    1. Chapter 11 Class 12 Three Dimensional Geometry (Term 2)
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 |■8(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 A(3, −1, 2) 𝑥_1= 3, 𝑦_1= −1, 𝑧_1= 2 B(5, 2, 4) 𝑥_2= 5, 𝑦_2 = 2, 𝑧_2= 4 C(–1, –1, 6) 𝑥_3= −1, 𝑦_3= −1, 𝑧_3= 6 Equation of plane is |■8(𝑥−3&𝑦−(−1)&𝑧−2@5−3&2−(−1)&4−2@−1−3&−1−(−1)&6−2)| = 0 |■8(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 |(𝑨𝒙_𝟏 + 𝑩𝒚_𝟏 + 𝑪𝒛_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 