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Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at April 16, 2024 by Teachoo

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(๐ฅ_1, ๐ฆ_1, ๐ง_1) B(๐ฅ_2, ๐ฆ_2, ๐ง_2) and C (๐ฅ_3, ๐ฆ_3, ๐ง_3) is |โ 8(๐โ๐_๐&๐โ๐_๐&๐โ๐_๐@๐_๐โ๐_๐&๐_๐โ๐_๐&๐_๐โ๐_๐@๐_๐โ๐_๐&๐_๐โ๐_๐&๐_๐โ๐_๐ )| = 0 Given, the three points are A(3, โ1, 2) ๐ฅ_1= 3, ๐ฆ_1= โ1, ๐ง_1= 2 B(5, 2, 4) ๐ฅ_2= 5, ๐ฆ_2 = 2, ๐ง_2= 4 C(โ1, โ1, 6) ๐ฅ_3= โ1, ๐ฆ_3= โ1, ๐ง_3= 6 Equation of plane is |โ 8(๐ฅโ3&๐ฆโ(โ1)&๐งโ2@5โ3&2โ(โ1)&4โ2@โ1โ3&โ1โ(โ1)&6โ2)| = 0 |โ 8(๐ฅโ3&๐ฆ+1&๐งโ2@2&3&2@โ4&0&4)| = 0 (x โ 3)[(3ร4)โ(0ร2)] โ (y + 1) [(2ร4)โ(โ4ร2)] + (z โ 2) [(2ร0)โ(โ4ร3)] (x โ 3)[12โ0] โ (y + 1) [8+8] +(๐งโ2) [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(๐ฅ_1, ๐ฆ_1, ๐ง_1) and the plane Ax + By + Cz = D is |(๐จ๐_๐ + ๐ฉ๐_๐ + ๐ช๐_๐โ ๐ซ)/โ(๐จ^๐ + ๐ฉ^๐ + ๐ช^๐ )| Given, the point is P(6, 5, 9) So, ๐ฅ_1= 6 , ๐ฆ_1= 5, ๐ง_1= 9 The equation of plane is 3x โ 4y + 3z = 19 Comparing with Ax + By + Cz = D, A = 3, B = โ4, C = 3, D = 19 Now, Distance of the point from the plane = |((3 ร 6) + (โ4 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/โ34| = 6/โ34 = 6/โ34 ร โ34/โ34 = (6โ34)/34 = (๐โ๐๐)/๐๐