{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 257 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 262 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 267 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 272 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 2 2 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 277 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 280 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 281 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 282 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 283 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 284 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 285 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 286 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 287 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 288 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 289 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 290 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 291 "" 1 14 0 0 0 0 0 1 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 292 "" 1 14 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 293 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 294 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 295 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 296 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 297 "" 1 14 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 298 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 299 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 300 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 301 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 302 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 303 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 304 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 305 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 306 "" 1 14 0 0 0 0 0 1 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 307 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 308 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 309 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 310 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 311 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 312 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 313 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 314 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 315 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 316 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 317 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 318 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 319 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 320 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 321 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 322 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 323 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 324 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 325 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 326 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 327 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 328 "" 1 14 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 329 "" 1 14 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 330 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 331 "" 0 1 0 0 0 0 2 2 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 332 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 333 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 334 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 335 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 336 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 337 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{PSTYLE "Normal " -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 257 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 1 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 258 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 1 1 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 259 1 {CSTYLE "" -1 -1 "Times " 1 12 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 0 "" }{TEXT 256 0 "" }{TEXT 257 16 "Lines and Planes" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT 292 25 "Parametrization of a line" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 120 "Find a vector equation t hat parametrizes the line that passes through the point P(5,1,2) and i s parallel to the vector 2" }{TEXT 258 1 "i" }{TEXT -1 2 "-3" }{TEXT 259 1 "j" }{TEXT -1 2 "+4" }{TEXT 260 3 "k, " }{TEXT 261 0 "" }{TEXT -1 20 "then graph the line." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 94 "First of all, we l oad a set of subroutines where we will find some subroutines that we n eed to" }}{PARA 0 "" 0 "" {TEXT -1 23 "graph vector functions." