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Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

 

Ex 11.3, 14 (c) - Find distance of (2, 3, -5) from plane x + 2y - 2z=9

Ex 11.3, 14 (c) - Chapter 11 Class 12 Three Dimensional Geometry - Part 2


Transcript

Question 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 |(𝑨𝒙_𝟏 + 〖𝑩𝒚〗_𝟏 +〖 𝑪𝒛〗_𝟏 − 𝑫)/√(𝑨^𝟐 + 𝑩^𝟐 + 𝑪^𝟐 )| Given, the point is (2, 3, −5) So, 𝑥_1= 2, 𝑦_1= 3, 𝑧_1= −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)/√(1^2 + 2^2 + 〖(−2)〗^2 )| = |(2 + 6 + 10 − 9)/√(1 + 4 + 4)| = |(18 − 9)/√9| = 9/3 = 3

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.