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

Last updated at Aug. 18, 2020 by Teachoo

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

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 (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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Ex 11.3, 14 (c) You are here

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Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

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