question_answer 22)

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. (a) 2x + 3y + 4z – 12 = 0 (b) 3y + 4z – 6 = 0 (c) z + y + z = 1 (d) 5y + 8 = 0  

Answer:

Let the co-ordinates of the foot of the perpendicular from the original to the plane is P(x1, y1, z1).       direction ratios of line OP are x1 ? 0, y1 ? 0, z1 ? 0, i.e., x1, y1, z1­,       (a) Equation of plane is 2x + 3y + 4z = 12       Divide it by             Its direction cosines are       As direction ratios area proportional to direction cosines                                 ? (1)       Also P(x1, y1, z1) lies on plane       2x + 3y + 4z =12                         Put              foot of perpendicular is       (b) Equation of plane is 3y + 4z ? 6 = 0       Dividing i.e. 5             Its direction cosines are       As direction ratios are proportional to direction cosines.                             .. (1)       Also P(x1, y1, z1) lies on the plane       3y + 4z ? 6 = 0                               Put              foot of perpendicular is       (c) Equation of plane is x + y + z = 1       Divide it by             Its direction cosines are                  As direction ratios are proportional to direction cosines.                   Also P(x1, y1, z1) lies on the plane            x + y + z = 1                                                          Put , we get              foot of perpendicular is (d) Equation of plane is 5y + 8 = 0, divide it by       Its direction cosines are 0, ?1, 0.                  As direction ratios are proportional to direction cosines       Also P(x1, y1, z1­) lies on the plane 5y + 8 = 0 Put K =  we get         foot of perpendicular is