MA1008. Calculus and Linear Algebra for Engineers. Course Notes for Section B. Stephen Wills. Department of Mathematics. University College Cork

Transcription

1 MA1008 Clculus nd Liner Algebr for Engineers Course Notes for Section B Stephen Wills Deprtment of Mthemtics University College Cork s.wills@ucc.ie /08

2 Contents 1 Vector Geometry Vector products Dot product/sclr product Cross product/vector product Lines nd plnes Resolving vectors Exercises

3 CONTENTS These notes cover the second hlf of MA1008, corresponding to Section B of the MA1008 finl exmintion. The notes re self-contined: no textbook is needed. However, suggestions of books nd web resources re provided on the course web pge. A number of the exmples covered in these notes hve ppered on exm ppers. This is indicted by expressions such s Exmple 1.14 (S03 8()) which indictes tht Exmple 1.14 ws question 8() from the Summer pper of Similrly Exmple 1.18 (A04 6()) ws question 6() from the Autumn pper of Ech section in the notes ends with collection of Exercises. The only wy to lern mthemtics is to do exercises. Assistnce is vilble t the tutorils. Plese inform me if you find ny mistkes in the notes or exercises; tht wy I cn pss on corrections to the entire clss nd sve everyone confusion or worry. Nottionl Preliminries The following symbols will be used throughout to denote logicl reltionships The symbol mens implies. For exmple we my write x = 3 x 0, nd similrly x = 3 x 2 = 9, but we cnnot write x 2 = 9 x = 3, since if x 2 = 9 then x = 3 or x = 3. The symbol mens if nd only if or is equivlent to. For exmple x = 2 2x+5 = 9, in other words x = 2 2x+5 = 9 nd 2x+5 = 9 x = 2 re both vlid sttements. Similrly x = 2 7 3x = 1. However we cnnot write x = 2 x 2 = 4, since lthough it is true tht x = 2 x 2 = 4, x 2 = 4 x = 2, since we could hve x = 2. Almost everything we do will be concerned with functions (of one or more vrible). These will usully be denoted f, g,, nd written f : A B if we wnt to specify wht cn be put into the function (things from the set A), nd wht cn come out of f (things from the set B). Often A = B = R; R denotes the rel numbers, i.e. the points on the coordinte xes. For exmple f : R R, f(x) = x 2 2 tkes ny rel number x, squres it nd subtrcts 2 from the result. We my need to tke cre with the domin A of the function to ensure tht the function is defined. The mp f(x) = 1 x 1 is defined when x 1, but f(1) is not defined since we cnnot divide by 0. We will consider functions tht depend on more thn one quntity. R 2 = {(x, y) : x, R, y R} denotes the set of ll points in the plne; similrly R 3 denotes the set of ll points in 3D spce. We cn define functions R 2 R, for exmple g(x, y) = x 2 sin(y x) is defined for ll points (x, y), but the function h(x, y) = ln(x y) is only defined when x y > 0 x > y. 1

4 1 Vector Geometry We consider two types of vectors: Position vectors These specify the coordintes of point reltive to fixed origin nd righthnded coordinte xes. For exmple r = (1, 7, 4) sys tht the point is reched by moving 1 unit in the x-direction, 7 units in the y-direction nd 4 units in the z-direction. Displcement vectors Given two points P 1 = (x 1, y 1, z 1 ) nd P 2 = (x 2, y 2, z 2 ), the displcement vector P 1 P 2 = (x 2 x 1, y 2 y 1, z 2 z 1 ) tkes you from P 1 to P 2. Note. The coordintes of point re the components its ssocited position vector, which is the displcement vector from the origin O = (0, 0, 0). z P 2 P 1 P 2 (1, 7, 4) P 1 OP 1 OP 2 x y O Vectors cn be dded together, nd multiplied by numbers/sclrs: (1, 0, 3) + (2, 1, 5) = (1 + 2, 0 + 1, 3 + 5) = (3, 1, 2), 4( 1, 0, 5) = (4 ( 1), 4 0, 4 5) = ( 4, 0, 20). Properties of vector ddition nd sclr multipliction include + b = b + ; + (b + c) = ( + b) + c; t( + b) = t + tb; (s + t) = s + t. The ddition is described by the prllelogrm lw; scling stretches the vector, nd reverses its direction if multiplying by negtive number. 2 b b + b b b Vectors nd b re prllel if we cn write = tb for some constnt t. Vectors in 3D-spce cn be written uniquely in terms of the stndrd bsis vectors i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1); in this cse r = (x, y, z) = (x, 0, 0) + (0, y, 0) + (0, 0, z) = xi + yj + zk. Similrly ny vector in the plne cn be written in terms of i = (1, 0) nd j = (0, 1). 1.1 Vector products There re two wys in which vectors cn be multiplied. 2

