Last updated at Dec. 24, 2019 by Teachoo

Transcript

Example 24 Find the distance of a point (2, 5, –3) from the plane 𝑟 . (6 𝑖 – 3 𝑗 + 2 𝑘) = 4 The distance of a point with position vector 𝑎 from the plane 𝑟. 𝑛 = d, where 𝑛 is the normal to the plane is 𝒂. 𝒏 − 𝒅 𝒏 Given, the point is (2, 5, −3) So, 𝑎 = 2 𝑖 + 5 𝑗 − 3 𝑘 The equation of plane is 𝑟.(6 𝑖 − 3 𝑗 + 2 𝑘) = 4 Comparing with 𝑟. 𝑛 = d, 𝑛 = 6 𝑖 − 3 𝑗 + 2 𝑘 & d = 4 Distance of point from plane = 𝑎. 𝑛 − 𝑑 𝑛 = 2 𝑖 + 5 𝑗 − 3 𝑘. 6 𝑖 − 3 𝑗 + 2 𝑘 − 4 62 + −32 + 22 = 2 × 6 + 5 × −3 + −3 × 2 − 4 36 + 9 + 4 = 12 − 15 − 6 − 4 49 = −137 = 𝟏𝟑𝟕

Examples

Example 1

Example, 2 Important

Example, 3

Example, 4 Important

Example, 5 Important

Example, 6 Important

Example, 7

Example 8

Example, 9 Deleted for CBSE Board 2022 Exams

Example 10 Important Deleted for CBSE Board 2022 Exams

Example 11

Example 12 Important

Example 13 Important

Example 14

Example 15

Example 16 Important

Example 17

Example 18

Example 19 Important

Example 20 Important

Example 21 Important

Example 22 Deleted for CBSE Board 2022 Exams

Example 23 Important Deleted for CBSE Board 2022 Exams

Example 24 You are here

Example, 25 Important Deleted for CBSE Board 2022 Exams

Example 26

Example 27 Important

Example 28 Important

Example 29 Important

Example 30 Important

Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

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 10 years. He provides courses for Maths and Science at Teachoo.