Ex 11.3, 4 (a) - Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

Transcript

Ex 11.3, 4 In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. (a) 2x + 3y + 4z 12 = 0 Assume a point P(x1, y1, z1) on the given plane Since perpendicular to plane is parallel to normal vector Vector is parallel to normal vector to the plane. Given equation of plane is 2x + 3y + 4z 12 = 0 2x + 3y + 4z = 12 Since, and are parallel their direction ratios are proportional. Finding direction ratios Direction ratios are proportional So, 1 2 = 1 2 = 1 2 = k 1 2 = 1 3 = 1 4 = k x1 = 2k , y1 = 3k , z1 = 4k Also, point P(x1, y1, z1) lies in the plane. Putting P (2k, 3k, 4k) in 2x + 3y + 4z = 12, 2(2k) + 3(3k) + 4(4k) = 12 4k + 9k + 16k = 12 29k = 12 k = 12 29 So, 1 = 2k = 2 12 29 = 24 29 1 = 3k = 3 12 29 = 36 29 & 1 = 4k = 4 12 29 = 48 29 Therefore, coordinate of foot of perpendicular are , , ,