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 |(𝑨𝒙_𝟏 + 〖𝑩𝒚〗_𝟏 +〖 𝑪𝒛〗_𝟏 − 𝑫)/√(𝑨^𝟐 + 𝑩^𝟐 + 𝑪^𝟐 )| Given, the point is (−6, 0, 0) So, 𝑥_1 = −6, 𝑦_1 = 0, 𝑧_1 = 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 )/√(2^2+(−3)^2+6^2 )| = |(−12 + 0 + 0 − 2)/√(4 + 9 + 36)| = |(−14)/√49| = |(−14)/7| = |−2| = 2

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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