5 VECTOR GEOMETRY Dot product/sclr product The dot product of = ( 1, 2, 3 ) nd b = (b 1, b 2, b 3 ) is.b = 1 b b b 3. The result is sclr. For exmple (3, 4, 0).( 2, 2, 5) = = 2. Note tht. = , the squre of the length of the vector, by Pythgors Theorem. The length of the vector is denoted, nd so = θ b 1 Properties of the sclr product re.b = b.;.(b + tc) =.b + t(.c);.b = b cosθ, where θ is the ngle between nd b. Since 1 cosθ 1, b.b b. If 0 nd b 0, then.b = 0 cosθ = 0 θ = π 2 nd b re orthogonl/perpendiculr. For exmple (1, 2, 1).(3, 1, 1) = = 0, so these vectors re orthogonl. Similrly the stndrd bsis vectors re mutully orthogonl: i.j = j.k = k.i = 0. Also, these three vectors re unit vectors they hve length 1, since i = j = k = 1. Exercise 1.1. Find three vectors orthogonl to = (1, 2, 2). Solution. 3

6 Lines nd plnes Cross product/vector product The cross product of = ( 1, 2, 3 ) nd b = (b 1, b 2, b 3 ) is b = ( 2 b 3 3 b 2, 3 b 1 1 b 3, 1 b 2 2 b 1 ) = 2 3 b 2 b 3 i 1 3 b 1 b 3 j b 1 b 2 k = i j k b 1 b 2 b 3 The expression on the second line involves determinnts which will be discussed in detil in Section 3. This time the result is vector. For exmple (3, 4, 0) ( 2, 2, 5) = (20 0, 0 15, 6+8) = (20, 15, 14). Properties of the cross product include b = b ; (b+c) = b+ c;.( b) = b.( b) = 0; b = b sin θ, where θ is the ngle between nd b. For = (3, 4, 0) nd b = ( 2, 2, 5) we hve ( ) ( 2, 2, 5). (3, 4, 0) ( 2, 2, 5) = ( 2, 2, 5).(20, 15, 14) = = 0, nd in generl b is lwys orthogonl to both nd b. The stndrd bsis vectors stisfy i j = k; j k = i; k i = j. Remrks. (i) The dot product mkes sense in dimensions other thn 3: (1, 2).(3, 5) = = 13; (1, 0, 1, 7).(2, 1, 1, 1) = = 6, but cross products only mke sense in 3D. (ii) The vector product is sometimes written b. 1.2 Lines nd plnes Two wys to specify line in the plne or in spce re (i) give two points on the line, or (ii) give one point on the line nd vector tht specifies the direction of the line. P 1 P 2 P P 0 OP 0 + t (t < 0) O Using the second method, if P 0 is the given point nd is vector in the direction, then the points P on the line re precisely those with position vectors OP = OP0 + t for some 4

7 VECTOR GEOMETRY number t. This is the prmetric eqution of the line; t is the prmeter. If P 0 = (x 0, y 0, z 0 ) nd = (, b, c), the point P = (x, y, z) lies on the line if nd only if (x, y, z) = (x 0, y 0, z 0 ) + t(, b, c) x = x 0 + t, y = y 0 + bt, nd z = z 0 + ct x x 0 = y y 0 b = z z 0 c for some t. This ltter eqution is the nonprmetric eqution of the line, nd bove we re ssuming tht 0, b 0 nd c 0. If, for exmple, = 0, then x = x 0 for ll t. Note tht t lest one of, b or c must be nonzero, since 0 in order to specify direction. In 2D pir of lines either intersect in one point or re prllel nd never meet. In 3D there is third possibility: the lines L 1 nd L 2 re skew if they do not intersect, but re not prllel. This mens tht they must live in prllel plnes. = t Prllel nd intersecting lines Skew lines If lines L 1 nd L 2 hve equtions r = OP 1 +s 1 nd r = OP 2 +t 2 then they intersect if nd only if OP 1 + s 1 = r = OP 2 + t 2 for some vlue of s nd t. In 3D our vectors hve three components, nd ll of these must gree so we hve three equtions in the two unknowns s nd t. Exercise 1.2. Find the prmetric nd nonprmetric equtions of the line through P 1 = (2, 3, 6) nd P 2 = (5, 3, 1). Find the point of intersection of this line with the line through (14, 1, 0) tht is prllel to the vector (0, 2, 11). Show tht the first line does not intersect the line Solution. x 1 4 = y = z 4 3 5

8 Lines nd plnes Consider the (x, y)-plne in 3D spce, i.e. the plne in which every point hs coordintes (x, y, 0). The position vector of ny point in this plne cn be written (uniquely) s xi + yj. Since the origin O is prt of this plne we cn write this s 0 + xi + yj. More generlly, one wy to specify ny plne is to give one point P 0 on the plne, nd two nonzero vectors nd b tht lie in the plne, nd re not prllel. The position vector of ny point P in the plne is then r = OP = OP0 + s + tb, for some s, t R. Alterntively, note tht there is vector n tht is perpendiculr to nd b, nd hence norml to the plne. Now point P is in the plne precisely if the displcement vector P 0 P lies in the plne, equivlently if P 0 P is orthogonl to n. So if c is the position vector of P 0, then the eqution for the position vector r = OP of point in the plne is P 0 P.n = 0 (r c).n = 0 r.n = c.n. 6

9 VECTOR GEOMETRY Note tht given the vectors nd b erlier, we cn tke n = b, since the cross product of nd b is orthogonl to nd to b, hence perpendiculr to the plne (note: since nd b re not prllel, we hve θ 0, π sinθ 0 n 0). n P b P 0 O r c r = c + s + tb (s < 0, t > 0) Exercise 1.3. Find the eqution of the plne tht contins the point c = (4, 5, 3) nd is perpendiculr to n = (2, 3, 1). Solution. Exercise 1.4. Find the eqution of the plne P tht contins the points P 1 = (1, 1, 1), P 2 = (2, 3, 4) nd P 3 = (0, 1, 2). Find lso the eqution of the line perpendiculr to P nd pssing through P 0 = (1, 2, 3). Find the coordintes of the point where the line intersects the plne. Solution. 7

10 Lines nd plnes Exmple 1.5. Find the eqution of the plne tht contins the lines x = 1 t, y = 2 + 3t, z = 2t nd x 1 = z 4 2, y = 5. Solution. ( 1, 3, 2) is vector prllel to the first line, nd (1, 0, 2) is vector prllel to the second line. So these vectors lie in the plne in question, nd thus norml vector to the plne is i j k ( 1, 3, 2) (1, 0, 2) = = (6, 4, 3) But (1, 2, 0) is point on the plne (tking t = 0 in the eqution of the first line), so the eqution of the plne is [ (x, y, z) (1, 2, 0) ] (6, 4, 3) = 0 6x + 4y 3z = 14. 8

11 VECTOR GEOMETRY Exmple 1.6. Find the eqution of the line L 1 through the points P 1 = (0, 3, 1) nd P 2 = (2, 1, 2), nd the eqution of the line L 2 through the point Q = ( 1, 1, 0) tht is norml to the plne P with eqution x + 2y 2z = 7. Find the point of intersection of the line L 1 with the plne P. Solution. P 1 P 2 = (2, 1, 2) (0, 3, 1) = (2, 2, 1) is prllel to L 1, nd the point P 1 = (0, 3, 1) lies on L 1, hence the eqution of the line is (x, y, z) = (0, 3, 1) + s(2, 2, 1) x = 2s, y = 3 2s, z = 1 + s. The vector (1, 2, 2) is norml to the given plne, nd hence prllel to the line L 2. Also Q = ( 1, 1, 0) is point on L 2, so the eqution of the line is (x, y, z) = ( 1, 1, 0) + t(1, 2, 2) x = 1 + t, y = 1 + 2t, z = 2t. When L 1 meets P, the point (x, y, z) = (2s, 3 2s, 1 + s) must lso stisfy the eqution of the plne, nd so x + 2y 2z = 2s + 2 (3 2s) 2 (1 + s) = 7 4s + 4 = 7 s = 3 4. Thus the point of intersection is ( 2 3 4, , 1 3 ) ( = 3 4 2, 9 2, 1 ) Resolving vectors It cn often be useful to write given vector s sum of two orthogonl vectors, one of which is in given direction. Let nd b be nonzero vectors; we wnt to write s multiple of b nd something perpendiculr to b. Tht is = tb + ( tb), nd lso b.( tb) = 0.b tb.b = 0 t =.b b.b. Note tht b/ b is unit vector (length= 1), nd tb = b t b b where b t = b.b b 2 = cosθ using.b = b cosθ, where θ is the ngle between nd b. Tht is, the pproprite multiple of the unit vector b/ b is given by stndrd trigonometry: tb θ tb b This ide llows us to clculte the distnce from point P 0 with position vector (x 0, y 0, z 0 ) to plne with eqution x + by + cz = d. Let r = (x 1, y 1, z 1 ) be ny point on the plne, so = (x 1 x 0, y 1 y 0, z 1 z 0 ) tkes us from P 0 to this point. But n = (, b, c) is the norml, 9

