69. The Shortest Distance Between Skew Lines

Transcription

1 69. The Shortest Distnce Between Skew Lines Find the ngle nd distnce between two given skew lines. (Skew lines re non-prllel non-intersecting lines.) This importnt problem is usully encountered in one of the following forms: I. Find the ngle nd distnce between two skew lines when point on ech line nd the direction of ech line re given - the former by coordintes nd the ltter by direction cosines. II. Find the ngle nd distnce between two opposite edges of tetrhedron whose six edges re known. The distnce between two skew lines is nturlly the shortest distnce between the lines, i.e., the length of perpendiculr to both lines. Solution of I. In the usul rectngulr xyz-coordinte system, let the two points be P 1 Ÿ 1,b 1, c 1 nd P 2 Ÿ 2, b 2, c 2 ; d P 1 P 2 2 " 1,b 2 " b 1,c 2 " c 1 is the direction vector from P 1 to P 2. Let u 1 l 1, m 1,n 1 nd u 2 l 2,m 2, n 2 be unit direction vectors for the given lines; then the components of u i re the direction cosines for the lines. Let F be the sought-for ngle nd k the sought-for mininum distnce between the two lines. P 2 u1 x u2 X 2 l 2 u2 u1 k d l 1 X 1 P 1 The solution to this problem becomes very simple with the introduction of the dot (or u sclr) product u 1 u 2 nd the cross product u 1 u 2. We hve cosf 1 u 2 u 1 u 2 l 1 l 2 m 1 m 2 n 1 n 2 from which F cn be found. (We cn ssume tht F is cute, thus the bsolute vlues.) u 1 u 2 is orthogonl (perpendiculr) to both lines, so the bsolute vlue of the (sclr) projection of d onto u 1 u 2 gives k. u 1 u 2 1

2 b θ Recll tht the vector projection of b on is b 1 b. Thus k du 1 u 2 u 1 u 2 du 1 u 2 sinf or nd the sclr projection is k det 2 " 1 b 2 " b 1 c 2 " c 1 l 1 m 1 n 1 /sinf. l 2 m 2 n 2 Note 1. Since skew lines re not prllel, u 1 u 2 p 0. Note 2. Let X 1 Ÿx 1, y 1, z 1 nd X 2 Ÿx 2,y 2, z 2 be closest points on the lines, nd let k X 1 X 2. Let r i be the unique numbers such tht X i P i r i u i. Since X 1 X 2 X 1 P 1 P 1 P 2 P 2 X 2, we get k "r 1 u 1 d r 2 u 2. Since k is orthogonl to both lines, tking the dot product with u 1 nd u 2 yields the system of liner equtions: u 1 u 1 r 1 " u 1 u 2 r 2 " u 1 d 0 u 1 u 2 r 1 " u 2 u 2 r 2 " u 2 d 0 in r 1 nd r 2. There is solution if u 1 u 2 p o1, nd this is the cse since the lines re not prllel. Then X i cn be found. Solution of II. Let the vertices of the tetrhedron be A, B,C,O, the six edges BC,CA,AB, OA, OB, OC hve lengths,b, c,p,q,r respectively, nd the vectors BC,CA,AB, OA, OB, OC be, b, c, p, q, r respectively. 2

3 C r B q b c O p A Let the ngle nd distnce between the two opposite edges c nd r be F nd k respectively. Determintion of F. First of ll, nd thus However, too, nd c r ABOC AOOBOA AC OBAC q " b, c r c r c r q " b c q q r " b c " b r. c r c r c c r r 2c r c 2 r 2 2crcosF 2c q c 2 q 2 " p 2, 2q r q 2 r 2 " 2, 2b c 2 " b 2 " c 2, 2b r p 2 " b 2 " r 2. Note tht when the lw of cosines is used with dabc, 2 b 2 c 2 " 2bc cos0bac, nd 0BAC nd the ngle 2 between b nd c re supplementl, so 2 b 2 c 2 2bc cos2 b 2 c 2 2b c. (Similrly for doac.) It follows tht 3

4 2crcosF c r c r " c 2 " r 2 c q q r " b c " b r " c 2 " r 2 2 b 2 c 2 " p 2 q 2 r 2 " c 2 " r 2 b 2 q 2 " 2 " p 2 nd F cn be found. (We cn lwys choose F in the rnge 0 ( to 90 (, since when two lines intersect, one of the verticl ngle pirs is in this rnge, the other being supplementl. Note tht b nd q, nd nd p re lengths of opposite sides of the tetrhedron.) Clcultion of k. Let the volume of the tetrhedron be T. By No. 68, we cn consider this quntity known. Trnslte vector r prllel to itself so itself so tht its strting point (initil point or til) is t A; cll the trnslted end point (or hed) Q. Then AQ 5 OC. C r B Q q b c O p A dcqa X daoc (SSS), nd thus tetrhedrons CQAB nd AOCB hve the sme volume T. Now consider dqab s the bse of tetrhedron CQAB nd C s its pex. The bse re is 1 AQABsin0QAB 1 rcsinf. (Remember tht F is the ngle 2 2 between c nd r.) To find the ltitude of CQAB s the (perpendiculr) distnce from C to the plne QAB, note tht the plne QAB cn be generted by trnslting line OC prllel to itself long line AB. Then line OC lies in plne prllel to plne QAB. It follows tht the ltitude from C to QAB is k, the shortest (perpendiculr) distnce between lines OC nd AB (nd between the two plnes). The volume of tetrhedron CQAB is then rcsinf 1 1 k, nd thus 6T kcr sinf or 3 2 k 6T crsinf. Corollry. c r c r sin F crsin F. With k denoting the shortest vector between 4

5 lines OC nd AB, we hve 6T k c r. This is sometimes expressed s the Theorem. The mixed product of the two opposite sides of tetrhedron nd the distnce between them (ll thought of s vectors) equls six times the volume of the tetrhedron. A direct consequence of this is the fmous Theorem of Steiner. All tetrhedrons hving two opposite edges of given length lying on two fixed lines hve the sme volume. 5