For these vectors to be mutually orthogonal, their dot products must each be equal to 
zero; that is: 
- - + - - 2 = 
(xy = x) (x, xt Gy YS) Gy=y +f 0 (24) 
= - = - 2 5 
(xy = x) (x, xt Gu 7 Yo) (y, Yo) +£ 0 (25) 
= = = = 2 = 
(xy - x) (x, - x) + Go Og ly, -y) +¢ 0 (26) 
Note that equations (24), (25) and (26) exclude the elements of the orientation 
matrix [M]. They simply relate the six image coordinates of the three vanishing points, 
ny» ny and n, with the three elements, XY, and f, of the interior orientation. 
The principal distance, f, can be eliminated by subtracting each of equations 
(25) and (26) from equation (24), thus yielding: 
- - - - = 2 
(xy x.) (xy x,) + E Yo) (y, y) 0 (27) 
(xy = x) (xy - X,) + (vy - yo) (vy - y2) = 0 (28) 
Equations (27) and (28) are obviously dot products equated to zero. Equation 
(27) indicates that line n.o is perpendicular to line nyn2, and equation (28) indicates 
that line nyo is perpendicular to line nyn,, as shown ín Figure 6. 
Ny 
     
  
o 
N 
Figure 6 
It is apparent from Figure 6 that the principal point must be located at the 
intersection of the perpendicular to line nyn, through ny and the perpendicular to line 
n,n, through ny. Analytically, the image coordinates of the principal point are deter- 
mined by solving equations (27) and (28) for x and Yor The solution is given bv: 
- 10 -