;:::;0]. It has length zero and no specific 4. a a 0 (additive inverse of a: a) direction. 5. c .a + b/ = ca + cb (Scalar multiplication distributes over • vector addition / vector sum: Add components slotwise. vector addition.) 6. .c + d/a = ca + da (another instance of distributivity) a + b = [a1;:::;an]+[b1;:::;bn]=[a1 +b1;:::;an +bn] 7. .cd/a = c.da/ (associativity of scalar multiples) • Triangle Law / Parallelogram Law: The geometric = (head-to-tail) interpretation of vector addition. 8. 1a a • scalar: a real number or symbol [more generally, a complex number or symbol or an element of a field] DOT PRODUCT: Definitions and facts • scalar multiplication: Given a scalar c and a vector a,the Let a = [a1;:::;an]andb=[b1;:::;bn]ben-D vectors. scalar multiple of c with a is obtained by multiplying each component of a by c. n • dot product (math definition): a · b = akbk ca = c [a ;:::;an]=[ca ;:::;can] 1 1 kX=1 1 • dot product (physics definition): a · b = kakkbkcos θ, Solution where θ is the angle between the vectors a and b; equivalent to math definition Recall the discussion of standard basis vectors in the Summary. Rewrite the vectors as a = [6; 0; 1] and b = [1; −2; 7]. (Make − a · b • angle between nonzero vectors: θ = cos 1 certain that you understand this.) We then have kakkbk √ kak = 62 + 02 + 12 = 37 • orthogonality: Vectors a and b are perpendicular if and a + b = [7p; −2; 8] only if a · b = 0. a − b = [5; 2; −6] · = ; ; • =a b 2a [12 0 2] scalar projection of b onto a:compab kak 3a + 4b = [18; 0; 3] + [4; −8; 28] = [22; −8; 31] : a·b a Compare this with the corresponding MATLAB example. (Also • vector projection of b onto a:proj b= ; i.e., a kak kak remind me to show you on a TI-89.) the scalar projection times the unit vector in the direction of a; also known as the parallel projection 664/14 • = − orthogonal projection of b onto a:orthab b projab, whence b is the vector sum of a vector parallel to a and a Find the unit vector that has the same direction as = − + vector perpendicular to a a 2i 4j 7k. • = · = k kk k θ work: W F d F d cos ,whereFis the (constant) Solution force vector and d is the displacement vector ; − ; a Rewrite a as [2 4 7]. Then • direction cosines: the components of aˆ = , the unit kak [2; −4; 7] 2 4 7 vector in the direction of a aˆ = √ = √ ; −√ ; √ 4 + 16 + 49 69 69 69 • direction angles: the function cos−1 = arccos mapped onto the direction cosine vector aˆ; gives the angles said vector 664/18 (and hence a itself) makes with the positive axes in Rn Given a = [−1; −2; −3] and b = [2; 8; −6], find the dot product a · b. DOT PRODUCT: Properties Solution Here a, b,andcare vectors and k is a scalar. We have 1. a · a = kak2 a · b = .−1/.2/ + .−2/.8/ + .−3/.−6/ =−2−16 + 18 = 0: 2. a · b = b · a (commutativity) Therefore, a and b are orthogonal (perpendicular to one another)! 3. a · .b + c/ = a · b + a · c (The dot product distributes over 664/16 vector addition.) Find a · b, given that kak = 6, kbk = 1 , and the angle between 4. .ka/ · b = k.a · b/ = a · .kb/ (Scalars roam freely across the π 3 a and b is θ = . dot product operation.) 4 5. 0 · a = 0 (The dot product of zero vector with any other Solution vector is zero.) Via the physics definition of dot product we have √ √ · = k kk k θ = . / 1 2 = : Hand Examples a b a b cos 6 3 2 2 664/10 665/34 Let a = 6i + k and b = i − 2j + 7k.Findkak,a+b,a−b,2a, Find the values of x such that the vectors v = [x; x; −1] and and 3a + 4b. w = [1; x; 6] are orthogonal. 2 Solution Solution For the given vectors to be orthogonal (perpendicular), their dot −→ −→ −→ product must equal zero. We have a = AB = B − A = [4; 4] − [1; 3] = [3; 1]. Here is a diary file and a diagram; arrow is a command Cooper wrote. v · w = 0 .x/.1/ + .x/.x/ + .−1/.6/ = 0 2 x + x − 6 = 0 % % Stewart 53/1 .x − 2/.x + 3/ = 0 % =−; origin = [0 0]; A = [1 3]; B = [4 4]; x 32 a=B-A a= 31 arrow(A, a); hold on; axis equal 665/48 arrow(origin, a, ’r’); grid on axis([-1 5 -1 5]) = + − plot(A(1), A(2), ’ms’, ’MarkerSize’, 12); Find the scalar and vector projections of b i 6j 2k onto plot(B(1), B(2), ’gs’, ’MarkerSize’, 12) a = 2i − 3j + k. plot(0, 0, ’ks’, ’MarkerSize’, 12) % echo off; diary off Solution Stewart 53/1 5 • The scalar projection is B 4 a a · b A comp b = 3 a kak − − = √2 18 2 y 2 4 + 9 + 1 √ 1 18 9 14 a =−√ or − : 14 7 0 • −1 The vector projection is −1 0 1 2 3 4 5 x a · b a proj b = a kak kak ; − ; =−√18 [2 √ 3 1] 14 14 s053x05 =−9 ; − ; 7[2 3 1] 18 27 9 = − ; ; − : Find the sum of the vectors v = [2; 3] and w = [3; −4], then 7 7 7 illustrate geometrically. 665/54 Find the work done by a force of 20 lb acting in the direction Solution N50◦W in moving an object 4 ft due west. + = ; − Solution We have v w [5 1]. Here is a diary file followed by a diagram that illustrates vector addition via the Triangle Law—the The angle between the force and displacement vectors is 40◦.The “head-to-tail” interpretation of vector addition. definition of work and our trusty TI-89 give ◦ W = F · d = kFkkdkcos θ = .20/.4/ cos 40 ≈ 61:28 ft-lb: % % Stewart 53/5 % v = [2 3]; w = [3 -4]; MATLAB Examples o = [0 0]; u = v + w u= 5-1 s053x01 arrow(o, v, ’b’); hold on; arrow(v, w, ’r’); arrow(o, u, ’m’) Find a vector a with representation given by the directed line axis equal; grid on −→ −→ axis([-1 6 -2 4]) segment AB connecting points A.1; 3/ and B.4; 4/.DrawAB % and the equivalent representation that starts at the origin. echo off; diary off 3 Stewart 53/5 37ˆ(1/2) 4 pretty(exact length of a) % ...that’s nicer! 3 1/2 37 2 % v w echo off; diary off y 1 0 s664x14 [664/14 revisited] u = v + w −1 −2 Find the unit vector that has the same direction as −1 0 1 2 3 4 5 6 = − + x a 2i 4j 7k. s664x10 [664/10 revisited] Solution Let a = 6i + k and b = i − 2j + 7k.Findkak,a+b,a−b,2a, and 3a + 4b.
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