Last updated at Sept. 25, 2018 by Teachoo

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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 𝑨 𝒙𝟏 + 𝑩𝒚𝟏 + 𝑪𝒛𝟏 − 𝑫 𝑨𝟐 + 𝑩𝟐 + 𝑪𝟐 Given, the point is (0, 0, 0) So, 𝑥1 = 0, 𝑦1 = 0, 𝑧1 = 0 and the equation of plane is 3x − 4y + 12z = 3 Comparing with Ax + By + Cz = D, A = 3, B = −4, C = 12, D = 3 Now, Distance of point from the plane is = 3 × 0 + −4 × 0 + 12 × 0 − 3 32 + −42 + 122 = 0 + 0 + 0 − 3 9 + 16 + 144 = 3 169 = 𝟑𝟏𝟑 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 𝑨 𝒙𝟏 + 𝑩𝒚𝟏 + 𝑪𝒛𝟏 − 𝑫 𝑨𝟐 + 𝑩𝟐 + 𝑪𝟐 Given, the point is (3, −2, 1) So, 𝑥1 = 3, 𝑦1 = −2, 𝑧1 = 1 And the equation of the plane is 2x − y + 2z + 3 = 0 2x − y + 2z = −3 −(2x − y + 2z) = 3 −2x + y − 2z = 3 Comparing with Ax + By + Cz = D, A = −2, B = 1, C = −2, D = 3 Now, Distance of the point form the plane = −2 × 3 + 1 × −2 + −2 × 1 − 3 −22 + 12 + 22 = −6 + −2 + −2 − 3 4 + 1 + 4 = −13 9 = −133 = 𝟏𝟑𝟑 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 𝑨 𝒙𝟏 + 𝑩𝒚𝟏 + 𝑪𝒛𝟏 − 𝑫 𝑨𝟐 + 𝑩𝟐 + 𝑪𝟐 Given, the point is (2, 3, −5) So, 𝑥1= 2, 𝑦1= 3, 𝑧1= −5 and the equation of the plane is 1x + 2y − 2z = 9 Comparing with Ax + By + Cz = D, A = 1, B = 2, C = −2, D = 9 Now, Distance of the point from the plane = 1 × 2 + 2 × 3 + −2 × −5 − 9 12 + 22 + (−2)2 = 2 + 6 + 10 − 9 1 + 4 + 4 = 18 − 9 9 = 93 = 3 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 𝑨 𝒙𝟏 + 𝑩𝒚𝟏 + 𝑪𝒛𝟏 − 𝑫 𝑨𝟐 + 𝑩𝟐 + 𝑪𝟐 Given, the point is (−6, 0, 0) So, 𝑥1 = −6, 𝑦1 = 0, 𝑧1 = 0 and the equation of plane is 2x − 3y + 6z − 2 = 0 2x − 3y + 6z = 2 Comparing with Ax + By + Cz = D, A = 2, B = −3, C = 6 D = 3 Now, Distance of the point from the plane = 2 × −6 + −3 × 0 + 6 × 0− 2 22+ −32+ 62 = −12 + 0 + 0 − 2 4 + 9 + 36 = −14 49 = −147 = −2 = 2

Chapter 11 Class 12 Three Dimensional Geometry

Example, 3
Important

Ex 11.1, 2 Important

Example, 6 Important

Example, 9 Important

Example 12 Important

Ex 11.2, 5 Important

Ex 11.2, 11 Important

Ex 11.2, 12 Important

Ex 11.2, 14 Important

Ex 11.2, 15 Important

Ex 11.2, 17 Important

Example 20 Important

Example 21 Important

Example 23 Important

Example 24 Important

Example, 25 Important

Ex 11.3, 4 Important

Ex 11.3, 11 Important

Ex 11.3, 12 Important

Ex 11.3, 14 Important You are here

Example 27 Important

Example 29 Important

Example 30 Important

Misc 6 Important

Misc 9 Important

Misc 14 Important

Misc 18 Important

Misc 20 Important

Misc 21 Important

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.