12 Resolving vectors nd.n = n cos θ, nd thus the required length is cos θ.n = = (x 1 x 0 ) + b(y 1 y 0 ) + c(z 1 z 0 ) n 2 + b 2 + c 2 = (x 1 + by 1 + cz 1 ) (x 0 + by 0 + cz 0 ) 2 + b 2 + c 2 = d (x 0 + by 0 + cz 0 ) 2 + b 2 + c 2 = r OP θ P 0 n = (, b, c) r Here we essentilly re resolving into something prllel to n (whose length is wht we relly wnt) nd vector in the plne itself. Exercise 1.7. Find the distnce from the point P 0 = (7, 0, 3) to the plne tht contins the points P 1 = (1, 2, 1), P 2 = ( 3, 4, 5) nd P 3 = (1, 0, 0). Solution. 10

13 VECTOR GEOMETRY Exercise 1.8. Find the distnce from P 0 = ( 1, 2, 1) to the line through P 1 = (1, 2, 3) in the direction of = (1, 1, 1). Find the point P 2 on this line tht is closest to P 0. Solution. 11

14 Exercises 1.4 Exercises 1. Consider the following vectors: = ( 1, 3), b = (2, 5), c = (4, 0, 1), d = (2, 1, 3), e = ( 4, 4, 4), f = (1, 5, 1). Clculte the following if possible: (i) b (ii) c (iii) ( + 3b) b (iv) (c + ) d (v) ( b)e (vi) d e (vii) d f (viii) e (c 4d) (ix) (c d) d (x) c (d d) 2. Compute the ngle between the following pirs of vectors using either the dot product, nd (if possible) the vector product: (i) = 2i 2k, b = 2j + 4k (ii) = (3, 1), b = ( 2, 2) (iii) = i + 4k, b = 2i + k (iv) = 3i + k, b = 4j + k 3. Find both the prmetric nd nonprmetric equtions of the given lines: (i) line through (1, 2, 3), prllel to (2, 1, 4) (ii) line through (2, 1, 3) nd (4, 0, 4) (iii) line through (1, 4, 1), prllel to the line x = 2 3t, y = 4, z = 6 + t (iv) line through (2, 0, 1), perpendiculr to the vectors (1, 0, 2) nd (0, 2, 1) 4. Determine if the following lines re prllel, skew, or intersect: (i) x = 4 + t, y = 2, z = 3 + 2t nd x = 2 + 2s, y = 2s, z = 1 + 4s (ii) x 3 = y 3 3 = z 4 1 nd x 2 1 = y 1 2 = z 6 2 (iii) x = 1 + 2t, y = 3, z = 1 4t nd x = 2 s, y = 2, z = 3 + 2s 5. Determine the eqution of the following plnes: (i) The plne contining the point (1, 3, 2) with norml vector (2, 1, 5) (ii) The plne contining the point ( 2, 1, 0) with norml vector ( 3, 0, 2) (iii) The plne contining the points (2, 0, 3), (1, 1, 0) nd (3, 2, 1) (iv) The plne contining the points ( 2, 2, 0), ( 2, 3, 2) nd (1, 2, 2) (v) The plne contining the point (2, 1, 1) nd prllel to the plne 3x y + 2z = 1 (vi) The plne contining the line (x, y, z) = (4 + t, 2, 3 + 2t) nd the line (x, y, z) = (2 + 2s, 2s, 1 + 4s) 6. Find the line of intersection of the following plnes: (i) 2x y z = 4 nd 3x 2y + z = 0 (ii) 3x + 4y = 1 nd x + y z = 3 7. Find the distnce from the point P 0 to the plne in ech of the following: (i) P 0 = (2, 0, 9), nd the plne 3x 2y + z = 4 (ii) P 0 = (1, 1, 1), nd the plne contining P 1 = (2, 1, 0) nd the line r = (t, 2t, 6 t) 8. In ech cse find the distnce from the point P 0 to the line, nd find the closest point on tht line to P 0 : (i) P 0 = ( 1, 4, 6), r = (3t, 1 t, 3 t) (ii) P 0 = (5, 12, 14), x 3 2 = y + 1 4, z = 5 12