Distance between (8,6,2) and (0,6,0)?

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Jorge Franklin · Accepted Answer

Let the length of the strait line be rYou can consider the points as a combination of triangles. First you work out the projection of the line on to the xy plain (the adjacent) using Pythagoras. You then work out the related triangle for the z plane again using Pythagoras where r is the hypotenuse (the line). You finish up with a 3 dimensional version of the standard form             r      2        =          x      2        +          y      2       except that in the 3d version you have             r      2        =          x      2        +          y      2        +          z      2      Given             (      x      ,      y      ,      z      )        →          (      8      ,      6      ,      2      )                     and                     (      0      ,      6      ,      0      )            ⇒          r      2        =                  (                  x          2                -                  x          1                )            2        +                  (                  y          2                -                  y          1                )            2        +                  (                  z          2                -                  z          1                )            2            ⇒    r    =                                        (            0            -            8            )                    2                +                              (            6            -            6            )                    2                +                              (            0            -            2            )                    2                          r    =                  64        +        0        +        4              =                  68              =                            2          2                ×        17                  r    =    2                  17

Distance between (8,6,2) and (0,6,0)?

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