Examples

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

### Transcript

Question 18 Find the distance between the point P(6, 5, 9) and the plane determined by the points A (3, โ 1, 2), B (5, 2, 4) and C(โ 1, โ 1, 6).The equation of a plane passing through points A(๐ฅ_1, ๐ฆ_1, ๐ง_1) B(๐ฅ_2, ๐ฆ_2, ๐ง_2) and C (๐ฅ_3, ๐ฆ_3, ๐ง_3) is |โ 8(๐โ๐_๐&๐โ๐_๐&๐โ๐_๐@๐_๐โ๐_๐&๐_๐โ๐_๐&๐_๐โ๐_๐@๐_๐โ๐_๐&๐_๐โ๐_๐&๐_๐โ๐_๐ )| = 0 Given, the three points are A(3, โ1, 2) ๐ฅ_1= 3, ๐ฆ_1= โ1, ๐ง_1= 2 B(5, 2, 4) ๐ฅ_2= 5, ๐ฆ_2 = 2, ๐ง_2= 4 C(โ1, โ1, 6) ๐ฅ_3= โ1, ๐ฆ_3= โ1, ๐ง_3= 6 Equation of plane is |โ 8(๐ฅโ3&๐ฆโ(โ1)&๐งโ2@5โ3&2โ(โ1)&4โ2@โ1โ3&โ1โ(โ1)&6โ2)| = 0 |โ 8(๐ฅโ3&๐ฆ+1&๐งโ2@2&3&2@โ4&0&4)| = 0 (x โ 3)[(3ร4)โ(0ร2)] โ (y + 1) [(2ร4)โ(โ4ร2)] + (z โ 2) [(2ร0)โ(โ4ร3)] (x โ 3)[12โ0] โ (y + 1) [8+8] +(๐งโ2) [0+12] = 0 12(x โ 3) โ 16(y + 1) + 12(z โ 2) = 0 3(x โ 3) โ 4(y + 1) + 3(z โ 2) = 0 3x โ 9 โ 4y โ 4 + 3z โ 6 = 0 3x โ 4y + 3z โ 19 = 0 Therefore, equation of plane is 3x โ 4y + 3z = 19 Now, the distance between a point P(๐ฅ_1, ๐ฆ_1, ๐ง_1) and the plane Ax + By + Cz = D is |(๐จ๐_๐ + ๐ฉ๐_๐ + ๐ช๐_๐โ ๐ซ)/โ(๐จ^๐ + ๐ฉ^๐ + ๐ช^๐ )| Given, the point is P(6, 5, 9) So, ๐ฅ_1= 6 , ๐ฆ_1= 5, ๐ง_1= 9 The equation of plane is 3x โ 4y + 3z = 19 Comparing with Ax + By + Cz = D, A = 3, B = โ4, C = 3, D = 19 Now, Distance of the point from the plane = |((3 ร 6) + (โ4 ร 5) + (3 ร 9) โ 19)/โ(3^2 + 4^2 + 3^2 )| =|(18 โ 20 + 27 โ 19)/โ(9 + 16 + 9)| =|6/โ34| = 6/โ34 = 6/โ34 ร โ34/โ34 = (6โ34)/34 = (๐โ๐๐)/๐๐