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "A line is r epresented by the vector equation " }}{PARA 0 "" 0 "" {TEXT 278 48 " \+ r" }{TEXT -1 6 "(t) = " } {TEXT 279 1 "r" }{TEXT -1 5 "0 + t" }{TEXT 280 1 "d" }{TEXT -1 3 ". \+ " }}{PARA 0 "" 0 "" {TEXT -1 14 "In this case, " }{TEXT 281 1 "r" } {TEXT -1 17 "0 is (5,1,2) and " }{TEXT 282 1 "d" }{TEXT -1 38 ", the d irection vector, is (2,-3,4). " }}{PARA 0 "" 0 "" {TEXT -1 35 "The eq uation for the line is then: " }}{PARA 0 "" 0 "" {TEXT -1 42 " \+ r(t) = (5" }{TEXT 287 1 "i" }{TEXT -1 3 " + \+ " }{TEXT 288 2 "j " }{TEXT -1 3 "+ 2" }{TEXT 289 1 "k" }{TEXT -1 7 ") \+ + t(2" }{TEXT 283 2 "i " }{TEXT -1 3 "- 3" }{TEXT 284 3 " j " }{TEXT -1 3 "+ 4" }{TEXT 285 1 "k" }{TEXT 286 0 "" }{TEXT -1 3 "). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 "Next, we will d efine Unit Vectors in the direction of the axes:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "i:=vector([ 1,0,0]); j:=vector([0,1,0]); k:=vector([0,0,1]);" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 "The above line is \+ the space curve corresponding to the following vector function." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "r1:=t->(5*i+j+2*k)+t*(2*i-3*j+4*k);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 92 "spacecurve(evalm(r1(t)),t=0..5,axes=normal,color=bl ue,labels=[x,y,z],orientation=[-117,69]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 78 "Any point on the line cor responds to a particular value of the parameter 't'. " }}{PARA 0 "" 0 "" {TEXT -1 24 "For example, the points " }}{PARA 0 "" 0 "" {TEXT -1 22 "(1,7,-6) and (7,-2,6) " }}{PARA 0 "" 0 "" {TEXT -1 92 "correspond \+ to t = -2 and t = 1. We can draw the segment on the line between these two points" }}{PARA 0 "" 0 "" {TEXT -1 51 "by restricting the values \+ of 't' between -2 and 1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "spacecurve(evalm(r1(t)),t=-2..1,axe s=normal,color=blue,labels=[x,y,z],orientation=[-117,69]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "Can you i dentify from this result the orientation on the line?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 291 29 "The Intersect ion of Two Lines" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 "Find the point and angle of intersection between the two \+ lines " }}{PARA 0 "" 0 "" {TEXT -1 7 "r(t)=(3" }{TEXT 262 1 "i" } {TEXT -1 3 " + " }{TEXT 263 1 "j" }{TEXT -1 4 " + 5" }{TEXT 264 1 "k" }{TEXT -1 7 ") + t( " }{TEXT 265 1 "i" }{TEXT -1 2 " -" }{TEXT 266 2 " j " }{TEXT -1 3 "+ 2" }{TEXT 267 1 "k" }{TEXT -1 6 ") and " }}{PARA 0 "" 0 "" {TEXT -1 7 "r(u)=( " }{TEXT 268 1 "i" }{TEXT -1 4 " + 4" } {TEXT 269 2 "j " }{TEXT -1 3 "+ 2" }{TEXT 270 1 "k" }{TEXT -1 6 ") + u (" }{TEXT 271 2 " j" }{TEXT -1 2 " +" }{TEXT 272 2 " k" }{TEXT -1 2 ") ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "r1:=t->(3*i+j+5*k)+t*(i-j+2*k); " }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 27 "r2:=u->(i+4*j+2*k)+u*(j+k);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "L1:=spacecurve(evalm(r1(t)),t=-5..5 ,color=blue): L2:=spacecurve(evalm(r2(u)),u=-5..5,color=red):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "display3d([L1,L2],axes=frame d,labels=([x,y,z]),orientation=[-40,79]);" }}}{EXCHG {PARA 257 "" 0 " " {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT 274 94 "To find the point of intersection, set r(t)=r(u) and solve the resulting equation for t an d u:" }}{PARA 0 "" 0 "" {TEXT 273 1 "i" }{TEXT -1 11 "(2 + t) + " } {TEXT 275 1 "j" }{TEXT -1 15 "(-3 - t - u) + " }{TEXT 276 1 "k" } {TEXT -1 14 "(3 + 2t - u)= " }{TEXT 277 1 "0" }{TEXT 290 0 "" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 108 "Solve as a system of equatio ns and then substitute back into the equations of the lines to obtain \+ the point." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "sys:=\{2+t=0,-3-t -u=0,3+2*t-u=0\};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "solve( sys);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "u:=-1;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 11 "Therefore, " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "r2(u);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "t:=-2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "r1(t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 101 "The point of intersection is (1,3,1) and this corresponds to the parameters t = -2 for the line r1(t)" }} {PARA 0 "" 0 "" {TEXT -1 30 "and u = -1 for the line r2(u)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "Try to visualiz e this by looking at the graph above." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 327 0 "" }{TEXT 328 0 "" } {TEXT 329 53 "Angle between two lines at its point of Intersection " } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 96 "The angl e at the intersection of two lines is the angle (acute) at which thei r unit direction " }}{PARA 0 "" 0 "" {TEXT -1 68 "vectors intersect ( a unit vector is a vector divided by its norm). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 87 "We need to load a set of \+ linear algebra subroutines where we will find the DOT PRODUCT " }} {PARA 0 "" 0 "" {TEXT -1 34 "and the CROSS PRODUCT for vectors." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "d1:= vector([1,-1,2]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "d2:=ve ctor([0,1,1]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "n1:=norm( array([1,-1,2]),2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "n2:= norm(array([0,1,1]),2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 78 "Now we can evaluate cos(theta), where the ta is the angle in between the lines." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "costheta:=abs(dotprod([ 1/n1,-1/n1,2/n1],[0,1/n2,1/n2]));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "theta:=evalf(arccos(costheta));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "thetadegrees:=evalf(theta*180/Pi);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "The angle of intersection is aprox . 1.28 radians or aprox. 73 degrees." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 336 0 "" }{TEXT 337 53 "Two lines that do not intersect and are not parallel:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "r12:=t 2->(3*i+j+5*k)+t2*(i-j+2*k); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "r22:=u2->(2*i+8*j+4*k)+u2*(j+k);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "L12:=spacecurve(evalm(r12(t2)),t2=-5..5,color=blue): L22:=spacecurve(evalm(r22(u2)),u2=-10..10,color=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "display3d([L12,L22],axes=framed,lab els=([x,y,z]),orientation=[-40,79]);" }}}{EXCHG {PARA 257 "" 0 "" {TEXT 331 100 "To find the point of intersection, set r12(t2)=r22(u2) \+ and solve the resulting equation for t and u:" }}{PARA 0 "" 0 "" {TEXT 330 1 "i" }{TEXT -1 9 "(1+t2) + " }{TEXT 332 1 "j" }{TEXT -1 17 "(-7 - t2 - u2) + " }{TEXT 333 1 "k" }{TEXT -1 16 "(1 + 2t2 - u2)= " } {TEXT 334 1 "0" }{TEXT 335 0 "" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 108 "Solve as a system of equations and then substitute back \+ into the equations of the lines to obtain the point." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "sys2:=\{1 +t2=0,-7-t2-u2=0,1+2*t2-u2=0\};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "solve(sys2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }{EXCHG {PARA 258 "" 0 "" {TEXT 326 19 "Equation of a Plane" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 55 "Find an equatio n for the plane that contains the point " }}{PARA 0 "" 0 "" {TEXT -1 54 " P(1,-2,0) " }}{PARA 0 "" 0 "" {TEXT -1 24 "and has a normal vector " }}{PARA 0 "" 0 "" {TEXT 293 42 " N" }{TEXT -1 4 " = 3" }{TEXT 294 1 "i" }{TEXT -1 4 " - 2" }{TEXT 295 1 "j" }{TEXT -1 3 " + " }{TEXT 296 1 "k" }{TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 29 "Graph the plane and the line." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 47 " The equation for the plane is the dot-pr oduct " }{TEXT 324 2 "N " }{TEXT 325 0 "" }{TEXT -1 8 "and the " }} {PARA 0 "" 0 "" {TEXT -1 48 "vector [(x - 1), (y + 2), z] set equal t o zero:" }}{PARA 0 "" 0 "" {TEXT -1 20 "3x - 2y + z - 7 = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "Another way to \+ define this plane is as a function of two variables." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "z1:=(x,y) ->2*y-3*x+7;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "plot3d(z1(x ,y),x=-10..10,y=-10..10,axes=framed,shading=zhue,labels=([x,y,z]));" } }}{EXCHG {PARA 0 "" 0 "" {TEXT 297 26 "Intersection of Two Planes" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "The plane s" }}{PARA 0 "" 0 "" {TEXT -1 74 " \+ P1: x + 2y + 3z = 0 and" }}{PARA 0 "" 0 "" {TEXT -1 70 " P2: -3x + 4y + z = \+ 0" }}{PARA 0 "" 0 "" {TEXT -1 75 "intersect at a line. Graph the plan es and find an equation for the line. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "z2:=(x,y)->(-1/3)*x-(2/3)*y;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "z3:=(x,y)->3*x+4*y;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 102 "Z2:=plot3d(z2(x,y),x=-10..10,y=-10..10,color=blue) : Z3:=plot3d(z3(x,y),x=-10..10,y=-10..10,color=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "display3d([Z2,Z3],axes=framed,labels=[x,y ,z],orientation=[154,42]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "The direction vector for the line at which the planes intersect is the vector" }}{PARA 0 "" 0 "" {TEXT -1 64 " \+ " }{TEXT 298 1 "N" }{TEXT -1 4 "1 x " }{TEXT 299 1 "N" }{TEXT -1 3 "2 , " }}{PARA 0 "" 0 "" {TEXT 300 1 "N" }{TEXT -1 6 "1 and " }{TEXT 301 1 "N" }{TEXT -1 53 "2 being the normal vectors to the planes P1 and P2 . " }}{PARA 0 "" 0 "" {TEXT -1 16 "This is because " }{TEXT 302 1 "N " }{TEXT -1 4 "1 x " }{TEXT 303 1 "N" }{TEXT -1 27 "2 is perpendicular to both " }{TEXT 304 1 "N" }{TEXT -1 6 "1 and " }{TEXT 305 1 "N" } {TEXT -1 2 "2 " }}{PARA 0 "" 0 "" {TEXT -1 50 "and is therefore parall el to both of the planes. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "N1:=vector([1,2,3]);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "N2:=vector([-3,4,1]);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "crossprod(N1,N2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "The direction vector of the line is the (-10,-1 0,10). " }}{PARA 0 "" 0 "" {TEXT -1 104 "In order to find a point on \+ the line, set one of the variables in the original equations equal to \+ zero, " }}{PARA 0 "" 0 "" {TEXT -1 73 "and then solve the system of eq uations for the remaining two variables. " }}{PARA 0 "" 0 "" {TEXT -1 42 "We will choose z = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "sys2:=\{x+2 *y=0,-3*x+4*y=0\};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "solve (sys2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "The point is (0,0,0)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "r5:=s->s*(-10*i-10*j+10*k); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "spacecurve(evalm(r5(s)) ,s=0..11,axes=framed,orientation=[154,42],color=blue);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "The equation for the line at which the pl anes intersects is then:" }}{PARA 0 "" 0 "" {TEXT -1 12 "r(t) = t(-10 " }{TEXT 307 2 "i " }{TEXT -1 4 "- 10" }{TEXT 308 1 "j" }{TEXT -1 5 " \+ + 10" }{TEXT 309 1 "k" }{TEXT -1 3 "), " }}{PARA 0 "" 0 "" {TEXT -1 9 "or simply" }}{PARA 0 "" 0 "" {TEXT -1 10 "r(t) = t(-" }{TEXT 310 2 "i " }{TEXT -1 2 "- " }{TEXT 311 1 "j" }{TEXT -1 3 " + " }{TEXT 312 1 "k " }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 306 39 "A Plane from Thre e Non-Collinear Points" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "Find an equation of a plane that contains the points A(1,0,1), B(2,1,0), and C(1,1,1)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 87 "First, we will obtain two vectors lying in the plane that is defined by these 3 points." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "v1:=vector([1,1,-1]); v2:=vector([0,1,0]);" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 82 "The crossproduct o f these two vectors will provide the normal vector to the plane." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "N:=crossprod(v1,v2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 94 "Pick an arbitrary point in the plane, P(x ,y,z). An arbitrary point P will be in the plane iff" }}{PARA 0 "" 0 "" {TEXT -1 97 "the vector, AP obtained by subtracting A (or B or C) f rom P is orthogonal to the normal vector N." }}{PARA 0 "" 0 "" {TEXT -1 16 "In other words, " }}{PARA 0 "" 0 "" {TEXT -1 74 "P is a point i n the plane if the dot product of N and AP is equal to zero." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "P :=dotprod(N,[x-1,y-0,z-1]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 30 "The equation for the plane is:" }}{PARA 0 "" 0 "" {TEXT -1 9 "x + z = 2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "z1:=(x,y)->2-x;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "plot3d(z1(x,y),x=-5..5,y=-5..5,axes =framed,shading=zhue,orientation=[65,78],labels=[x,y,z]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 259 "" 0 "" {TEXT -1 32 "Work on the following exercises:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 145 "1. Obtain a vector parametrization for the line that passes through (5,3,-2) and is parallel to the plane 2x + y - 2z = -5 , then graph the line." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 66 "2. Graph the line that contains the points (3,-4,0) and (1,2,-1)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "3. Graph the planes" }}{PARA 0 "" 0 "" {TEXT -1 22 "P1: \+ x -4y + 3z = 2 and" }}{PARA 0 "" 0 "" {TEXT -1 18 "P2: 2x + y + 3z =5 " }}{PARA 0 "" 0 "" {TEXT -1 45 " and find the angle at which they in tersect." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 114 "4. Obtain an equation for and the graph of the plane defined by \+ the two intersecting lines from Example 2 above: " }}{PARA 0 "" 0 "" {TEXT -1 8 " r(t)=(3" }{TEXT 313 1 "i" }{TEXT -1 3 " + " }{TEXT 314 1 "j" }{TEXT -1 4 " + 5" }{TEXT 315 1 "k" }{TEXT -1 7 ") + t( " }{TEXT 316 1 "i" }{TEXT -1 2 " -" }{TEXT 317 2 "j " }{TEXT -1 3 "+ 2" }{TEXT 318 1 "k" }{TEXT -1 6 ") and " }}{PARA 0 "" 0 "" {TEXT -1 7 "r(u)=( " }{TEXT 319 1 "i" }{TEXT -1 4 " + 4" }{TEXT 320 2 "j " }{TEXT -1 3 "+ 2 " }{TEXT 321 1 "k" }{TEXT -1 6 ") + u(" }{TEXT 322 2 " j" }{TEXT -1 2 " +" }{TEXT 323 2 " k" }{TEXT -1 2 ")." }}}}{MARK "79" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }