1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise
  3. Examples
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Question 14 - Examples - Chapter 11 Class 12 Three Dimensional Geometry

Last updated at Dec. 16, 2024 by Teachoo

Transcript

Question 14 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﷮ ﷮ 6﷮2﷯ + −3﷯﷮2﷯ + 2﷮2﷯﷯﷯﷯ = 2 × 6﷯ + 5 × −3﷯ + −3 × 2﷯ − 4﷮ ﷮36 + 9 + 4﷯﷯﷯ = 12 − 15 − 6 − 4﷮ ﷮49﷯﷯﷯ = −13﷮7﷯﷯ = 𝟏𝟑﷮𝟕﷯
  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise

    3. Examples

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Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo