anonymous
  • anonymous
Hi there, Im having troube with understanding the solution to the example 5 in cartesian-coordinates-and-vector course notes:"Show that if /A - B/ = /A + B/ , then A is perpendicular to B.". I dont see how if /A - B/ = ((A.A - 2A.B + B.B)) and if /A + B/ = ((A + B)!(A + B)) = ((A.A + 2A.B + B.B) would imply that A.B=0. Cheers!
OCW Scholar - Physics I: Classical Mechanics
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anonymous
  • anonymous
Hi there, Im having troube with understanding the solution to the example 5 in cartesian-coordinates-and-vector course notes:"Show that if /A - B/ = /A + B/ , then A is perpendicular to B.". I dont see how if /A - B/ = ((A.A - 2A.B + B.B)) and if /A + B/ = ((A + B)!(A + B)) = ((A.A + 2A.B + B.B) would imply that A.B=0. Cheers!
OCW Scholar - Physics I: Classical Mechanics
katieb
  • katieb
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anonymous
  • anonymous
first its |A-B|^2 = A.A -2A.B + B.B and similarly for |A+B| but that's not the essential idea here. What is essential is that if |A-B| = |A+B| then A.A -2A.B + B.B = A.A + 2A.B + B.B => that -A.B = A.B which can only mean that A.B=0 therefore A perp to B

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