Last updated at Dec. 16, 2024 by Teachoo
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 |(๐จ๐_๐ + ใ๐ฉ๐ใ_๐ +ใ ๐ช๐ใ_๐ โ ๐ซ)/โ(๐จ^๐ + ๐ฉ^๐ + ๐ช^๐ )| Given, the point is (โ6, 0, 0) So, ๐ฅ_1 = โ6, ๐ฆ_1 = 0, ๐ง_1 = 0 and the equation of plane is 2x โ 3y + 6z โ 2 = 0 2x โ 3y + 6z = 2 Comparing with Ax + By + Cz = D, A = 2, B = โ3, C = 6 D = 3 Now, Distance of the point from the plane = |((2 ร โ6) + (โ3 ร 0) + (6 ร 0)โ 2 )/โ(2^2+(โ3)^2+6^2 )| = |(โ12 + 0 + 0 โ 2)/โ(4 + 9 + 36)| = |(โ14)/โ49| = |(โ14)/7| = |โ2| = 2
Plane
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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 and Computer Science at Teachoo