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Ex 11.3, 14 (d) - Find distance of (-6, 0, 0) from plane 2x-3y+6z-2=0

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


Transcript

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 (βˆ’6, 0, 0) So, π‘₯_1 = βˆ’6, 𝑦_1 = 0, 𝑧_1 = 0 and the equation of plane is 2x βˆ’ 3y + 6z βˆ’ 2 = 0 2x βˆ’ 3y + 6z = 2 Comparing with Ax + By + Cz = D, A = 2, B = βˆ’3, C = 6 D = 3 Now, Distance of the point from the plane = |((2 Γ— βˆ’6) + (βˆ’3 Γ— 0) + (6 Γ— 0)βˆ’ 2 )/√(2^2+(βˆ’3)^2+6^2 )| = |(βˆ’12 + 0 + 0 βˆ’ 2)/√(4 + 9 + 36)| = |(βˆ’14)/√49| = |(βˆ’14)/7| = |βˆ’2| = 2

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