Vectors and the Geometry of Space

Transcription

1 12 Vectors nd the Geometr of Spce Emples of the surfces nd solids we stud in this chpter re proloids (used for stellite dishes) nd hperoloids (used for cooling towers of nucler rectors). Mrk C. Burnett / Photo Reserchers, Inc Dvid Frier / Coris In this chpter we introduce vectors nd coordinte sstems for three-dimensionl spce. This will e the setting for our stud of the clculus of functions of two vriles in Chpter 14 ecuse the grph of such function is surfce in spce. In this chpter we will see tht vectors provide prticulrl simple descriptions of lines nd plnes in spce. 785 Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

2 786 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE 12.1 Three-Dimensionl Coordinte Sstems O FIGURE 1 Coordinte es FIGURE 2 Right-hnd rule To locte point in plne, two numers re necessr. We know tht n point in the plne cn e represented s n ordered pir, of rel numers, where is the -coordinte nd is the -coordinte. For this reson, plne is clled two-dimensionl. To locte point in spce, three numers re required. We represent n point in spce n ordered triple,, c of rel numers. In order to represent points in spce, we first choose fied point O (the origin) nd three directed lines through O tht re perpendiculr to ech other, clled the coordinte es nd leled the -is, -is, nd -is. Usull we think of the - nd -es s eing horiontl nd the -is s eing verticl, nd we drw the orienttion of the es s in Figure 1. The direction of the -is is determined the right-hnd rule s illustrted in Figure 2: If ou curl the fingers of our right hnd round the -is in the direction of 9 counterclockwise rottion from the positive -is to the positive -is, then our thum points in the positive direction of the -is. The three coordinte es determine the three coordinte plnes illustrted in Figure 3(). The -plne is the plne tht contins the - nd -es; the -plne contins the - nd -es; the -plne contins the - nd -es. These three coordinte plnes divide spce into eight prts, clled octnts. The first octnt, in the foreground, is determined the positive es. -plne O -plne -plne left wll O floor right wll FIGURE 3 () Coordinte plnes () FIGURE 4 O P(,, c) c Becuse mn people hve some difficult visuliing digrms of three-dimensionl figures, ou m find it helpful to do the following [see Figure 3()]. Look t n ottom corner of room nd cll the corner the origin. The wll on our left is in the -plne, the wll on our right is in the -plne, nd the floor is in the -plne. The -is runs long the intersection of the floor nd the left wll. The -is runs long the intersection of the floor nd the right wll. The -is runs up from the floor towrd the ceiling long the intersection of the two wlls. You re situted in the first octnt, nd ou cn now imgine seven other rooms situted in the other seven octnts (three on the sme floor nd four on the floor elow), ll connected the common corner point O. Now if P is n point in spce, let e the (directed) distnce from the -plne to P, let e the distnce from the -plne to P, nd let c e the distnce from the -plne to P. We represent the point P the ordered triple,, c of rel numers nd we cll,, nd cthe coordintes of P ; is the -coordinte, is the -coordinte, nd cis the -coordinte. Thus, to locte the point,, c, we cn strt t the origin O nd move units long the -is, then units prllel to the -is, nd then c units prllel to the -is s in Figure 4. Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

3 SECTION 12.1 THREE-DIMENSIONAL COORDINATE SYSTEMS 787 S(,, c) (,, ) (,, c) R(,, c) P(,, c) (,, ) Q(,, ) The point P,, c determines rectngulr o s in Figure 5. If we drop perpendiculr from P to the -plne, we get point Q with coordintes,, clled the projection of P onto the -plne. Similrl, R,, c nd S,, c re the projections of P onto the -plne nd -plne, respectivel. As numericl illustrtions, the points 4, 3, 5 nd 3, 2, 6 re plotted in Figure 6. 3 _4 3 _2 _5 (_4, 3, _5) _6 (3, _2, _6) FIGURE 5 FIGURE 6 The Crtesin product,,,, is the set of ll ordered triples of rel numers nd is denoted 3. We hve given one-to-one correspondence etween points P in spce nd ordered triples,, c in 3. It is clled threedimensionl rectngulr coordinte sstem. Notice tht, in terms of coordintes, the first octnt cn e descried s the set of points whose coordintes re ll positive. In two-dimensionl nltic geometr, the grph of n eqution involving nd is curve in 2. In three-dimensionl nltic geometr, n eqution in,, nd represents surfce in 3. v () EXAMPLE 1 3 Wht surfces in 3 re represented the following equtions? () 5 SOLUTION () The eqution 3 represents the set,, 3, which is the set of ll points in 3 whose -coordinte is 3. This is the horiontl plne tht is prllel to the -plne nd three units ove it s in Figure 7() FIGURE 7 () =3, plne in R# () =5, plne in R# (c) =5, line in R@ () The eqution 5 represents the set of ll points in 3 whose -coordinte is 5. This is the verticl plne tht is prllel to the -plne nd five units to the right of it s in Figure 7(). Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

4 788 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE NOTE When n eqution is given, we must understnd from the contet whether it represents curve in 2 or surfce in 3. In Emple 1, 5 represents plne in 3, ut of course 5 cn lso represent line in 2 if we re deling with two-dimensionl nltic geometr. See Figure 7() nd (c). In generl, if k is constnt, then k represents plne prllel to the -plne, k is plne prllel to the -plne, nd k is plne prllel to the -plne. In Figure 5, the fces of the rectngulr o re formed the three coordinte plnes (the -plne), (the -plne), nd (the -plne), nd the plnes,, nd c. EXAMPLE 2 () Which points,, stisf the equtions nd 3 () Wht does the eqution represent s surfce in 3? SOLUTION () Becuse 3, the points lie in the horiontl plne 3 from Emple 1(). Becuse 2 2 1, the points lie on the circle with rdius 1 nd center on the -is. See Figure 8. () Given tht 2 2 1, with no restrictions on, we see tht the point,, could lie on circle in n horiontl plne k. So the surfce in 3 consists of ll possile horiontl circles 2 2 1, k, nd is therefore the circulr clinder with rdius 1 whose is is the -is. See Figure 9. 3 FIGURE 8 The circle + =1, =3 FIGURE 9 The clinder + =1 v EXAMPLE 3 Descrie nd sketch the surfce in 3 represented the eqution. SOLUTION The eqution represents the set of ll points in 3 whose - nd -coordintes re equl, tht is,,,,. This is verticl plne tht intersects the -plne in the line,. The portion of this plne tht lies in the first octnt is sketched in Figure 1. The fmilir formul for the distnce etween two points in plne is esil etended to the following three-dimensionl formul. FIGURE 1 The plne = P 1P 2 Distnce Formul in Three Dimensions The distnce etween the points P 1 1, 1, 1 nd P 2 2, 2, 2 is P 1P 2 s Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

5 SECTION 12.1 THREE-DIMENSIONAL COORDINATE SYSTEMS 789 P (,, ) P (, fi, ) To see wh this formul is true, we construct rectngulr o s in Figure 11, where P 1 nd P 2 re opposite vertices nd the fces of the o re prllel to the coordinte plnes. If A 2, 1, 1 nd B 2, 2, 1 re the vertices of the o indicted in the figure, then P 1A 2 1 AB 2 1 BP B(, fi, ) A(,, ) Becuse tringles nd P 1 AB re oth right-ngled, two pplictions of the Pthgo - ren Theorem give P 1 BP 2 P1P2 2 P1B 2 BP2 2 FIGURE 11 nd P 1B 2 P 1A 2 AB 2 Comining these equtions, we get P 1P 2 2 P 1A 2 AB 2 BP Therefore EXAMPLE 4 P 1P 2 s The distnce from the point P2, 1, 7 to the point Q1, 3, 5 is PQ s s v EXAMPLE 5 Find n eqution of sphere with rdius r nd center Ch, k, l. r P (,, ) SOLUTION B definition, sphere is the set of ll points P,, whose distnce from C is r. (See Figure 12.) Thus P is on the sphere if nd onl if PC r. Squring oth sides, we hve PC 2 r 2 or h 2 k 2 l 2 r 2 C (h, k, l ) The result of Emple 5 is worth rememering. Eqution of Sphere An eqution of sphere with center Ch, k, l nd rdius r is h 2 k 2 l 2 r 2 FIGURE 12 In prticulr, if the center is the origin O, then n eqution of the sphere is r 2 EXAMPLE 6 Show tht is the eqution of sphere, nd find its center nd rdius. SOLUTION We cn rewrite the given eqution in the form of n eqution of sphere if we complete squres: Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

6 79 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE Compring this eqution with the stndrd form, we see tht it is the eqution of sphere with center 2, 3, 1 nd rdius s8 2s2. EXAMPLE 7 Wht region in 3 is represented the following inequlities? SOLUTION The inequlities cn e rewritten s 2 FIGURE s so the represent the points,, whose distnce from the origin is t lest 1 nd t most 2. But we re lso given tht, so the points lie on or elow the -plne. Thus the given inequlities represent the region tht lies etween (or on) the spheres nd nd eneth (or on) the -plne. It is sketched in Figure Eercises 1. Suppose ou strt t the origin, move long the -is distnce of 4 units in the positive direction, nd then move downwrd distnce of 3 units. Wht re the coordintes of our position? 2. Sketch the points, 5, 2, 4,, 1, 2, 4, 6, nd 1, 1, 2 on single set of coordinte es. 3. Which of the points A4,, 1, B3, 1, 5, nd C2, 4, 6 is closest to the -plne? Which point lies in the -plne? 4. Wht re the projections of the point (2, 3, 5) on the -, -, nd -plnes? Drw rectngulr o with the origin nd 2, 3, 5 s opposite vertices nd with its fces prllel to the coordinte plnes. Lel ll vertices of the o. Find the length of the digonl of the o. 5. Descrie nd sketch the surfce in represented the eqution () Wht does the eqution 4 represent in 2? Wht does it represent in 3? Illustrte with sketches. () Wht does the eqution 3 represent in 3? Wht does 5 represent? Wht does the pir of equtions 3, 5 represent? In other words, descrie the set of points,, such tht 3 nd 5. Illustrte with sketch Determine whether the points lie on stright line. () A2, 4, 2, B3, 7, 2, C1, 3, 3 () D, 5, 5, E1, 2, 4, F3, 4, 2 1. Find the distnce from 4, 2, 6 to ech of the following. () The -plne () The -plne (c) The -plne (d) The -is (e) The -is (f) The -is 11. Find n eqution of the sphere with center 3, 2, 5 nd rdius 4. Wht is the intersection of this sphere with the -plne? 12. Find n eqution of the sphere with center 2, 6, 4 nd rdius 5. Descrie its intersection with ech of the coordinte plnes. 13. Find n eqution of the sphere tht psses through the point 4, 3, 1nd hs center 3, 8, Find n eqution of the sphere tht psses through the origin nd whose center is 1, 2, Show tht the eqution represents sphere, nd find its center nd rdius Find the lengths of the sides of the tringle PQR. Is it right tringle? Is it n isosceles tringle? 7. P3, 2, 3, Q7,, 1, R1, 2, 1 8. P2, 1,, Q4, 1, 1, R4, 5, Homework Hints ville t stewrtclculus.com Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

