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

Serial order wise

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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.