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Example 24 - Chapter 11 Class 12 Three Dimensional Geometry
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
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Example, 3
Example, 4 Important
Example, 5 Important
Example, 6 Important
Example, 7
Example 8
Example, 9 Not in Syllabus - CBSE Exams 2021
Example 10 Not in Syllabus - CBSE Exams 2021
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 Not in Syllabus - CBSE Exams 2021
Example 23 Important Not in Syllabus - CBSE Exams 2021
Example 24 You are here
Example, 25 Important
Example 26
Example 27 Important
Example 28 Important
Example 29 Important
Example 30 Important
Chapter 11 Class 12 Three Dimensional Geometry
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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.