7 SECTION 12.2 VECTORS () Prove tht the midpoint of the line segment from P 1 1, 1, 1 to P 2 2, 2, 2 is ,, () Find the lengths of the medins of the tringle with ver tices A1, 2, 3, B2,, 5, nd C4, 1, Find n eqution of sphere if one of its dimeters hs end - points 2, 1, 4 nd 4, 3, Find equtions of the spheres with center 2, 3, 6 tht touch () the -plne, () the -plne, (c) the -plne. 22. Find n eqution of the lrgest sphere with center (5, 4, 9) tht is contined in the first octnt Descrie in words the region of represented the equtions or inequlities , Write inequlities to descrie the region. 35. The region etween the -plne nd the verticl plne The solid clinder tht lies on or elow the plne 8 nd on or ove the disk in the -plne with center the origin nd rdius The region consisting of ll points etween (ut not on) the spheres of rdius r nd R centered t the origin, where r R 38. The solid upper hemisphere of the sphere of rdius 2 centered t the origin 39. The figure shows line L 1 in spce nd second line L 2, which is the projection of on the -plne. (In other words, L 1 the points on re directl eneth, or ove, the points on L 1.) () Find the coordintes of the point P on the line L 1. () Locte on the digrm the points A, B, nd C, where the line L 1 intersects the -plne, the -plne, nd the -plne, respectivel. 4. Consider the points P such tht the distnce from P to A1, 5, 3 is twice the distnce from P to B6, 2, 2. Show tht the set of ll such points is sphere, nd find its center nd rdius. 41. Find n eqution of the set of ll points equidistnt from the points A1, 5, 3 nd B6, 2, 2. Descrie the set. 42. Find the volume of the solid tht lies inside oth of the spheres nd L 2 L P Find the distnce etween the spheres nd Descrie nd sketch solid with the following properties. When illuminted rs prllel to the -is, its shdow is circulr disk. If the rs re prllel to the -is, its shdow is squre. If the rs re prllel to the -is, its shdow is n isosceles tringle. L 12.2 Vectors A v C B FIGURE 1 Equivlent vectors u D The term vector is used scientists to indicte quntit (such s displcement or velocit or force) tht hs oth mgnitude nd direction. A vector is often represented n rrow or directed line segment. The length of the rrow represents the mgnitude of the vector nd the rrow points in the direction of the vector. We denote vector printing letter in oldfce v or putting n rrow ove the letter v l. For instnce, suppose prticle moves long line segment from point A to point B. The corresponding displcement vector v, shown in Figure 1, hs initil point (the til) nd terminl point (the tip) nd we indicte this writing AB l A B v. Notice tht the vec- Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

8 792 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE tor u CD l hs the sme length nd the sme direction s v even though it is in different position. We s tht u nd v re equivlent (or equl) nd we write u v. The ero vector, denoted, hs length. It is the onl vector with no specific direction. Comining Vectors C B Suppose prticle moves from A to B, so its displcement vector is AB l. Then the prticle chnges direction nd moves from B to C, with displcement vector BC l s in Figure 2. The comined effect of these displcements is tht the prticle hs moved from. The resulting displcement vector AC l is clled the sum of AB l nd BC l A to C nd we write A AC l AB l BC l FIGURE 2 In generl, if we strt with vectors u nd v, we first move v so tht its til coincides with the tip of u nd define the sum of u nd v s follows. Definition of Vector Addition If u nd v re vectors positioned so the initil point of v is t the terminl point of u, then the sum u v is the vector from the initil point of u to the terminl point of v. The definition of vector ddition is illustrted in Figure 3. You cn see wh this defi nition is sometimes clled the Tringle Lw. u u+v v v v+u u+v v u u FIGURE 3 The Tringle Lw FIGURE 4 The Prllelogrm Lw FIGURE 5 In Figure 4 we strt with the sme vectors u nd v s in Figure 3 nd drw nother cop of v with the sme initil point s u. Completing the prllelogrm, we see tht u v v u. This lso gives nother w to construct the sum: If we plce u nd v so the strt t the sme point, then u v lies long the digonl of the prllelogrm with u nd v s sides. (This is clled the Prllelogrm Lw.) v EXAMPLE 1 Drw the sum of the vectors nd shown in Figure 5. SOLUTION First we trnslte nd plce its til t the tip of, eing creful to drw cop of tht hs the sme length nd direction. Then we drw the vector [see Figure 6()] strting t the initil point of nd ending t the terminl point of the cop of. Alterntivel, we could plce so it strts where strts nd construct the Prllelogrm Lw s in Figure 6(). TEC Visul 12.2 shows how the Tringle nd Prllelogrm Lws work for vrious vectors nd. + + FIGURE 6 () () Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

9 SECTION 12.2 VECTORS 793 It is possile to multipl vector rel numer c. (In this contet we cll the rel numer c sclr to distinguish it from vector.) For instnce, we wnt 2v to e the sme vector s v v, which hs the sme direction s vut is twice s long. In generl, we multipl vector sclr s follows. Definition of Sclr Multipliction If c is sclr nd v is vector, then the sclr multiple cv is the vector whose length is c times the length of v nd whose direction is the sme s v if c nd is opposite to v if c. If c or v, then cv. v 2v 1 2 v This definition is illustrted in Figure 7. We see tht rel numers work like scling fctors here; tht s wh we cll them sclrs. Notice tht two nonero vectors re prllel if the re sclr multiples of one nother. In prticulr, the vector v 1v hs the sme length s v ut points in the opposite direction. We cll it the negtive of v. B the difference u v of two vectors we men u v u v _v _1.5v FIGURE 7 Sclr multiples of v So we cn construct u v first drwing the negtive of v, v, nd then dding it to u the Prllelogrm Lw s in Figure 8(). Alterntivel, since v u v u, the vector u v, when dded to v, gives u. So we could construct u v s in Fig ure 8() mens of the Tringle Lw. v u u-v u-v FIGURE 8 Drwing u-v _v () v () u FIGURE 9 _2-2 FIGURE 1 EXAMPLE 2 If nd re the vectors shown in Figure 9, drw 2. SOLUTION We first drw the vector 2 pointing in the direction opposite to nd twice s long. We plce it with its til t the tip of nd then use the Tringle Lw to drw 2 s in Figure 1. Components For some purposes it s est to introduce coordinte sstem nd tret vectors lgericll. If we plce the initil point of vector t the origin of rectngulr coordinte sstem, then the terminl point of hs coordintes of the form 1, 2 or 1, 2, 3, depending on whether our coordinte sstem is two- or three-dimensionl (see Figure 11). (, ) O (,, ) O FIGURE 11 =k, l =k,, l Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

10 794 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE (4, 5) These coordintes re clled the components of nd we write (1, 3) P(3, 2) 1, 2 or 1, 2, 3 FIGURE 12 Representtions of the vector =k3, 2l O A(,, ) position vector of P FIGURE 13 Representtions of =k,, l P(,, ) B(+, +, + ) We use the nottion 1, 2 for the ordered pir tht refers to vector so s not to confuse it with the ordered pir 1, 2 tht refers to point in the plne. For instnce, the vectors shown in Figure 12 re ll equivlent to the vector OP l 3, 2 whose terminl point is P3, 2. Wht the hve in common is tht the terminl point is reched from the initil point displcement of three units to the right nd two upwrd. We cn think of ll these geometric vectors s representtions of the lgeric vector 3, 2. The prticulr representtion OP l from the origin to the point P3, 2 is clled the position vector of the point P. In three dimensions, the vector OP l 1, 2, 3 is the position vector of the point P. (See Figure 13.) Let s consider n other representtion AB l 1, 2, 3 of, where the initil point is A 1, 1, 1 nd the terminl point is B 2, 2, 2. Then we must hve 1 1 2, 1 2 2, nd nd so 1 2 1, 2 2 1, nd Thus we hve the following result. 1 Given the points nd, the vector with representtion AB l A 1, 1, 1 B 2, 2, 2 is 2 1, 2 1, 2 1 v EXAMPLE 3 Find the vector represented the directed line segment with initil point A2, 3, 4) nd terminl point B2, 1, 1. SOLUTION B 1, the vector corresponding to AB l is 2 2, 1 3, 1 4 4, 4, 3 The mgnitude or length of the vector v is the length of n of its representtions nd is denoted the smol v or v. B using the distnce formul to compute the length of segment OP, we otin the following formuls. ( +, + ) The length of the two-dimensionl vector 1, 2 is s The length of the three-dimensionl vector 1, 2, 3 is + s FIGURE 14 How do we dd vectors lgericll? Figure 14 shows tht if 1, 2 nd 1, 2, then the sum is 1 1, 2 2, t lest for the cse where the components re positive. In other words, to dd lgeric vectors we dd their components. Similrl, to sutrct vectors we sutrct components. From the similr tringles in Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

11 SECTION 12.2 VECTORS 795 Figure 15 we see tht the components of c re c 1 nd c 2. So to multipl vector sclr we multipl ech component tht sclr. FIGURE 15 c c c If 1, 2 nd 1, 2, then 1 1, , 2 2 c c 1, c 2 Similrl, for three-dimensionl vectors, 1, 2, 3 1, 2, 3 1 1, 2 2, 3 3 1, 2, 3 1, 2, 3 1 1, 2 2, 3 3 c 1, 2, 3 c 1, c 2, c 3 v EXAMPLE 4 If 4,, 3 nd 2, 1, 5, find nd the vectors,, 3, nd 2 5. SOLUTION s s25 5 4,, 3 2, 1, 5 4 2, 1, 3 5 2, 1, 8 4,, 3 2, 1, 5 4 2, 1, 3 5 6, 1, , 1, 5 32, 31, 35 6, 3, ,, 3 52, 1, 5 8,, 6 1, 5, 25 2, 5, 31 We denote V 2 the set of ll two-dimensionl vectors nd V 3 the set of ll threedimensionl vectors. More generll, we will lter need to consider the set V n of ll n-dimensionl vectors. An n-dimensionl vector is n ordered n-tuple: Vectors in n dimensions re used to list vrious quntities in n orgnied w. For instnce, the components of si-dimensionl vector p p 1, p 2, p 3, p 4, p 5, p 6 might represent the prices of si dif ferent ingredients required to mke prticulr product. Four-dimensionl vectors,,, t re used in reltivit theor, where the first three components specif position in spce nd the fourth represents time. 1, 2,..., n where 1, 2,..., n re rel numers tht re clled the components of. Addition nd sclr multipliction re defined in terms of components just s for the cses n 2 nd n 3. Properties of Vectors If,, nd c re vectors in nd c nd d re sclrs, then c c c c c 6. c d c d 7. cd cd 8. 1 V n Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

12 796 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE These eight properties of vectors cn e redil verified either geometricll or lgericll. For instnce, Propert 1 cn e seen from Figure 4 (it s equivlent to the Prllelogrm Lw) or s follows for the cse n 2: 1, 2 1, 2 1 1, 2 2 Q c 1 1, 2 2 1, 2 1, 2 (+)+c =+(+c) P + +c We cn see wh Propert 2 (the ssocitive lw) is true looking t Figure 16 nd ppling the Tringle Lw severl times: The vector PQ l is otined either first constructing nd then dding c or dding to the vector c. Three vectors in pl specil role. Let V 3 FIGURE 16 i 1,, j, 1, k,, 1 These vectors i, j, nd k re clled the stndrd sis vectors. The hve length 1 nd point in the directions of the positive -, -, nd -es. Similrl, in two dimensions we define i 1, nd j, 1. (See Figure 17.) (, 1) FIGURE 17 Stndrd sis vectors in V nd V j () i (1, ) k i () j (, ) If 1, 2, 3, then we cn write j 1, 2, 3 1,,, 2,,, 3 i 1 1,, 2, 1, 3,, 1 () = i+ j 2 1 i 2 j 3 k Thus n vector in V 3 cn e epressed in terms of i, j, nd k. For instnce, i (,, ) k Similrl, in two dimensions, we cn write 1, 2, 6 i 2j 6k j () = i+ j+ k FIGURE , 2 1 i 2 j See Figure 18 for the geometric interprettion of Equtions 3 nd 2 nd compre with Figure 17. Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

