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

Distance of point from plane

Chapter 11 Class 12 Three Dimensional Geometry

Concept wise

- Direction cosines and ratios
- Equation of line - given point and //vector
- Equation of line - given 2 points
- Angle between two lines - Vector
- Angle between two lines - Cartisian
- Angle between two lines - Direction ratios or cosines
- Shortest distance between two skew lines
- Shortest distance between two parallel lines
- Equation of plane - In Normal Form
- Equation of plane - Prependicular to Vector & Passing Through Point
- Equation of plane - Passing Through 3 Non Collinear Points
- Equation of plane - Intercept Form
- Equation of plane - Passing Through Intersection Of Planes
- Coplanarity of 2 lines
- Angle between two planes
- Distance of point from plane
- Angle between Line and Plane
- Equation of line under planes condition
- Point with Lines and Planes

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.