Last updated at March 11, 2017 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 24 Important You are here

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Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.