13 SECTION 12.2 VECTORS 797 EXAMPLE 5 If i 2j 3k nd 4i 7 k, epress the vector 2 3 in terms of i, j, nd k. SOLUTION Using Properties 1, 2, 5, 6, nd 7 of vectors, we hve Gis Josih Willrd Gis ( ), professor of mthemticl phsics t Yle College, pulished the first ook on vectors, Vector Anlsis, in More complicted ojects, clled quternions, hd erlier een invented Hmilton s mthemticl tools for descriing spce, ut the weren t es for scientists to use. Quternions hve sclr prt nd vector prt. Gi s ide ws to use the vector prt seprtel. Mwell nd Heviside hd similr ides, ut Gi s pproch hs proved to e the most convenient w to stud spce i 2j 3k 34i 7k 2i 4j 6k 12i 21k 14i 4j 15k A unit vector is vector whose length is 1. For instnce, i, j, nd k re ll unit vectors. In generl, if, then the unit vector tht hs the sme direction s is 4 u 1 In order to verif this, we let c 1. Then u c nd c is positive sclr, so u hs the sme direction s. Also u c c 1 1 EXAMPLE 6 Find the unit vector in the direction of the vector 2i j 2k. SOLUTION The given vector hs length 2i j 2k s s9 3 so, Eqution 4, the unit vector with the sme direction is 1 3 2i j 2k 2 3 i 1 3 j 2 3 k 5 32 T T 1 FIGURE T T 5 32 w FIGURE 2 Applictions Vectors re useful in mn spects of phsics nd engineering. In Chpter 13 we will see how the descrie the velocit nd ccelertion of ojects moving in spce. Here we look t forces. A force is represented vector ecuse it hs oth mgnitude (mesured in pounds or newtons) nd direction. If severl forces re cting on n oject, the resultnt force eperienced the oject is the vector sum of these forces. EXAMPLE 7 A 1-l weight hngs from two wires s shown in Figure 19. Find the tensions (forces) T 1 nd T 2 in oth wires nd the mgnitudes of the tensions. SOLUTION We first epress T 1 nd T 2 in terms of their horiontl nd verticl components. From Figure 2 we see tht 5 6 The resultnt T 1 T 2 of the tensions counterlnces the weight w nd so we must hve Thus T 1 T 1 cos 5 i T 1 sin 5 j T 2 T 2 cos 32 i T 2 sin 32 j T 1 T 2 w 1 j ( T 1 cos 5 T 2 cos 32 ) i ( T 1 sin 5 T 2 sin 32 ) j 1 j Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

14 798 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE Equting components, we get Solving the first of these equtions for So the mgnitudes of the tensions re T 1 cos 5 T 2 cos 32 T 1 sin 5 T 2 sin 32 1 T 2 T 1 sin 5 T 1 cos 5 sin 32 1 cos 32 nd sustituting into the second, we get T l sin 5 tn 32 cos 5 nd T 2 T 1 cos l cos 32 Sustituting these vlues in 5 nd 6, we otin the tension vectors T i 65.6 j T i 34.4 j 12.2 Eercises 1. Are the following quntities vectors or sclrs? Eplin. () The cost of theter ticket () The current in river (c) The initil flight pth from Houston to Dlls (d) The popultion of the world 2. Wht is the reltionship etween the point (4, 7) nd the vector 4, 7? Illustrte with sketch. 3. Nme ll the equl vectors in the prllelogrm shown. A E B 5. Cop the vectors in the figure nd use them to drw the following vectors. () u v () u w (c) v w (d) u v (e) v u w (f) u w v u v w 6. Cop the vectors in the figure nd use them to drw the following vectors. () () 1 (c) 2 (d) 3 (e) 2 (f) 2 D C 4. Write ech comintion of vectors s single vector. () AB l BC l () CD l DB l (c) DB l AB l (d) DC l CA l AB l A B D C 7. In the figure, the tip of c nd the til of d re oth the midpoint of QR. Epress c nd d in terms of nd. Q c P d R 1. Homework Hints ville t stewrtclculus.com Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

15 SECTION 12.2 VECTORS If the vectors in the figure stisf nd u v w, wht is? 9 14 Find vector with representtion given the directed line segment AB l. Drw AB l nd the equivlent representtion strting t the origin. 9. A1, 1, B3, 2 1. A4, 1, B1, A1, 3, B2, A2, 1, 13. A, 3, 1, B2, 3, A4,, 2, Find the sum of the given vectors nd illustrte geometricll , 4, 6, , 1, 17. 3,, 1,, 8, 18. 1, 3, 2, Find, 2 3,, nd , 12, 2. 4 i j, 21. i 2 j 3k, i 4 j 4 k, Find unit vector tht hs the sme direction s the given vector. 25. w u v 3, 6 i 2 j u v Find vector tht hs the sme direction s 2, 4, 2 ut hs length Wht is the ngle etween the given vector nd the positive direction of the -is? 29. If v lies in the first qudrnt nd mkes n ngle 3 with the positive -is nd, find v in component form. 3. If child pulls sled through the snow on level pth with force of 5 N eerted t n ngle of 38 ove the horiontl, find the horiontl nd verticl components of the force. 31. A qurterck throws footll with ngle of elevtion 4 nd speed 6 fts. Find the horiontl nd verticl components of the velocit vector. w 2 i j 5k 2 j k 23. 3i 7j 24. 4, 2, 4 8i j 4k 27. i s3 j 28. 8i 6j v 4 B, 6 B4, 2, 1 1, 5,, Find the mgnitude of the resultnt force nd the ngle it mkes with the positive -is l l 3 N 2 N 34. The mgnitude of velocit vector is clled speed. Suppose tht wind is lowing from the direction N45W t speed of 5 kmh. (This mens tht the direction from which the wind lows is 45 west of the northerl direction.) A pilot is steering plne in the direction N6E t n irspeed (speed in still ir) of 25 kmh. The true course, or trck, of the plne is the direction of the resul tnt of the velocit vectors of the plne nd the wind. The ground speed of the plne is the mgnitude of the resultnt. Find the true course nd the ground speed of the plne. 35. A womn wlks due west on the deck of ship t 3 mih. The ship is moving north t speed of 22 mih. Find the speed nd direction of the womn reltive to the surfce of the wter. 36. Ropes 3 m nd 5 m in length re fstened to holid decortion tht is suspended over town squre. The decortion hs mss of 5 kg. The ropes, fstened t different heights, mke ngles of 52 nd 4 with the horiontl. Find the tension in ech wire nd the mgnitude of ech tension m 5 m 37. A clothesline is tied etween two poles, 8 m prt. The line is quite tut nd hs negligile sg. When wet shirt with mss of.8 kg is hung t the middle of the line, the mid point is pulled down 8 cm. Find the tension in ech hlf of the clothesline. 38. The tension T t ech end of the chin hs mgnitude 25 N (see the figure). Wht is the weight of the chin? 39. A otmn wnts to cross cnl tht is 3 km wide nd wnts to lnd t point 2 km upstrem from his strting point. The current in the cnl flows t 3.5 kmh nd the speed of his ot is 13 kmh. () In wht direction should he steer? () How long will the trip tke? Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

16 8 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE 4. Three forces ct on n oject. Two of the forces re t n ngle of 1 to ech other nd hve mgnitudes 25 N nd 12 N. The third is perpendiculr to the plne of these two forces nd hs mgnitude 4 N. Clculte the mgnitude of the force tht would ectl counterlnce these three forces. 41. Find the unit vectors tht re prllel to the tngent line to the prol 2 t the point 2, () Find the unit vectors tht re prllel to the tngent line to the curve 2 sin t the point 6, 1. () Find the unit vectors tht re perpendiculr to the tngent line. (c) Sketch the curve 2 sin nd the vectors in prts () nd (), ll strting t 6, If,, nd re the vertices of tringle, find AB l A B BC l C CA l. 44. Let C e the point on the line segment tht is twice s fr from s it is from. If OA l AB B A, OB l, nd c OC l, show tht c () Drw the vectors 3, 2, 2, 1, nd c 7, 1. () Show, mens of sketch, tht there re sclrs s nd t such tht c s t. (c) Use the sketch to estimte the vlues of s nd t. (d) Find the ect vlues of s nd t. 46. Suppose tht nd re nonero vectors tht re not prllel nd c is n vector in the plne determined nd. Give geometric rgument to show tht c cn e written s c s t for suitle sclrs s nd t. Then give n rgument using components. 47. If r,, nd r,,, descrie the set of ll points,, such tht. r r If r,, r 1 1, 1, nd r 2 2, 2, descrie the set of ll points, such tht r r1 r r2 k, where. k r1 r2 49. Figure 16 gives geometric demonstrtion of Propert 2 of vectors. Use components to give n lgeric proof of this fct for the cse n Prove Propert 5 of vectors lgericll for the cse n 3. Then use similr tringles to give geometric proof. 51. Use vectors to prove tht the line joining the midpoints of two sides of tringle is prllel to the third side nd hlf its length. 52. Suppose the three coordinte plnes re ll mirrored nd light r given the vector 1, 2, 3 first strikes the -plne, s shown in the figure. Use the fct tht the ngle of incidence equls the ngle of reflection to show tht the direction of the reflected r is given 1, 2, 3. Deduce tht, fter eing reflected ll three mutull perpendiculr mirrors, the resulting r is prllel to the initil r. (Americn spce scientists used this principle, together with lser ems nd n rr of corner mirrors on the moon, to clculte ver precisel the distnce from the erth to the moon.) 12.3 The Dot Product So fr we hve dded two vectors nd multiplied vector sclr. The question rises: Is it possile to multipl two vectors so tht their product is useful quntit? One such product is the dot product, whose definition follows. Another is the cross product, which is discussed in the net section. 1 Definition If 1, 2, 3 nd 1, 2, 3, then the dot product of nd is the numer given Thus, to find the dot product of nd, we multipl corresponding components nd dd. The result is not vector. It is rel numer, tht is, sclr. For this reson, the dot product is sometimes clled the sclr product (or inner product). Although Definition 1 is given for three-dimensionl vectors, the dot product of two-dimensionl vectors is defined in similr fshion: 1, 2 1, Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

17 v EXAMPLE 1 SECTION 12.3 THE DOT PRODUCT 81 2, 4 3, , 7, 4 6, 2, ( 1 2 ) 6 i 2j 3k 2j k The dot product oes mn of the lws tht hold for ordinr products of rel numers. These re stted in the following theorem. 2 Properties of the Dot Product If,, nd c re vectors in V 3 nd c is sclr, then c c 4. c c c 5. These properties re esil proved using Definition 1. For instnce, here re the proofs of Properties 1 nd 3: c 1, 2, 3 1 c 1, 2 c 2, 3 c c c c c c c c 1 2 c 2 3 c 3 c The proofs of the remining properties re left s eercises. B - A FIGURE 1 The dot product cn e given geometric interprettion in terms of the ngle etween nd, which is defined to e the ngle etween the representtions of nd tht strt t the origin, where. In other words, is the ngle etween the line segments OA l nd OB l in Figure 1. Note tht if nd re prllel vectors, then or. The formul in the following theorem is used phsicists s the definition of the dot product. 3 Theorem If is the ngle etween the vectors nd, then cos PROOF If we ppl the Lw of Cosines to tringle OAB in Figure 1, we get 4 AB 2 OA 2 OB 2 2 OA OB cos (Oserve tht the Lw of Cosines still pplies in the limiting cses when or, or or.) But OA, OB, nd AB, so Eqution 4 ecomes cos Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

18 82 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE Using Properties 1, 2, nd 3 of the dot product, we cn rewrite the left side of this eqution s follows: Therefore Eqution 5 gives Thus or cos 2 2 cos cos EXAMPLE 2 If the vectors nd hve lengths 4 nd 6, nd the ngle etween them is 3, find. SOLUTION Using Theorem 3, we hve cos The formul in Theorem 3 lso enles us to find the ngle etween two vectors. 6 Corollr If is the ngle etween the nonero vectors nd, then cos v EXAMPLE 3 Find the ngle etween the vectors 2, 2, 1 nd 5, 3, 2. SOLUTION Since nd since s we hve, from Corollr 6, So the ngle etween nd is nd cos 2 3s38 s s38 cos 1 2 3s or 84 Two nonero vectors nd re clled perpendiculr or orthogonl if the ngle etween them is 2. Then Theorem 3 gives cos2 Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

