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

Last updated at Aug. 24, 2021 by Teachoo

Last updated at Aug. 24, 2021 by Teachoo

Transcript

Ex 11.3, 4 In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin. (c) x + y + z = 1 Assume a point P (x1, y1, z1) on the plane. Since perpendicular to plane is parallel to normal vector Vector is parallel to normal vector to the plane. Given, equation of the plane is x + y + z = 1 1x + 1y + 1z = 1 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 1 = 1 1 = 1 1 = k x1 = y1 = z1 = k Also, point P(x1, y1, z1) lies in the given plane. Putting P (k, k, k) in x + y + z = 1, k + k + k = 1 3k = 1 k = 1 3 So, 1 = k = 1 3 , 1 = k = 1 3 , 1 = k = 1 3 Therefore, coordinate of foot of perpendicular are , ,

Ex 11.3

Ex 11.3, 1 (a)

Ex 11.3, 1 (b)

Ex 11.3, 1 (c) Important

Ex 11.3, 1 (d) Important

Ex 11.3, 2

Ex 11.3, 3 (a)

Ex 11.3, 3 (b)

Ex 11.3, 3 (c) Important

Ex 11.3, 4 (a) Important

Ex 11.3, 4 (b)

Ex 11.3, 4 (c) You are here

Ex 11.3, 4 (d) Important

Ex 11.3, 5 (a) Important

Ex 11.3, 5 (b)

Ex 11.3, 6 (a) Important

Ex 11.3, 6 (b)

Ex 11.3, 7

Ex 11.3, 8

Ex 11.3, 9

Ex 11.3, 10 Important

Ex 11.3, 11 Important

Ex 11.3, 12 Important Deleted for CBSE Board 2022 Exams

Ex 11.3, 13 (a) Important Deleted for CBSE Board 2022 Exams

Ex 11.3, 13 (b) Important

Ex 11.3, 13 (c)

Ex 11.3, 13 (d)

Ex 11.3, 13 (e) Deleted for CBSE Board 2022 Exams

Ex 11.3, 14 (a) Important

Ex 11.3, 14 (b)

Ex 11.3, 14 (c)

Ex 11.3, 14 (d) Important

Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.