# Unit vector in 3-D

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 August 18, 2003, 05:16 Unit vector in 3-D #1 beggin Guest   Posts: n/a Hi, how many projections (2 or 3) it is necessary to define an unit vector in 3-D space? Three projections (ax, ay, az) it is obviously enough. But since ax^2 +ay^2+ az^2=1 it is may be enough two values? Thnx.

 August 18, 2003, 05:40 Re: Unit vector in 3-D #2 versi Guest   Posts: n/a only 2 on spherical surface (\theta, \phi).

 August 18, 2003, 06:09 Re: Unit vector in 3-D #3 Mario Guest   Posts: n/a ax^2 +ay^2+ az^2=1 is enought to determine the absolute value of 3rd component only. the direction must be specified additionally. so, you need all 3 components in 3D

 August 18, 2003, 07:00 Re: Unit vector in 3-D #4 beggin Guest   Posts: n/a Thank you, but why on sphere enough two numbers (latitude and longitude)? My be spherical coordinates give "more information" than the cartesian coordinates? I always considered that they are equivalent.

 August 18, 2003, 08:07 Re: Unit vector in 3-D #5 Kevin Guest   Posts: n/a You need two points to define a vector. In 3D, you need the 3 coordinates of both points, i.e. x1,y1,z1 and x2,y2,z2. In your case, point 1 is at the axes origin and the norm of your vector is 1. Only two more informations are to be provided : - two angles, in cylindrical coordinates, the norm being known - two of the cartesian coordinates, the last one being detremined by the norm relation (x2**2 +y2**2+z2**2 = 1).

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