19 SECTION 12.3 THE DOT PRODUCT 83 nd conversel if, then cos, so 2. The ero vector is considered to e perpendiculr to ll vectors. Therefore we hve the following method for determining whether two vectors re orthogonl. 7 Two vectors nd re orthogonl if nd onl if. EXAMPLE 4 Show tht 2i 2j k is perpendiculr to 5i 4j 2k. SOLUTION Since 2i 2j k 5i 4j 2k FIGURE 2 > cute = =π/2 < otuse these vectors re perpendiculr 7. Becuse cos if 2 nd cos if 2, we see tht is positive for 2 nd negtive for 2. We cn think of s mesuring the etent to which nd point in the sme direction. The dot product is positive if nd point in the sme generl direction, if the re perpendiculr, nd negtive if the point in generll opposite directions (see Figure 2). In the etreme cse where nd point in ectl the sme direction, we hve, so cos 1 nd TEC Visul 12.3A shows n nimtion of Figure 2. If nd point in ectl opposite directions, then nd so cos 1 nd. Direction Angles nd Direction Cosines FIGURE 3 ç å The direction ngles of nonero vector re the ngles,, nd (in the intervl, tht mkes with the positive -, -, nd -es. (See Figure 3.) The cosines of these direction ngles, cos, cos, nd cos, re clled the direction cosines of the vector. Using Corollr 6 with replced i, we otin 8 cos (This cn lso e seen directl from Figure 3.) Similrl, we lso hve i i 1 9 cos 2 cos 3 B squring the epressions in Equtions 8 nd 9 nd dding, we see tht 1 cos 2 cos 2 cos 2 1 We cn lso use Equtions 8 nd 9 to write 1, 2, 3 cos, cos, cos cos, cos, cos Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

20 84 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE Therefore 11 which ss tht the direction cosines of re the components of the unit vector in the direction of. EXAMPLE 5 SOLUTION Since nd so 1 cos, cos, cos Find the direction ngles of the vector 1, 2, 3. s s14 cos 1 s14 cos 2 s14, Equtions 8 nd 9 give cos 3 s14 cos 1 1 s1474 cos 1 2 s1458 cos 1 3 s1437 TEC Visul 12.3B shows how Figure 4 chnges when we vr nd. P R S R S proj P proj Q FIGURE 4 Vector projections Q Projections Figure 4 shows representtions PQ l nd PR l of two vectors nd with the sme initil point P. If S is the foot of the perpendiculr from to the line contining PQ l, then the vector with representtion PS l R is clled the vector projection of onto nd is denoted proj. (You cn think of it s shdow of ). The sclr projection of onto (lso clled the component of long ) is defined to e the signed mgnitude of the vector projection, which is the numer cos, where is the ngle etween nd. (See Figure 5.) This is denoted comp. Oserve tht it is negtive if 2. The eqution cos ( cos ) shows tht the dot product of nd cn e interpreted s the length of times the sclr projection of onto. Since cos the component of long cn e computed tking the dot product of with the unit vector in the direction of. We summrie these ides s follows. R Sclr projection of onto : comp P Q S cos = comp Vector projection of onto : proj 2 FIGURE 5 Sclr projection Notice tht the vector projection is the sclr projection times the unit vector in the direction of. Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

21 SECTION 12.3 THE DOT PRODUCT 85 v EXAMPLE 6 Find the sclr projection nd vector projection of 1, 1, 2 onto 2, 3, 1. s s14 SOLUTION Since, the sclr projection of onto is comp s14 3 s14 The vector projection is this sclr projection times the unit vector in the direction of : proj 3 s , , 3 F R One use of projections occurs in phsics in clculting work. In Section 6.4 we defined the work done constnt force Fin moving n oject through distnce ds W Fd, ut this pplies onl when the force is directed long the line of motion of the oject. Suppose, however, tht the constnt force is vector F PR l pointing in some other direction, s in Figure 6. If the force moves the oject from to, then the displcement vector is PQ l P Q D. The work done this force is defined to e the product of the component of the force long D nd the distnce moved: P FIGURE 6 D S Q But then, from Theorem 3, we hve 12 W ( F cos ) D W F D cos F D Thus the work done constnt force F is the dot product F D, where D is the displcement vector. 35 EXAMPLE 7 A wgon is pulled distnce of 1 m long horiontl pth constnt force of 7 N. The hndle of the wgon is held t n ngle of 35 ove the horiontl. Find the work done the force. F 35 D SOLUTION If F nd D re the force nd displcement vectors, s pictured in Figure 7, then the work done is W F D F D cos 35 FIGURE 7 71 cos Nm 5734 J EXAMPLE 8 A force is given vector F 3i 4j 5k nd moves prticle from the point P2, 1, to the point Q4, 6, 2. Find the work done. SOLUTION The displcement vector is D PQ l 2, 5, 2, so Eqution 12, the work done is W F D 3, 4, 5 2, 5, If the unit of length is meters nd the mgnitude of the force is mesured in newtons, then the work done is 36 J. Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

22 86 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE 12.3 Eercises 1. Which of the following epressions re meningful? Which re meningless? Eplin. () c () c (c) c (d) c (e) c (f) 2 1 Find. 2. 2, 3,.7, , 1 3, 5, , 2, 3, 2, 5, , 1, 1 4, 6, 3, 8 6. p, p, 2p, 2q, q, q 7. 2i j, i j k 8. 3i 2j k, 4i 5k c 9.,, the ngle etween nd is 23 1., s6, the ngle etween nd is If u is unit vector, find u v nd u w u Find, correct to the nerest degree, the three ngles of the tringle with the given vertices. 21. P2,, Q, 3, R3, A1,, 1, B3, 2,, C1, 3, Determine whether the given vectors re orthogonl, prllel, or neither. 23. () 5, 3, 7, 6, 8, 2 () 4, 6, 3, 2 (c) i 2 j 5k, 3i 4 j k (d) 2i 6 j 4k, 3i 9 j 6k 24. () u 3, 9, 6, () u i j 2k, v 4, 12, 8 v 2i j k (c) u,, c, v,, 25. Use vectors to decide whether the tringle with vertices P1, 3, 2, Q2,, 4, nd R6, 2, 5 is right-ngled. 26. Find the vlues of such tht the ngle etween the vectors 2, 1, 1, nd 1,, is Find unit vector tht is orthogonl to oth i jnd i k. u w v w v 28. Find two unit vectors tht mke n ngle of 6 with v 3, Find the cute ngle etween the lines , , () Show tht i j j k k i. () Show tht i i j j k k A street vendor sells hmurgers, hot dogs, nd c soft drinks on given d. He chrges $2 for hmurger, $1.5 for hot dog, nd $1 for soft drink. If A,, c nd P 2, 1.5, 1, wht is the mening of the dot product A P? 15 2 Find the ngle etween the vectors. (First find n ect epression nd then pproimte to the nerest degree.) 15. 4, 3, 16. 2, 5, 17. 3, 1, 5, 18. 4,, 2, 19. 4i 3j k, 2. i 2j 2k, 2, 1 5, 12 2, 4, 3 2, 1, 2i k 4i 3k Find the cute ngles etween the curves t their points of intersection. (The ngle etween two curves is the ngle etween their tngent lines t the point of intersection.) 31. 2, sin, cos, Find the direction cosines nd direction ngles of the vector. (Give the direction ngles correct to the nerest degree.) 33. 2, 1, , 3, i 2j 3k c, c, c, where c 1 2 i j k 38. If vector hs direction ngles 4 nd 3, find the third direction ngle. 1. Homework Hints ville t stewrtclculus.com Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

23 SECTION 12.3 THE DOT PRODUCT Find the sclr nd vector projections of onto , 12, 4. 1, 4, 41. 3, 6, 2, 42. 2, 3, 6, 43. 2i j 4k, 44. i j k, 4, 6 2, 3 1, 2, 3 5, 1, 4 j 1 2 k i j k 45. Show tht the vector orth proj is orthogonl to. (It is clled n orthogonl projection of.) 46. For the vectors in Eercise 4, find orth nd illustrte drwing the vectors,, proj, nd orth. 47. If 3,, 1, find vector such tht comp Suppose tht nd re nonero vectors. () Under wht circumstnces is comp comp? () Under wht circumstnces is proj proj? 49. Find the work done force F 8 i 6 j 9k tht moves n oject from the point, 1, 8 to the point 6, 12, 2 long stright line. The distnce is mesured in meters nd the force in newtons. 5. A tow truck drgs stlled cr long rod. The chin mkes n ngle of 3 with the rod nd the tension in the chin is 15 N. How much work is done the truck in pulling the cr 1 km? 51. A sled is pulled long level pth through snow rope. A 3-l force cting t n ngle of 4 ove the horiontl moves the sled 8 ft. Find the work done the force. 52. A ot sils south with the help of wind lowing in the direction S36E with mgnitude 4 l. Find the work done the wind s the ot moves 12 ft. 53. Use sclr projection to show tht the distnce from point P 1 1, 1 to the line c is 1 1 c s 2 2 Use this formul to find the distnce from the point 2, 3 to the line If r,,, 1, 2, 3, nd 1, 2, 3, show tht the vector eqution r r represents sphere, nd find its center nd rdius. 55. Find the ngle etween digonl of cue nd one of its edges. 56. Find the ngle etween digonl of cue nd digonl of one of its fces. 57. A molecule of methne, CH 4, is structured with the four hdrogen toms t the vertices of regulr tetrhedron nd the cron tom t the centroid. The ond ngle is the ngle formed the H C H comintion; it is the ngle etween the lines tht join the cron tom to two of the hdrogen toms. Show tht the ond ngle is out [ Hint: Tke the vertices of the tetrhedron to e the points 1,,,, 1,,,, 1, nd 1, 1, 1, s shown in the figure. Then the centroid is ( 1 2, 1 2, 1 2 )]. H 58. If c, where,, nd c re ll nonero vectors, show tht c isects the ngle etween nd. 59. Prove Properties 2, 4, nd 5 of the dot product (Theorem 2). 6. Suppose tht ll sides of qudrilterl re equl in length nd opposite sides re prllel. Use vector methods to show tht the digonls re perpendiculr. 61. Use Theorem 3 to prove the Cuch-Schwr Inequlit: 62. The Tringle Inequlit for vectors is () Give geometric interprettion of the Tringle Inequlit. () Use the Cuch-Schwr Inequlit from Eercise 61 to prove the Tringle Inequlit. [Hint: Use the fct tht 2 nd use Propert 3 of the dot product.] 63. The Prllelogrm Lw sttes tht H C () Give geometric interprettion of the Prllelogrm Lw. () Prove the Prllelogrm Lw. (See the hint in Eercise 62.) 64. Show tht if u v nd u v re orthogonl, then the vectors und v must hve the sme length. H H Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

24 88 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE 12.4 The Cross Product Given two nonero vectors 1, 2, 3 nd 1, 2, 3, it is ver useful to e le to find nonero vector c tht is perpendiculr to oth nd, s we will see in the net section nd in Chpters 13 nd 14. If c c 1, c 2, c 3 is such vector, then c nd c nd so c 1 2 c 2 3 c 3 1 c 1 2 c 2 3 c 3 To eliminte we multipl 1 nd 2 nd sutrct: c c c 2 Eqution 3 hs the form pc 1 qc 2, for which n ovious solution is c 1 q nd c 2 p. So solution of 3 is c c Sustituting these vlues into 1 nd 2, we then get c This mens tht vector perpendiculr to oth nd is c 1, c 2, c , , Hmilton The cross product ws invented the Irish mthemticin Sir Willim Rown Hmilton ( ), who hd creted precursor of vectors, clled quternions. When he ws five ers old Hmilton could red Ltin, Greek, nd Herew. At ge eight he dded French nd Itlin nd when ten he could red Aric nd Snskrit. At the ge of 21, while still n undergrdute t Trinit College in Dulin, Hmilton ws ppointed Professor of Astronom t the universit nd Rol Astronomer of Irelnd! The resulting vector is clled the cross product of nd nd is denoted. 4 Definition If 1, 2, 3 nd 1, 2, 3, then the cross product of nd is the vector , , Notice tht the cross product of two vectors nd, unlike the dot product, is vector. For this reson it is lso clled the vector product. Note tht is defined onl when nd re three-dimensionl vectors. In order to mke Definition 4 esier to rememer, we use the nottion of determinnts. A determinnt of order 2 is defined c d d c For emple, A determinnt of order 3 cn e defined in terms of second-order determinnts s follows: c c c c 2 c c 1 c c 1 c 2 Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

