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Ex 11.3, 14 (b) - Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

Last updated at Aug. 18, 2020 by

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Ex 11.3, 14 (b) - Find distance of (3, -2, 1) from plane 2x-y+2z+3=0

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

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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 |(𝑨𝒙_𝟏 + γ€–π‘©π’šγ€—_𝟏 +γ€– π‘ͺ𝒛〗_𝟏 βˆ’ 𝑫)/√(𝑨^𝟐 + 𝑩^𝟐 + π‘ͺ^𝟐 )| Given, the point is (3, βˆ’2, 1) So, π‘₯_1 = 3, 𝑦_1 = βˆ’2, 𝑧_1 = 1 a And the equation of the plane is 2x βˆ’ y + 2z + 3 = 0 2x βˆ’ y + 2z = βˆ’3 βˆ’(2x βˆ’ y + 2z) = 3 βˆ’2x + y βˆ’ 2z = 3 Comparing with Ax + By + Cz = D, A = βˆ’2, B = 1, C = βˆ’2, D = 3 Now, Distance of the point form the plane = |((βˆ’2 Γ— 3) + (1 Γ— βˆ’2) + (βˆ’2 Γ— 1) βˆ’ 3)/√((βˆ’2)^2 + 1^2 + (2)^2 )| = |((βˆ’6) + (βˆ’2) + (βˆ’2) βˆ’ 3)/√(4 + 1 + 4)| = |(βˆ’13)/√9| = |(βˆ’13)/3| = πŸπŸ‘/πŸ‘

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