72 PRACTICAL MATHEMATICS 



OS I/ 2> 



Now -+r+- = lis the equation to the plane. 

 a b c 



Then x - + y ^ + % * ^ = P 



a 6 c 



or a? cos a + ?/ cos [5 + z cos y = p 



Ix + my + nz = p 



thus expressing the equation to a plane in terms of the direction 

 cosines and the length of the perpendicular drawn from the origin 

 to the plane. 



Also a, (3, and y are the angles made by the line OP with the 

 axes OX, OY, and OZ respectively, but OP is perpendicular to 

 the plane. Hence the angles the plane makes with these axes 

 will be the complements of these angles, and therefore the plane 

 makes angles (90- a), (90- p), and (90- y) with the axes 

 OX, OY, and OZ respectively. 



Example. For the plane 2lx + 35y + I5z 105 = 0. Find : 



(1) The intercepts on the axes of reference. 



(2) The length of OP, the perpendicular drawn from the origin 

 to the plane. 



(3) The angles which OP makes with the axes of reference. 



(4) The angles which the plane makes with the axes of reference. 



(5) The co-ordinates of P. 



2lx + S5y + I5z = 105 



T-i x y z 



Then 5 + I+" 1 



Hence the plane makes intercepts of 5, 3, and 7 on the axes OX, 

 OY, and OZ respectively. 



If p is the length of OP and a, (3, and y are the angles OP 

 makes with the axes OX, OY, and OZ, 



then / = cos a = ^ 



5 



rt 

 m = cos (3 = ^ 



P 

 n = cos Y = - 



i 7 



But l 2 +m z + n 2 = 1 



p* p z p* 

 Hence | + | + I_ = 1 



9 x 25 x 49 

 *" 

 p = 2-414 