25 SECTION 12.4 THE CROSS PRODUCT 89 Oserve tht ech term on the right side of Eqution 5 involves numer i in the first row of the determinnt, nd i is multiplied the second-order determinnt otined from the left side deleting the row nd column in which i ppers. Notice lso the minus sign in the second term. For emple, If we now rewrite Definition 4 using second-order determinnts nd the stndrd sis vectors i, j, nd k, we see tht the cross product of the vectors 1 i 2 j 3 k nd 1 i 2 j 3 k is i j k In view of the similrit etween Equtions 5 nd 6, we often write i j k Although the first row of the smolic determinnt in Eqution 7 consists of vectors, if we epnd it s if it were n ordinr determinnt using the rule in Eqution 5, we otin Eqution 6. The smolic formul in Eqution 7 is prol the esiest w of rememering nd computing cross products. v EXAMPLE 1 If 1, 3, 4 nd 2, 7, 5, then i j k i 2 5 j 2 7 k i 5 8 j 7 6 k 43i 13j k v EXAMPLE 2 Show tht for n vector in. SOLUTION If 1, 2, 3, then i j k i j k i j k V 3 Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

26 Since vector is completel determined its mgnitude nd direction, we cn now s tht is the vector tht is perpendiculr to oth nd, whose orienttion is deter- 81 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE We constructed the cross product so tht it would e perpendiculr to oth nd. This is one of the most importnt properties of cross product, so let s emphsie nd verif it in the following theorem nd give forml proof. 8 Theorem The vector is orthogonl to oth nd. n PROOF In order to show tht is orthogonl to, we compute their dot product s follows: A similr computtion shows tht. Therefore is orthogonl to oth nd. If nd re represented directed line segments with the sme initil point (s in Figure 1), then Theorem 8 ss tht the cross product points in direction perpendiculr to the plne through nd. It turns out tht the direction of is given the right-hnd rule: If the fingers of our right hnd curl in the direction of rottion (through n ngle less thn 18) from to, then our thum points in the direction of. Now tht we know the direction of the vector, the remining thing we need to complete its geometric description is its length. This is given the following theorem FIGURE 1 The right-hnd rule gives the direction of. 9 Theorem If is the ngle etween nd (so ), then sin TEC Visul 12.4 shows how chnges s chnges. PROOF From the definitions of the cross product nd length of vector, we hve cos cos sin 2 ( Theorem ) Geometric chrcterition of Tking squre roots nd oserving tht ssin 2 sin ecuse sin when, we hve sin Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

27 mined the right-hnd rule, nd whose length is phsicists define. SECTION 12.4 THE CROSS PRODUCT 811 sin. In fct, tht is ectl how 1 Corollr Two nonero vectors nd re prllel if nd onl if sin FIGURE 2 PROOF Two nonero vectors nd re prllel if nd onl if or. In either cse sin, so nd therefore. The geometric interprettion of Theorem 9 cn e seen looking t Figure 2. If nd re represented directed line segments with the sme initil point, then the determine prllelogrm with se, ltitude sin, nd re A ( sin ) Thus we hve the following w of interpreting the mgnitude of cross product. The length of the cross product is equl to the re of the prllelogrm determined nd. EXAMPLE 3 Find vector perpendiculr to the plne tht psses through the points P1, 4, 6, Q2, 5, 1, nd R1, 1, 1. SOLUTION The vector PQ l PR l is perpendiculr to oth PQ l nd PR l nd is therefore perpendiculr to the plne through P, Q, nd R. We know from (12.2.1) tht PQ l 2 1 i 5 4 j 1 6 k 3i j 7k PR l 1 1 i 1 4 j 1 6 k 5 j 5k We compute the cross product of these vectors: PQ l PR l i j k i 15 j 15 k 4 i 15 j 15k So the vector 4, 15, 15 is perpendiculr to the given plne. An nonero sclr multiple of this vector, such s 8, 3, 3, is lso perpendiculr to the plne. EXAMPLE 4 Find the re of the tringle with vertices P1, 4, 6, Q2, 5, 1, nd R1, 1, 1. SOLUTION In Emple 3 we computed tht PQ l PR l 4, 15, 15. The re of the prllelogrm with djcent sides PQ nd PR is the length of this cross product: PQ l PR l s s82 The re A of the tringle PQR is hlf the re of this prllelogrm, tht is, 2 s82. 5 Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

28 812 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE If we ppl Theorems 8 nd 9 to the stndrd sis vectors i, j, nd k using 2, we otin i j k j k i k i j j i k k j i i k j Oserve tht i j j i Thus the cross product is not commuttive. Also i i j i k j wheres i i j j So the ssocitive lw for multipliction does not usull hold; tht is, in generl, c c However, some of the usul lws of lger do hold for cross products. The following theorem summries the properties of vector products. 11 Theorem If,, nd c re vectors nd c is sclr, then (c) c( ) (c) 3. ( c) c 4. ( ) c c c c c c c c These properties cn e proved writing the vectors in terms of their components nd using the definition of cross product. We give the proof of Propert 5 nd leve the remining proofs s eercises. PROOF OF PROPERTY 5 If 1, 2, 3, 1, 2, 3, nd c c 1, c 2, c 3, then 12 c 1 2 c 3 3 c c 1 1 c c 2 2 c c c c c c c c c c 3 c Triple Products The product c tht occurs in Propert 5 is clled the sclr triple product of the vectors,, nd c. Notice from Eqution 12 tht we cn write the sclr triple product s determinnt: c c 1 c 2 c 3 Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

29 c h c FIGURE 3 SECTION 12.4 THE CROSS PRODUCT 813 The geometric significnce of the sclr triple product cn e seen considering the prllelepiped determined the vectors,, nd c. (See Figure 3.) The re of the se prllelogrm is A c. If is the ngle etween nd c, then the height h of the prllelepiped is h cos. (We must use cos insted of cos in cse 2.) Therefore the volume of the prllelepiped is V Ah c cos c Thus we hve proved the following formul. 14 The volume of the prllelepiped determined the vectors,, nd c is the mgnitude of their sclr triple product: V c If we use the formul in 14 nd discover tht the volume of the prllelepiped determined,, nd c is, then the vectors must lie in the sme plne; tht is, the re coplnr. v EXAMPLE 5 Use the sclr triple product to show tht the vectors 1, 4, 7, 2, 1, 4, nd c, 9, 18 re coplnr. SOLUTION We use Eqution 13 to compute their sclr triple product: c Therefore, 14, the volume of the prllelepiped determined,, nd c is. This mens tht,, nd c re coplnr. The product c tht occurs in Propert 6 is clled the vector triple product of,, nd c. Propert 6 will e used to derive Kepler s First Lw of plnetr motion in Chpter 13. Its proof is left s Eercise 5. Torque The ide of cross product occurs often in phsics. In prticulr, we consider force F cting on rigid od t point given position vector r. (For instnce, if we tighten olt ppling force to wrench s in Figure 4, we produce turning effect.) The torque (reltive to the origin) is defined to e the cross product of the position nd force vectors r r F FIGURE 4 F nd mesures the tendenc of the od to rotte out the origin. The direction of the torque vector indictes the is of rottion. According to Theorem 9, the mgnitude of the torque vector is r F r F sin Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

30 814 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE where is the ngle etween the position nd force vectors. Oserve tht the onl component of F tht cn cuse rottion is the one perpendiculr to r, tht is, F sin. The mgnitude of the torque is equl to the re of the prllelogrm determined r nd F..25 m FIGURE N EXAMPLE 6 A olt is tightened ppling 4-N force to.25-m wrench s shown in Figure 5. Find the mgnitude of the torque out the center of the olt. SOLUTION The mgnitude of the torque vector is r F r F sin sin 75 1 sin Nm If the olt is right-threded, then the torque vector itself is n 9.66 n where n is unit vector directed down into the pge Eercises 1 7 Find the cross product nd verif tht it is orthogonl to oth nd. 1. 6,, 2, 2. 1, 1, 1, 3. i 3j 2k, 4. j 7k, 5. i j k, 6. ti cos tj sin tk, 7. t, 1, 1t,, 8, 2, 4, 6 i 5k 2i j 4k 1 2 i j 1 2 k t 2, t 2, 1 i sin tj cos tk 8. If i 2k nd j k, find. Sketch,, nd s vectors strting t the origin. u v Find nd determine whether u v is directed into the pge or out of the pge u =4 45 v =5 v =16 u = The figure shows vector in the -plne nd vector in the direction of k. Their lengths re 3 nd 2. () Find. () Use the right-hnd rule to decide whether the com ponents of re positive, negtive, or Find the vector, not with determinnts, ut using properties of cross products. 9. i j k j k k i 12. k i 2j i j i j 13. Stte whether ech epression is meningful. If not, eplin wh. If so, stte whether it is vector or sclr. () c () c (c) c (d) c (e) c d (f) c d 17. If 2, 1, 3 nd 4, 2, 1, find nd. 18. If 1,, 1, 2, 1, 1, nd c, 1, 3, show tht c c. 19. Find two unit vectors orthogonl to oth 3, 2, 1 nd 1, 1,. 1. Homework Hints ville t stewrtclculus.com Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

31 SECTION 12.4 THE CROSS PRODUCT Find two unit vectors orthogonl to oth j k nd i j. 21. Show tht for n vector in. 22. Show tht for ll vectors nd in V Prove Propert 1 of Theorem Prove Propert 2 of Theorem Prove Propert 3 of Theorem Prove Propert 4 of Theorem Find the re of the prllelogrm with vertices A2, 1, B, 4, C4, 2, nd D2, Find the re of the prllelogrm with vertices K1, 2, 3, L1, 3, 6, M3, 8, 6, nd N3, 7, () Find nonero vector orthogonl to the plne through the points P, Q, nd R, nd () find the re of tringle PQR. 29. P1,, 1, Q2, 1, 3, R4, 2, 5 3. P,, 3, Q4, 2,, R3, 3, P, 2,, Q4, 1, 2, R5, 3, P1, 3, 1, Q, 5, 2, R4, 3, Find the volume of the prllelepiped determined the vectors,, nd c , 2, 3, 1, 1, 2, c 2, 1, i j, j k, c i j k Find the volume of the prllelepiped with djcent edges PQ, PR, nd PS. 35. P2, 1,, Q2, 3, 2, R1, 4, 1, S3, 6, P3,, 1, Q1, 2, 5, R5, 1, 1, S, 4, Use the sclr triple product to verif tht the vectors u i 5 j 2 k, v 3i j, nd w 5i 9 j 4 k re coplnr. 38. Use the sclr triple product to determine whether the points A1, 3, 2, B3, 1, 6, C5, 2,, nd D3, 6, 4 lie in the sme plne. 39. A iccle pedl is pushed foot with 6-N force s shown. The shft of the pedl is 18 cm long. Find the mgnitude of the torque out P. 6 N 7 1 P V 3 4. Find the mgnitude of the torque out P if 36-l force is pplied s shown. 41. A wrench 3 cm long lies long the positive -is nd grips olt t the origin. A force is pplied in the direction, 3, 4 t the end of the wrench. Find the mgnitude of the force needed to suppl 1 Nm of torque to the olt. 42. Let v 5j nd let u e vector with length 3 tht strts t the origin nd rottes in the -plne. Find the mimum nd minimum vlues of the length of the vector u v. In wht direction does u v point? 43. If s3 nd 1, 2, 2, find the ngle etween nd. 44. () Find ll vectors v such tht () Eplin wh there is no vector v such tht 45. () Let P e point not on the line L tht psses through the points Q nd R. Show tht the distnce d from the point P to the line L is where QR l nd QP l. () Use the formul in prt () to find the distnce from the point P1, 1, 1 to the line through Q, 6, 8 nd R1, 4, () Let P e point not on the plne tht psses through the points Q, R, nd S. Show tht the distnce d from P to the plne is where QR l l l, QS, nd c QP. () Use the formul in prt () to find the distnce from the point P2, 1, 4 to the plne through the points Q1,,, R, 2,, nd S,, Show tht. 48. If c, show tht P 4 ft 3 36 l 4 ft 1, 2, 1 v 3, 1, 5 1, 2, 1 v 3, 1, 5 d d c c c Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

