5 
But the triangle N'LN, right-angled at L, gives 
M'K 2 = N'N 2 = O - x') 2 + {y- y'y ; 
MM' 2 = (x — ,v') a + (y - y) 2 + (z — %'y, 
or d = \/\x — iv’y + (y — r/') 2 + (# - #') 2 . 
Both in this formula, and in that of Art. 5, we take the 
radical with a positive sign ; as the question only relates to 
the absolute distance of the points. 
Cor. Let a, /3, y be the angles which MM' makes with 
the edges of the parallelepiped, and which are manifestly the 
same as those made by a line OP, drawn through the origin 
parallel to MM', with the co-ordinate axes. Then from the 
triangle MM'H, right-angled at H, we have 
cos« = 
M'H 
MM' 
similarly, 
cos /3 = 
y-y 
d 
cos y — 
7- To find the equation to a plane. 
We may consider a plane as a surface generated by a 
straight line which moves so as always to remain parallel 
to itself, and to intersect a straight line given in position. 
Let AC (fig. 5) in the plane of zx be the line to which 
the generating line PQ is always parallel, and BC in the plane 
of yz, intersecting the former in C, the line along which it 
moves. Let z = Ax + c be the equation to the line AC 
referred to the axes Ox, Oz, in the plane in which it lies; 
similarly, let z = By + c be the equation to BC referred to 
the axes Oy, Oz, where in bdth cases c = OC. From P any 
point in the generating line, and from Q where it meets BC, 
draw PN, QM, parallel to the axis of z, and join MN. Then 
the plane QN is parallel to zx, and therefore MN is parallel 
to Ox (Euc. xi, 15, 16) ; and consequently PQ is inclined 
to MN at the same angle that AC is to Ox; and since MN, 
NP, are the co-ordinates of P in the plane QN, 
PN = A . MN + QM.