    1. Chapter 11 Class 12 Three Dimensional Geometry
2. Serial order wise
3. Examples

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﷯

Examples 