32 816 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE 49. Prove tht Prove Propert 6 of Theorem 11, tht is, 51. Use Eercise 5 to prove tht 52. Prove tht c c c c c c c d c d c d 53. Suppose tht. () If c, does it follow tht c? () If c, does it follow tht c? (c) If c nd c, does it follow tht c? v 1 v 2 v If,, nd re noncoplnr vectors, let k 1 v 2 v 3 v 1 v 2 v 3 k 3 k 2 v 1 v 2 v 1 v 2 v 3 v 3 v 1 v 1 v 2 v 3 (These vectors occur in the stud of crstllogrph. Vectors of the form n 1v 1 n 2v 2 n 3v 3, where ech n i is n integer, form lttice for crstl. Vectors written similrl in terms of k 1, k 2, nd k 3 form the reciprocl lttice.) () Show tht k i is perpendiculr to v j if i j. () Show tht k i v i 1 for i 1, 2, 3. 1 (c) Show tht k 1 k 2 k 3. v 1 v 2 v 3 DISCOVERY PROJECT THE GEOMETRY OF A TETRAHEDRON A tetrhedron is solid with four vertices, P, Q, R, nd S, nd four tringulr fces, s shown in the figure. P 1. Let v 1, v 2, v 3, nd v 4 e vectors with lengths equl to the res of the fces opposite the vertices P, Q, R, nd S, respectivel, nd directions perpendiculr to the respective fces nd pointing outwrd. Show tht v 1 v 2 v 3 v 4 Q S R 2. The volume V of tetrhedron is one-third the distnce from verte to the opposite fce, times the re of tht fce. () Find formul for the volume of tetrhedron in terms of the coordintes of its vertices P, Q, R, nd S. () Find the volume of the tetrhedron whose vertices re P1, 1, 1, Q1, 2, 3, R1, 1, 2, nd S3, 1, Suppose the tetrhedron in the figure hs trirectngulr verte S. (This mens tht the three ngles t S re ll right ngles.) Let A, B, nd C e the res of the three fces tht meet t S, nd let D e the re of the opposite fce PQR. Using the result of Prolem 1, or otherwise, show tht D 2 A 2 B 2 C 2 (This is three-dimensionl version of the Pthgoren Theorem.) 12.5 Equtions of Lines nd Plnes A line in the -plne is determined when point on the line nd the direction of the line (its slope or ngle of inclintion) re given. The eqution of the line cn then e written using the point-slope form. Likewise, line L in three-dimensionl spce is determined when we know point P,, on L nd the direction of L. In three dimensions the direction of line is convenientl descried vector, so we let v e vector prllel to L. Let P,, e n ritrr point on L nd let nd r e the position vectors of nd P (tht is, the hve r P Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

33 SECTION 12.5 EQUATIONS OF LINES AND PLANES 817 L O P (,, ) P(,, ) r r v representtions OPA nd OPA). If is the vector with representtion A, P P s in Figure 1, then the Tringle Lw for vector ddition gives r r. But, since nd v re prllel vectors, there is sclr t such tht tv. Thus 1 r r tv FIGURE 1 t< t= t> r L which is vector eqution of L. Ech vlue of the prmeter t gives the position vector r of point on L. In other words, s tvries, the line is trced out the tip of the vector r. As Figure 2 indictes, positive vlues of t correspond to points on L tht lie on one side of P, wheres negtive vlues of t correspond to points tht lie on the other side of P. If the vector v tht gives the direction of the line L is written in component form s v,, c, then we hve tv t, t, tc. We cn lso write r,, nd r,,, so the vector eqution 1 ecomes,, t, t, tc FIGURE 2 Two vectors re equl if nd onl if corresponding components re equl. Therefore we hve the three sclr equtions: 2 t t ct where t. These equtions re clled prmetric equtions of the line L through the point P,, nd prllel to the vector v,, c. Ech vlue of the prmeter t gives point,, on L. Figure 3 shows the line L in Em ple 1 nd its reltion to the given point nd to the vector tht gives its direction. EXAMPLE 1 () Find vector eqution nd prmetric equtions for the line tht psses through the point 5, 1, 3 nd is prllel to the vector i 4 j 2k. () Find two other points on the line. L (5, 1, 3) r v=i+4j-2k SOLUTION () Here r 5, 1, 3 5i j 3k nd v i 4 j 2k, so the vector equ - tion 1 ecomes r 5i j 3k ti 4 j 2k or r 5 t i 1 4t j 3 2t k Prmetric equtions re FIGURE 3 5 t 1 4t 3 2t () Choosing the prmeter vlue t 1 gives 6, 5, nd 1, so 6, 5, 1 is point on the line. Similrl, t 1 gives the point 4, 3, 5. The vector eqution nd prmetric equtions of line re not unique. If we chnge the point or the prmeter or choose different prllel vector, then the equtions chnge. For instnce, if, insted of 5, 1, 3, we choose the point 6, 5, 1 in Emple 1, then the prmetric equtions of the line ecome 6 t 5 4t 1 2t Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

34 818 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE Or, if we st with the point 5, 1, 3 ut choose the prllel vector 2i 8j 4k, we rrive t the equtions 5 2t 1 8t 3 4t In generl, if vector v,, c is used to descrie the direction of line L, then the numers,, nd cre clled direction numers of L. Since n vector prllel to v could lso e used, we see tht n three numers proportionl to,, nd c could lso e used s set of direction numers for L. Another w of descriing line L is to eliminte the prmeter t from Equtions 2. If none of,, or c is, we cn solve ech of these equtions for t, equte the results, nd otin 3 c These equtions re clled smmetric equtions of L. Notice tht the numers,, nd c tht pper in the denomintors of Equtions 3 re direction numers of L, tht is, components of vector prllel to L. If one of,, or c is, we cn still eliminte t. For instnce, if, we could write the equtions of L s c This mens tht L lies in the verticl plne. Figure 4 shows the line L in Emple 2 nd the point P where it intersects the -plne. 1 B P _1 L EXAMPLE 2 () Find prmetric equtions nd smmetric equtions of the line tht psses through the points A2, 4, 3 nd B3, 1, 1. () At wht point does this line intersect the -plne? SOLUTION () We re not eplicitl given vector prllel to the line, ut oserve tht the vector v with representtion AB l is prllel to the line nd v 3 2, 1 4, 1 3 1, 5, 4 FIGURE 4 A Thus direction numers re 1, 5, nd c 4. Tking the point 2, 4, 3 s, we see tht prmetric equtions 2 re P 2 t 4 5t 3 4t nd smmetric equtions 3 re () The line intersects the -plne when, so we put in the smmetric equtions nd otin ( 11 4, 1 4, ) This gives 11 nd 1 4 4, so the line intersects the -plne t the point. Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

35 SECTION 12.5 EQUATIONS OF LINES AND PLANES 819 In generl, the procedure of Emple 2 shows tht direction numers of the line L through the points P,, nd P 1 1, 1, 1 re 1, 1, nd 1 nd so smmetric equtions of L re Often, we need description, not of n entire line, ut of just line segment. How, for instnce, could we descrie the line segment AB in Emple 2? If we put t in the prmetric equtions in Emple 2(), we get the point 2, 4, 3 nd if we put t 1 we get 3, 1, 1. So the line segment AB is descried the prmetric equtions 2 t 4 5t 3 4t t 1 or the corresponding vector eqution rt 2 t, 4 5t, 3 4t t 1 In generl, we know from Eqution 1 tht the vector eqution of line through the (tip of the) vector r in the direction of vector vis r r tv. If the line lso psses through (the tip of) r 1, then we cn tke v r 1 r nd so its vector eqution is r r tr 1 r 1 tr tr 1 The line segment from r to r 1 is given the prmeter intervl t 1. 4 The line segment from r to r 1 is given the vector eqution rt 1 tr tr 1 t 1 The lines L 1 nd L 2 in Emple 3, shown in Figure 5, re skew lines. 5 L L _5 FIGURE 5 v EXAMPLE 3 Show tht the lines L 1 nd L 2 with prmetric equtions 1 t 2 3t 4 t 2s 3 s 3 4s re skew lines; tht is, the do not intersect nd re not prllel (nd therefore do not lie in the sme plne). SOLUTION The lines re not prllel ecuse the corresponding vectors 1, 3, 1 nd 2, 1, 4 re not prllel. (Their components re not proportionl.) If L 1 nd L 2 hd point of intersection, there would e vlues of t nd s such tht 1 t 2s 2 3t 3 s 4 t 3 4s But if we solve the first two equtions, we get t 11 5 nd s 8 5, nd these vlues don t stisf the third eqution. Therefore there re no vlues of t nd s tht stisf the three equtions, so nd do not intersect. Thus nd re skew lines. L 1 L 2 Plnes Although line in spce is determined point nd direction, plne in spce is more difficult to descrie. A single vector prllel to plne is not enough to conve the direction of the plne, ut vector perpendiculr to the plne does completel specif L 1 L 2 Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

36 82 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE r P(,, ) r-r n its direction. Thus plne in spce is determined point P,, in the plne nd vector n tht is orthogonl to the plne. This orthogonl vector n is clled norml vector. Let P,, e n ritrr point in the plne, nd let r nd r e the position vectors of P nd P. Then the vector r r is represented A. P P(See Figure 6.) The norml vector n is orthogonl to ever vector in the given plne. In prticulr, n is orthogonl to r r nd so we hve r P (,, ) 5 n r r FIGURE 6 which cn e rewritten s 6 n r n r Either Eqution 5 or Eqution 6 is clled vector eqution of the plne. To otin sclr eqution for the plne, we write n,, c, r,,, nd r,,. Then the vector eqution 5 ecomes or,, c,, 7 c Eqution 7 is the sclr eqution of the plne through P,, with norml vector n,, c. v EXAMPLE 4 Find n eqution of the plne through the point 2, 4, 1 with norml vector n 2, 3, 4. Find the intercepts nd sketch the plne. (,, 3) SOLUTION Putting 2, 3, c 4, 2, 4, nd 1 in Eqution 7, we see tht n eqution of the plne is (6,, ) FIGURE 7 (, 4, ) or To find the -intercept we set in this eqution nd otin 6. Similrl, the -intercept is 4 nd the -intercept is 3. This enles us to sketch the portion of the plne tht lies in the first octnt (see Figure 7). B collecting terms in Eqution 7 s we did in Emple 4, we cn rewrite the eqution of plne s 8 c d where d c. Eqution 8 is clled liner eqution in,, nd. Conversel, it cn e shown tht if,, nd c re not ll, then the liner eqution 8 represents plne with norml vector,, c. (See Eercise 81.) Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

37 SECTION 12.5 EQUATIONS OF LINES AND PLANES 821 Figure 8 shows the portion of the plne in Emple 5 tht is enclosed tringle PQR. Q(3, _1, 6) EXAMPLE 5 Find n eqution of the plne tht psses through the points P1, 3, 2, Q3, 1, 6, nd R5, 2,. SOLUTION The vectors nd corresponding to PQ l nd PR l re 2, 4, 4 4, 1, 2 FIGURE 8 R(5, 2, ) P(1, 3, 2) Since oth nd lie in the plne, their cross product is orthogonl to the plne nd cn e tken s the norml vector. Thus i j k n i 2 j 14 k With the point P1, 3, 2 nd the norml vector n, n eqution of the plne is or EXAMPLE 6 Find the point t which the line with prmetric equtions 2 3t, 4t, 5 t intersects the plne SOLUTION We sustitute the epressions for,, nd from the prmetric equtions into the eqution of the plne: 42 3t 54t 25 t 18 FIGURE 9 n n Figure 1 shows the plnes in Emple 7 nd their line of intersection L. ++=1-2+3=1 6 L 4 2 _2 _4 _2 2 _2 2 FIGURE 1 This simplifies to 1t 2, so t 2. Therefore the point of intersection occurs when the prmeter vlue is t 2. Then , 42 8, nd so the point of intersection is 4, 8, 3. Two plnes re prllel if their norml vectors re prllel. For instnce, the plnes nd re prllel ecuse their norml vectors re n 1 1, 2, 3 nd n 2 2, 4, 6 nd n 2 2n 1. If two plnes re not prllel, then the intersect in stright line nd the ngle etween the two plnes is defined s the cute ngle etween their norml vectors (see ngle in Figure 9). v EXAMPLE 7 () Find the ngle etween the plnes 1 nd () Find smmetric equtions for the line of intersection L of these two plnes. SOLUTION () The norml vectors of these plnes re n 1 1, 1, 1 n 2 1, 2, 3 nd so, if is the ngle etween the plnes, Corollr gives cos n 1 n 2 n 1 n 2 cos 1 2 s s1 1 1 s s42 () We first need to find point on L. For instnce, we cn find the point where the line intersects the -plne setting in the equtions of oth plnes. This gives the Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

