question_answer 19)

In each of the following, determine the direction cosines of the normal to the plane and the distance from the origin: (a) z = 2 (b) x + y + z = 1 (c) 2x + 3y – z = 5 (d) 5y + 8 = 0  

Answer:

(a) z = 2       It can be written as 0x + 0y + 1z = 2. Compare it with lx + my + nz = d       We get l = 0, m = 0, n = 1 and d = 2        direction cosines of normal to the plane are 0, 0, 1 and distance from origin = 2       (b) x + y = z = 1       Here direction ratios of normal to the plane are 1, 1, 1 direction cosines of it are             i.e.,       On dividing x + y + z = 1 by             It is of the form lx + my + nz = d             i.e., distance of the plane from origin  units.       (c) 2x + 3y ? z = 5       Here direction ratios of normal to the plane are 2, 3, ?1 direction cosines of it are                   Also on dividing 2x + 3y ? z = 5 by , we get             It is the form lx + my + nz =d             i.e., distance of the plane from origin             (d) 5y + 8 = 0, it can be written as       0x + 5y + 0z + 8 = 0       Here direction ratios of normal to the plane are 0, 5, 0.       Direction cosines of it are             On  dividing 5y + 8 = 0 by 5, we get             or ?y + 0z =       It is of the form lx + my + nz = d,       i.e. distance of the plane from origin  units.