Subscribe to our Youtube Channel - https://you.tube/teachoo

Last updated at Aug. 18, 2020 by Teachoo

Transcript

Ex 11.3, 14 In the following cases, find the distance of each of the given points from the corresponding given plane. The distance of the point (x1, y1, z1) from the plane Ax + By + Cz = D is |(๐จ๐_๐ + ใ๐ฉ๐ใ_๐ +ใ ๐ช๐ใ_๐ โ ๐ซ)/โ(๐จ^๐ + ๐ฉ^๐ + ๐ช^๐ )| Given, the point is (โ6, 0, 0) So, ๐ฅ_1 = โ6, ๐ฆ_1 = 0, ๐ง_1 = 0 and the equation of plane is 2x โ 3y + 6z โ 2 = 0 2x โ 3y + 6z = 2 Comparing with Ax + By + Cz = D, A = 2, B = โ3, C = 6 D = 3 Now, Distance of the point from the plane = |((2 ร โ6) + (โ3 ร 0) + (6 ร 0)โ 2 )/โ(2^2+(โ3)^2+6^2 )| = |(โ12 + 0 + 0 โ 2)/โ(4 + 9 + 36)| = |(โ14)/โ49| = |(โ14)/7| = |โ2| = 2

Ex 11.3

Ex 11.3, 1
Important

Ex 11.3, 2

Ex 11.3, 3

Ex 11.3, 4 Important

Ex 11.3, 5 Important

Ex 11.3, 6 Important

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 Not in Syllabus - CBSE Exams 2021

Ex 11.3, 13 (a) Important

Ex 11.3, 13 (b)

Ex 11.3, 13 (c)

Ex 11.3, 13 (d)

Ex 11.3, 13 (e)

Ex 11.3, 14 (a) Important

Ex 11.3, 14 (b)

Ex 11.3, 14 (c)

Ex 11.3, 14 (d) You are here

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.