38 822 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE Another w to find the line of intersection is to solve the equtions of the plnes for two of the vriles in terms of the third, which cn e tken s the prmeter. 2 1 = 2 3 _1 _2 _1 FIGURE = _ _1 _2 Figure 11 shows how the line L in Emple 7 cn lso e regrded s the line of intersection of plnes derived from its smmetric equtions. L equtions 1 nd 2 1, whose solution is 1,. So the point 1,, lies on L. Now we oserve tht, since L lies in oth plnes, it is perpendiculr to oth of the norml vectors. Thus vector v prllel to L is given the cross product i j k v n 1 n i 2 j 3 k nd so the smmetric equtions of L cn e written s 1 5 NOTE Since liner eqution in,, nd represents plne nd two nonprllel plnes intersect in line, it follows tht two liner equtions cn represent line. The points,, tht stisf oth 1 1 c 1 d 1 nd 2 2 c 2 d 2 lie on oth of these plnes, nd so the pir of liner equtions represents the line of intersection of the plnes (if the re not prllel). For instnce, in Emple 7 the line L ws given s the line of intersection of the plnes 1 nd The smmetric equtions tht we found for L could e written s which is gin pir of liner equtions. The ehiit L s the line of intersection of the plnes 15 2 nd 2 3. (See Figure 11.) In generl, when we write the equtions of line in the smmetric form we cn regrd the line s the line of intersection of the two plnes 2 nd 3 2 c 3 nd c EXAMPLE 8 Find formul for the distnce D from point P 1 1, 1, 1 to the plne c d. SOLUTION Let P,, e n point in the given plne nd let e the vector corresponding to P PA. 1 Then 1, 1, 1 P FIGURE 12 n P D From Figure 12 ou cn see tht the distnce D from P 1 to the plne is equl to the solute vlue of the sclr projection of onto the norml vector n,, c. (See Section 12.3.) Thus D comp n n n 1 1 c 1 s 2 2 c c 1 c s 2 2 c 2 Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

39 SECTION 12.5 EQUATIONS OF LINES AND PLANES 823 Since P lies in the plne, its coordintes stisf the eqution of the plne nd so we hve c d. Thus the formul for D cn e written s 9 D 1 1 c 1 d s 2 2 c 2 EXAMPLE 9 Find the distnce etween the prllel plnes nd 5 1. SOLUTION First we note tht the plnes re prllel ecuse their norml vectors 1, 2, 2 nd 5, 1, 1 re prllel. To find the distnce D etween the plnes, we choose n point on one plne nd clculte its distnce to the other plne. In prticulr, if we put in the eqution of the first plne, we get 1 5 nd so ( 1 2,, ) is point in this plne. B Formul 9, the distnce etween ( 1 2,, ) nd the plne 5 1 is D 5( 1 2 ) s So the distnce etween the plnes is s s3 s3 6 EXAMPLE 1 In Emple 3 we showed tht the lines L 1 : 1 t 2 3t 4 t L 2 : 2s 3 s 3 4s re skew. Find the distnce etween them. SOLUTION Since the two lines L 1 nd L 2 re skew, the cn e viewed s ling on two prllel plnes P 1 nd P 2. The distnce etween L 1 nd L 2 is the sme s the distnce etween P 1 nd P 2, which cn e computed s in Emple 9. The common norml vector to oth plnes must e orthogonl to oth v 1 1, 3, 1 (the direction of L 1 ) nd v 2 2, 1, 4 (the direction of L 2 ). So norml vector is i j k n v 1 v i 6 j 5k If we put s in the equtions of L 2, we get the point, 3, 3 on L 2 nd so n eqution for is P or If we now set t in the equtions for L 1, we get the point 1, 2, 4 on P 1. So the distnce etween L 1 nd L 2 is the sme s the distnce from 1, 2, 4 to B Formul 9, this distnce is D s s23.53 Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

40 824 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE 12.5 Eercises 1. Determine whether ech sttement is true or flse. () Two lines prllel to third line re prllel. () Two lines perpendiculr to third line re prllel. (c) Two plnes prllel to third plne re prllel. (d) Two plnes perpendiculr to third plne re prllel. (e) Two lines prllel to plne re prllel. (f) Two lines perpendiculr to plne re prllel. (g) Two plnes prllel to line re prllel. (h) Two plnes perpendiculr to line re prllel. ( i) Two plnes either intersect or re prllel. (j) Two lines either intersect or re prllel. (k) A plne nd line either intersect or re prllel. 2 5 Find vector eqution nd prmetric equtions for the line. 2. The line through the point 6, 5, 2 nd prllel to the vector 1, 3, The line through the point 2, 2.4, 3.5 nd prllel to the vector 3i 2j k 4. The line through the point, 14, 1 nd prllel to the line 1 2t, 6 3t, 3 9t 5. The line through the point (1,, 6) nd perpendiculr to the plne Find prmetric equtions nd smmetric equtions for the line. 6. The line through the origin nd the point 4, 3, 1 7. The line through the points (, 1 2, 1) nd 2, 1, 3 8. The line through the points 1., 2.4, 4.6 nd 2.6, 1.2,.3 9. The line through the points 8, 1, 4 nd 3, 2, 4 1. The line through 2, 1, nd perpendiculr to oth i j nd j k 11. The line through 1, 1, 1 nd prllel to the line The line of intersection of the plnes nd Is the line through 4, 6, 1 nd 2,, 3 prllel to the line through 1, 18, 4 nd 5, 3, 14? 14. Is the line through 2, 4, nd 1, 1, 1 perpendiculr to the line through 2, 3, 4 nd 3, 1, 8? 15. () Find smmetric equtions for the line tht psses through the point 1, 5, 6 nd is prllel to the vector 1, 2, 3. () Find the points in which the required line in prt () intersects the coordinte plnes. 16. () Find prmetric equtions for the line through 2, 4, 6 tht is perpendiculr to the plne 3 7. () In wht points does this line intersect the coordinte plnes? 17. Find vector eqution for the line segment from 2, 1, 4 to 4, 6, Find prmetric equtions for the line segment from 1, 3, 1 to 5, 6, Determine whether the lines L 1 nd L 2 re prllel, skew, or intersecting. If the intersect, find the point of intersection. 19. L 1 : 3 2t, 4 t, 1 3t L 2 : 1 4s, 3 2s, 4 5s 2. L 1 : 5 12t, 3 9t, 1 3t L 2 : 3 8s, 6s, 7 2s L 1 : L 2 : L 1 : L 2 : Find n eqution of the plne. 23. The plne through the origin nd perpendiculr to the vector 1, 2, The plne through the point 5, 3, 5 nd with norml vector 2i j k 25. The plne through the point (1, 1 2, 3) nd with norml vector i 4j k 26. The plne through the point 2,, 1 nd perpendiculr to the line 3t, 2 t, 3 4t 27. The plne through the point 1, 1, 1 nd prllel to the plne The plne through the point 2, 4, 6 nd prllel to the plne 29. The plne through the point (1, 1 2, 1 3) nd prllel to the plne 3. The plne tht contins the line 1 t, 2 t, 4 3t nd is prllel to the plne The plne through the points, 1, 1, 1,, 1, nd 1, 1, 32. The plne through the origin nd the points 2, 4, 6 nd 5, 1, 3 1. Homework Hints ville t stewrtclculus.com Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

41 SECTION 12.5 EQUATIONS OF LINES AND PLANES The plne through the points 3, 1, 2, 8, 2, 4, nd 1, 2, The plne tht psses through the point 1, 2, 3 nd contins the line 3t, 1 t, 2 t 35. The plne tht psses through the point 6,, 2 nd contins the line 4 2t, 3 5t, 7 4t 36. The plne tht psses through the point 1, 1, 1 nd contins the line with smmetric equtions The plne tht psses through the point 1, 2, 1 nd contins the line of intersection of the plnes 2 nd The plne tht psses through the points, 2, 5 nd 1, 3, 1 nd is perpendiculr to the plne The plne tht psses through the point 1, 5, 1 nd is perpendiculr to the plnes nd The plne tht psses through the line of intersection of the plnes 1 nd 2 3 nd is perpendiculr to the plne Use intercepts to help sketch the plne Find the point t which the line intersects the given plne t, 2 t, 5t; t, 4t, 2 3t; ; 48. Where does the line through 1,, 1 nd 4, 2, 2 intersect the plne 6? 49. Find direction numers for the line of intersection of the plnes 1 nd. 5. Find the cosine of the ngle etween the plnes nd Determine whether the plnes re prllel, perpendiculr, or neither. If neither, find the ngle etween them , , 53. 1, , , , () Find prmetric equtions for the line of intersection of the plnes nd () find the ngle etween the plnes , , Find smmetric equtions for the line of intersection of the plnes , , Find n eqution for the plne consisting of ll points tht re equidistnt from the points 1,, 2 nd 3, 4,. 62. Find n eqution for the plne consisting of ll points tht re equidistnt from the points 2, 5, 5 nd 6, 3, Find n eqution of the plne with -intercept, -intercept, nd -intercept c. 64. () Find the point t which the given lines intersect: r 1, 1, t 1, 1, 2 r 2,, 2 s1, 1, () Find n eqution of the plne tht contins these lines. 65. Find prmetric equtions for the line through the point, 1, 2 tht is prllel to the plne 2 nd perpendiculr to the line 1 t, 1 t, 2t. 66. Find prmetric equtions for the line through the point, 1, 2 tht is perpendiculr to the line 1 t, 1 t, 2t nd intersects this line. 67. Which of the following four plnes re prllel? Are n of them identicl? P 1: P 2: P 3: P 4: Which of the following four lines re prllel? Are n of them identicl? L 1: 1 6t, 1 3t, 12t 5 L 2: 1 2t, t, 1 4t L 3: L 4: r 3, 1, 5 t 4, 2, Use the formul in Eercise 45 in Section 12.4 to find the distnce from the point to the given line , 1, 2; 1 t, 3 2t, 4 3t 7., 1, 3; 2t, 6 2t, 3 t Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

