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0/19/3/3/3 . Beispiel 20 - 214 Umwandlung in eine Normalenvektor-Darstellung . Gegeben sei eine Ebene .
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n → = a → × b → =| x y z −5 −6 7 −1 −5 0 |=( 0+35 −7+0 25−6 )=( 35 −7 19 ) MathType@MTEF@5@5@+= feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbb a9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr 0=vqpWqaaeaabaGaaiaacaqaaeaadaqaaqaaaOqaamaaFiaabaGaam OBaaGaay51GaGaeyypa0Zaa8HaaeaacaWGHbaacaGLxdcacqGHxdaT daWhcaqaaiaadkgaaiaawEniaiabg2da9maaemaabaqbaeqabmWaaa qaaiaadIhaaeaacaWG5baabaGaamOEaaqaaiabgkHiTiaaiwdaaeaa cqGHsislcaaI2aaabaGaaG4naaqaaiabgkHiTiaaigdaaeaacqGHsi slcaaI1aaabaGaaGimaaaaaiaawEa7caGLiWoacqGH9aqpdaqadaqa auaabeqadeaaaeaacaaIWaGaey4kaSIaaG4maiaaiwdaaeaacqGHsi slcaaI3aGaey4kaSIaaGimaaqaaiaaikdacaaI1aGaeyOeI0IaaGOn aaaaaiaawIcacaGLPaaacqGH9aqpdaqadaqaauaabeqadeaaaeaaca aIZaGaaGynaaqaaiabgkHiTiaaiEdaaeaacaaIXaGaaGyoaaaaaiaa wIcacaGLPaaaaaa@63AF@
n→ ⋅ (r→ -r1 →) = 0 . . 35-7 19 ⋅rx - 3 ry - 5 rz - 1 = 0. . Umwandlung in Achsenabschnittsform: . 35(rx - 3) - 7(ry - 5) + 19(rz - 1) = 0 . 35rx - 7ry + 19rz = 105 - 35 + 19 = 99 .