Plane

Question 1 (a)

Question 1 (b)

Question 1 (c) Important

Question 1 (d) Important

Question 2

Question 3 (a)

Question 3 (b)

Question 3 (c) Important

Question 4 (a) Important

Question 4 (b) You are here

Question 4 (c)

Question 4 (d) Important

Question 5 (a) Important

Question 5 (b)

Question 6 (a) Important

Question 6 (b)

Question 7

Question 8

Question 9

Question 10 Important

Question 11 Important

Question 12 Important

Question 13 (a) Important

Question 13 (b) Important

Question 13 (c)

Question 13 (d)

Question 13 (e)

Question 14 (a) Important

Question 14 (b)

Question 14 (c)

Question 14 (d) Important

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at April 16, 2024 by Teachoo

Question 4 In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. (b) 3y + 4z 6 = 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 3y + 4z 6 = 0 3y + 4z = 6 0x + 3y + 4z = 6 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 0 = 1 3 = 1 4 = k x1 = 0 , y1 = 3k , z1 = 4k Also, point P(x1, y1, z1) lies in the given plane. Putting P (0, 3k, 4k) in 0x + 3y + 4z = 6, 0 (k) + 3(3k) + 4(4k) = 6 25k = 6 k = 6 25 So, 1 = 0 1 = 3k = 3 6 25 = 18 25 1 = 4k = 4 6 25 = 24 25 Therefore, coordinates of foot of perpendicular are ,