42 826 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE Find the distnce from the point to the given plne , 2, 4, 72. 6, 3, 5, Find the distnce etween the given prllel plnes , , Show tht the distnce etween the prllel plnes c d 1 nd c d 2 is D d1 d2 s 2 2 c Find equtions of the plnes tht re prllel to the plne nd two units w from it. 77. Show tht the lines with smmetric equtions nd re skew, nd find the distnce etween these lines. 78. Find the distnce etween the skew lines with prmetric equtions 1 t, 1 6t, 2t, nd 1 2s, 5 15s, 2 6s. L Let e the line through the origin nd the point 2,, 1. Let L 2 e the line through the points 1, 1, 1 nd 4, 1, 3. Find the distnce etween L 1 nd L Let L 1 e the line through the points 1, 2, 6 nd 2, 4, 8. Let L 2 e the line of intersection of the plnes 1 nd 2, where 1 is the plne 2 1 nd 2 is the plne through the points 3, 2, 1,,, 1, nd 1, 2, 1. Clculte the distnce etween L 1 nd L If,, nd c re not ll, show tht the eqution c d represents plne nd,, c is norml vector to the plne. Hint: Suppose nd rewrite the eqution in the form d c 82. Give geometric description of ech fmil of plnes. () c () c 1 (c) cos sin 1 LABORATORY PROJECT PUTTING 3D IN PERSPECTIVE Computer grphics progrmmers fce the sme chllenge s the gret pinters of the pst: how to represent three-dimensionl scene s flt imge on two-dimensionl plne ( screen or cnvs). To crete the illusion of perspective, in which closer ojects pper lrger thn those frther w, three-dimensionl ojects in the computer s memor re projected onto rectngulr screen window from viewpoint where the ee, or cmer, is locted. The viewing volume the portion of spce tht will e visile is the region contined the four plnes tht pss through the viewpoint nd n edge of the screen window. If ojects in the scene etend eond these four plnes, the must e truncted efore piel dt re sent to the screen. These plnes re therefore clled clipping plnes. 1. Suppose the screen is represented rectngle in the -plne with vertices, 4, nd, 4, 6, nd the cmer is plced t 1,,. A line L in the scene psses through the points 23, 285, 12 nd 86, 15, 264. At wht points should L e clipped the clipping plnes? 2. If the clipped line segment is projected on the screen window, identif the resulting line segment. 3. Use prmetric equtions to plot the edges of the screen window, the clipped line segment, nd its projection on the screen window. Then dd sight lines connecting the viewpoint to ech end of the clipped segments to verif tht the projection is correct. 4. A rectngle with vertices 621, 147, 26, 563, 31, 242, 657, 111, 86, nd 599, 67, 122 is dded to the scene. The line L intersects this rectngle. To mke the rectngle pper opque, progrmmer cn use hidden line rendering, which removes portions of ojects tht re ehind other ojects. Identif the portion of L tht should e removed. Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

43 SECTION 12.6 CYLINDERS AND QUADRIC SURFACES Clinders nd Qudric Surfces We hve lred looked t two specil tpes of surfces: plnes (in Section 12.5) nd spheres (in Section 12.1). Here we investigte two other tpes of surfces: clinders nd qudric surfces. In order to sketch the grph of surfce, it is useful to determine the curves of intersection of the surfce with plnes prllel to the coordinte plnes. These curves re clled trces (or cross-sections) of the surfce. FIGURE 1 The surfce = is prolic clinder. Clinders A clinder is surfce tht consists of ll lines (clled rulings) tht re prllel to given line nd pss through given plne curve. v EXAMPLE 1 Sketch the grph of the surfce 2. SOLUTION Notice tht the eqution of the grph, 2, doesn t involve. This mens tht n verticl plne with eqution k (prllel to the -plne) intersects the grph in curve with eqution 2. So these verticl trces re prols. Figure 1 shows how the grph is formed tking the prol 2 in the -plne nd moving it in the direction of the -is. The grph is surfce, clled prolic clinder, mde up of infinitel mn shifted copies of the sme prol. Here the rulings of the clinder re prllel to the -is. We noticed tht the vrile is missing from the eqution of the clinder in Em ple 1. This is tpicl of surfce whose rulings re prllel to one of the coordinte es. If one of the vriles,, or is missing from the eqution of surfce, then the surfce is clinder. FIGURE 2 + =1 EXAMPLE 2 Identif nd sketch the surfces. () () SOLUTION () Since is missing nd the equtions 2 2 1, k represent circle with rdius 1 in the plne k, the surfce is circulr clinder whose is is the -is. (See Figure 2.) Here the rulings re verticl lines. () In this cse is missing nd the surfce is circulr clinder whose is is the -is. (See Figure 3.) It is otined tking the circle 2 2 1, in the -plne nd moving it prllel to the -is. NOTE When ou re deling with surfces, it is importnt to recognie tht n eqution like represents clinder nd not circle. The trce of the clinder in the -plne is the circle with equtions 2 2 1,. FIGURE 3 +@=1 Qudric Surfces A qudric surfce is the grph of second-degree eqution in three vriles,, nd. The most generl such eqution is A 2 B 2 C 2 D E F G H I J where A, B, C,..., J re constnts, ut trnsltion nd rottion it cn e rought into one of the two stndrd forms A 2 B 2 C 2 J or A 2 B 2 I Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

44 828 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE Qudric surfces re the counterprts in three dimensions of the conic sections in the plne. (See Section 1.5 for review of conic sections.) EXAMPLE 3 Use trces to sketch the qudric surfce with eqution SOLUTION B sustituting, we find tht the trce in the -plne is , which we recognie s n eqution of n ellipse. In generl, the horiontl trce in the plne k is k 1 4 k (,, 2) which is n ellipse, provided tht k 2 4, tht is, 2 k 2. Similrl, the verticl trces re lso ellipses: k 2 k if 1 k 1 (, 3, ) (1,, ) The ellipsoid + + = k 1 9 k if 3 k 3 Figure 4 shows how drwing some trces indictes the shpe of the surfce. It s clled n ellipsoid ecuse ll of its trces re ellipses. Notice tht it is smmetric with respect to ech coordinte plne; this is reflection of the fct tht its eqution involves onl even powers of,, nd. EXAMPLE 4 Use trces to sketch the surfce SOLUTION If we put, we get 2, so the -plne intersects the surfce in prol. If we put k ( constnt), we get 2 4k 2. This mens tht if we slice the grph with n plne prllel to the -plne, we otin prol tht opens upwrd. Similrl, if k, the trce is 4 2 k 2, which is gin prol tht opens upwrd. If we put k, we get the horiontl trces k, which we recognie s fmil of ellipses. Knowing the shpes of the trces, we cn sketch the grph in Figure 5. Becuse of the ellipticl nd prolic trces, the qudric surfce is clled n elliptic proloid. FIGURE 5 The surfce =4 + is n elliptic proloid. Horiontl trces re ellipses; verticl trces re prols. Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

45 SECTION 12.6 CYLINDERS AND QUADRIC SURFACES 829 FIGURE 6 Verticl trces re prols; horiontl trces re hperols. All trces re leled with the vlue of k. v EXAMPLE 5 Sketch the surfce 2 2. SOLUTION The trces in the verticl plnes k re the prols 2 k 2, which open upwrd. The trces in k re the prols 2 k 2, which open downwrd. The horiontl trces re 2 2 k, fmil of hperols. We drw the fmilies of trces in Figure 6, nd we show how the trces pper when plced in their correct plnes in Figure 7. Trces in =k re = -k@ Trces in =k re =_ +k@ _1 _1 Trces in =k re - =k FIGURE 7 Trces moved to their correct plnes Trces in =k 1 _1 _1 1 Trces in =k Trces in =k _1 TEC In Module 12.6A ou cn investigte how trces determine the shpe of surfce. In Figure 8 we fit together the trces from Figure 7 to form the surfce 2 2, hperolic proloid. Notice tht the shpe of the surfce ner the origin resemles tht of sddle. This surfce will e investigted further in Section 14.7 when we discuss sddle points. FIGURE 8 The surfce = - is hperolic proloid. EXAMPLE 6 Sketch the surfce SOLUTION The trce in n horiontl plne k is the ellipse k 2 4 k Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

46 83 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE ut the trces in the - nd -plnes re the hperols (2,, ) (, 1, ) nd This surfce is clled hperoloid of one sheet nd is sketched in Figure 9. FIGURE 9 The ide of using trces to drw surfce is emploed in three-dimensionl grphing softwre for computers. In most such softwre, trces in the verticl plnes k nd k re drwn for equll spced vlues of k, nd prts of the grph re eliminted using hidden line removl. Tle 1 shows computer-drwn grphs of the si sic tpes of qudric surfces in stndrd form. All surfces re smmetric with respect to the -is. If qudric surfce is smmetric out different is, its eqution chnges ccordingl. TABLE 1 Grphs of qudric surfces Surfce Eqution Surfce Eqution Ellipsoid c 1 2 All trces re ellipses. If c, the ellipsoid is sphere. Cone 2 c Horiontl trces re ellipses. Verticl trces in the plnes k nd k re hperols if k ut re pirs of lines if k. Elliptic Proloid c Horiontl trces re ellipses. Verticl trces re prols. The vrile rised to the first power indictes the is of the proloid. Hperoloid of One Sheet c 1 2 Horiontl trces re ellipses. Verticl trces re hperols. The is of smmetr corresponds to the vrile whose coefficient is negtive. Hperolic Proloid c Horiontl trces re hperols. Verticl trces re prols. The cse where c is illustrted. Hperoloid of Two Sheets c 1 2 Horiontl trces in k re ellipses if k c or k c. Verticl trces re hperols. The two minus signs indicte two sheets. Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

47 SECTION 12.6 CYLINDERS AND QUADRIC SURFACES 831 TEC In Module 12.6B ou cn see how chnging,, nd c in Tle 1 ffects the shpe of the qudric surfce. v EXAMPLE 7 Identif nd sketch the surfce SOLUTION Dividing 4, we first put the eqution in stndrd form: Compring this eqution with Tle 1, we see tht it represents hperoloid of two sheets, the onl difference eing tht in this cse the is of the hperoloid is the -is. The trces in the - nd -plnes re the hperols nd FIGURE @+4= (, _2, ) (, 2, ) The surfce hs no trce in the -plne, ut trces in the verticl plnes k for re the ellipses k 2 which cn e written s k k These trces re used to mke the sketch in Figure 1. EXAMPLE 8 2 k Clssif the qudric surfce SOLUTION B completing the squre we rewrite the eqution s 2 2 k k Compring this eqution with Tle 1, we see tht it represents n elliptic proloid. Here, however, the is of the proloid is prllel to the -is, nd it hs een shifted so tht its verte is the point 3, 1,. The trces in the plne k k 1 re the ellipses k 1 k The trce in the -plne is the prol with eqution 1 3 2,. The proloid is sketched in Figure 11. FIGURE 11 +2@-6-+1= (3, 1, ) Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.

48 832 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE Applictions of Qudric Surfces Emples of qudric surfces cn e found in the world round us. In fct, the world itself is good emple. Although the erth is commonl modeled s sphere, more ccurte model is n ellipsoid ecuse the erth s rottion hs cused flttening t the poles. (See Eercise 47.) Circulr proloids, otined rotting prol out its is, re used to collect nd reflect light, sound, nd rdio nd television signls. In rdio telescope, for instnce, signls from distnt strs tht strike the owl re ll reflected to the receiver t the focus nd re therefore mplified. (The ide is eplined in Prolem 2 on pge 271.) The sme principle pplies to microphones nd stellite dishes in the shpe of proloids. Cooling towers for nucler rectors re usull designed in the shpe of hperoloids of one sheet for resons of structurl stilit. Pirs of hperoloids re used to trnsmit rottionl motion etween skew es. (The cogs of the gers re the generting lines of the hperoloids. See Eercise 49.) Dvid Frier / Coris Mrk C. Burnett / Photo Reserchers, Inc A stellite dish reflects signls to the focus of proloid. Nucler rectors hve cooling towers in the shpe of hperoloids. Hperoloids produce ger trnsmission Eercises 1. () Wht does the eqution 2 represent s curve in 2? () Wht does it represent s surfce in 3? (c) Wht does the eqution 2 represent? 2. () Sketch the grph of e s curve in 2. () Sketch the grph of e s surfce in 3. (c) Descrie nd sketch the surfce e. 3 8 Descrie nd sketch the surfce sin 9. () Find nd identif the trces of the qudric surfce nd eplin wh the grph looks like the grph of the hperoloid of one sheet in Tle 1. () If we chnge the eqution in prt () to , how is the grph ffected? (c) Wht if we chnge the eqution in prt () to ? ; Grphing clcultor or computer required 1. Homework Hints ville t stewrtclculus.com Copright 21 Cengge Lerning. All Rights Reserved. M not e copied, scnned, or duplicted, in whole or in prt. Due to electronic rights, some third prt content m e suppressed from the ebook nd/or echpter(s). Editoril review hs deemed tht n suppressed content does not mterill ffect the overll lerning eperience. Cengge Lerning reserves the right to remove dditionl content t n time if susequent rights restrictions require it.