Misc 21 - Chapter 11 Class 12 Three Dimensional Geometry

Transcript

Misc 21 Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then 1﷮𝑎2﷯ + 1﷮𝑏2﷯ + 1﷮𝑐2﷯ = 1﷮𝑝2﷯ . Distance of the point ( 𝑥﷮1﷯, 𝑦﷮1﷯, 𝑧﷮1﷯) from the plane Ax + By + Cz = D is 𝑨 𝒙﷮𝟏﷯ + 𝑩 𝒚﷮𝟏﷯ + 𝑪 𝒛﷮𝟏﷯ − 𝑫﷮ ﷮ 𝑨﷮𝟐﷯ + 𝑩﷮𝟐﷯ + 𝑪﷮𝟐﷯﷯﷯﷯ The equation of a plane having intercepts 𝑎, b, c on the x −, y− and z − axis respectively is 𝒙﷮𝒂﷯ + 𝒚﷮𝒃﷯ + 𝒛﷮𝒄﷯ = 1 Comparing with Ax + By + Cz = D, A = 1﷮𝑎﷯ , B = 1﷮𝑏﷯ , C = 1﷮𝑐﷯ , D = 1 Given, the plane is at a distance of ‘𝑝’ units from the origin. So, The point is O(0, 0, 0) So, 𝑥﷮1﷯ = 0, 𝑦﷮1﷯= 0, 𝑧﷮1﷯= 0 Now, Distance = 𝐴 𝑥﷮1﷯ + 𝐵 𝑦﷮1﷯ + 𝐶 𝑧﷮1﷯ − 𝐷﷮ ﷮ 𝐴﷮2﷯ + 𝐵﷮2﷯ + 𝐶﷮2﷯﷯﷯﷯ Putting values 𝑝 = 1﷮𝑎﷯ × 0 + 1﷮𝑏﷯×0 + 1﷮𝑐﷯ ×0−1﷮ ﷮ 1﷮𝑎﷯﷯﷮2﷯+ 1﷮𝑏﷯﷯﷮2﷯+ 1﷮𝑐﷯﷯﷮2﷯﷯﷯﷯ 𝑝 = 0 + 0 + 0 − 1﷮ ﷮ 1﷮ 𝑎﷮2﷯﷯ + 1﷮ 𝑏﷮2﷯﷯ + 1﷮ 𝑐﷮2﷯﷯﷯ ﷯﷯ 𝑝 = −1﷮ ﷮ 1﷮ 𝑎﷮2﷯﷯ + 1﷮ 𝑏﷮2﷯﷯ + 1﷮ 𝑐﷮2﷯﷯﷯ ﷯﷯ 𝑝 = 1﷮ ﷮ 1﷮ 𝑎﷮2﷯﷯ + 1﷮ 𝑏﷮2﷯﷯ + 1﷮ 𝑐﷮2﷯﷯﷯ ﷯ 1﷮𝑝﷯ = ﷮ 1﷮ 𝑎﷮2﷯﷯ + 1﷮ 𝑏﷮2﷯﷯ + 1﷮ 𝑐﷮2﷯﷯﷯ Squaring both sides 1﷮ 𝑝﷮2﷯﷯ = 1﷮ 𝑎﷮2﷯﷯ + 1﷮ 𝑏﷮2﷯﷯ + 1﷮ 𝑐﷮2﷯﷯ Hence proved.