Example 28 - Chapter 11 Class 12 Three Dimensional Geometry

Transcript

Example 28 Find the distance between the point P(6, 5, 9) and the plane determined by the points A (3, – 1, 2), B (5, 2, 4) and C(– 1, – 1, 6). The equation of a plane passing through points A( 𝑥﷮1﷯, 𝑦﷮1﷯, 𝑧﷮1﷯) B( 𝑥﷮2﷯, 𝑦﷮2﷯, 𝑧﷮2﷯) and C ( 𝑥﷮3﷯, 𝑦﷮3﷯, 𝑧﷮3﷯) is 𝒙− 𝒙﷮𝟏﷯﷮𝒚− 𝒚﷮𝟏﷯﷮𝒛− 𝒛﷮𝟏﷯﷮ 𝒙﷮𝟐﷯− 𝒙﷮𝟏﷯﷮ 𝒚﷮𝟐﷯− 𝒚﷮𝟏﷯﷮ 𝒛﷮𝟐﷯− 𝒛﷮𝟏﷯﷮ 𝒙﷮𝟑﷯− 𝒙﷮𝟏﷯﷮ 𝒚﷮𝟑﷯− 𝒚﷮𝟏﷯﷮ 𝒛﷮𝟑﷯− 𝒛﷮𝟏﷯﷯﷯ = 0 Given, the three points are Equation of plane is 𝑥−3﷮𝑦−(−1)﷮𝑧−2﷮5−3﷮2−(−1)﷮4−2﷮−1−3﷮−1−(−1)﷮6−2﷯﷯ = 0 𝑥−3﷮𝑦+1﷮𝑧−2﷮2﷮3﷮2﷮−4﷮0﷮4﷯﷯ = 0 (x − 3) 3×4﷯−(0×2)﷯ − (y + 1) (2×4)− −4×2﷯﷯ + (z − 2) 2×0﷯−(−4×3)﷯ (x − 3) 12−0﷯ − (y + 1) 8+8﷯ +(𝑧−2) 0+12﷯ = 0 12(x − 3) – 16(y + 1) + 12(z − 2) = 0 3(x − 3) – 4(y + 1) + 3(z − 2) = 0 3x − 9 – 4y – 4 + 3z − 6 = 0 3x – 4y + 3z − 19 = 0 Therefore, equation of plane is 3x – 4y + 3z = 19 Now, the distance between a point P( 𝑥﷮1﷯, 𝑦﷮1﷯, 𝑧﷮1﷯) and the plane Ax + By + Cz = D is 𝑨 𝒙﷮𝟏﷯ + 𝑩 𝒚﷮𝟏﷯ + 𝑪 𝒛﷮𝟏﷯− 𝑫﷮ ﷮ 𝑨﷮𝟐﷯ + 𝑩﷮𝟐﷯ + 𝑪﷮𝟐﷯﷯﷯﷯ Given, the point is P(6, 5, 9) So, 𝑥﷮1﷯= 6 , 𝑦﷮1﷯= 5 𝑧﷮1﷯= 9 The equation of plane is 3x – 4y + 3z = 19 Comparing with Ax + By + Cz = D, A = 3, B = –4, C = 3 d = 19 Now, Distance of the point from the plane = 3 × 6﷯ + −4 × 5﷯ + 3 × 9﷯ − 19﷮ ﷮ 3﷮2﷯ + 4﷮2﷯ + 3﷮2﷯﷯﷯﷯ = 18 − 20 + 27 − 19﷮ ﷮9 + 16 + 9﷯﷯﷯ = 6﷮ ﷮34﷯﷯﷯ = 6﷮ ﷮34﷯﷯ = 6﷮ ﷮34﷯﷯ × ﷮34﷯﷮ ﷮34﷯﷯ = 6 ﷮34﷯﷮34﷯ = 𝟑 ﷮𝟑𝟒﷯﷮𝟏𝟕﷯