Example 28 - Find distance between point P(6, 5, 9) and plane

Example  28 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Example  28 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3 Example  28 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4

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

Transcript

Example 28 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’™โˆ’๐’™_๐Ÿ&๐’šโˆ’๐’š_๐Ÿ&๐’›โˆ’๐’›_๐Ÿ@๐’™_๐Ÿโˆ’๐’™_๐Ÿ&๐’š_๐Ÿโˆ’๐’š_๐Ÿ&๐’›_๐Ÿโˆ’๐’›_๐Ÿ@๐’™_๐Ÿ‘โˆ’๐’™_๐Ÿ&๐’š_๐Ÿ‘โˆ’๐’š_๐Ÿ&๐’›_๐Ÿ‘โˆ’๐’›_๐Ÿ )| = 0 Given, the three points are A(3, โˆ’1, 2) ๐‘ฅ_1= 3, ๐‘ฆ_1= โˆ’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= โˆ’1, ๐‘ฆ_3= โˆ’1, ๐‘ง_3= 6 Equation of plane is |โ– 8(๐‘ฅโˆ’3&๐‘ฆโˆ’(โˆ’1)&๐‘งโˆ’2@5โˆ’3&2โˆ’(โˆ’1)&4โˆ’2@โˆ’1โˆ’3&โˆ’1โˆ’(โˆ’1)&6โˆ’2)| = 0 |โ– 8(๐‘ฅโˆ’3&๐‘ฆ+1&๐‘งโˆ’2@2&3&2@โˆ’4&0&4)| = 0 (x โˆ’ 3)[(3ร—4)โˆ’(0ร—2)] โˆ’ (y + 1) [(2ร—4)โˆ’(โˆ’4ร—2)] + (z โˆ’ 2) [(2ร—0)โˆ’(โˆ’4ร—3)] (x โˆ’ 3)[12โˆ’0] โˆ’ (y + 1) [8+8] +(๐‘งโˆ’2) [0+12] = 0 12(x โˆ’ 3) โ€“ 16(y + 1) + 12(z โˆ’ 2) = 0 3(x โˆ’ 3) โ€“ 4(y + 1) + 3(z โˆ’ 2) = 0 3x โˆ’ 9 โ€“ 4y โ€“ 4 + 3z โˆ’ 6 = 0 3x โ€“ 4y + 3z โˆ’ 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Ÿ + ๐‘ฉ๐’š_๐Ÿ + ๐‘ช๐’›_๐Ÿโˆ’ ๐‘ซ)/โˆš(๐‘จ^๐Ÿ + ๐‘ฉ^๐Ÿ + ๐‘ช^๐Ÿ )| 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) + (โˆ’4 ร— 5) + (3 ร— 9) โˆ’ 19)/โˆš(3^2 + 4^2 + 3^2 )| =|(18 โˆ’ 20 + 27 โˆ’ 19)/โˆš(9 + 16 + 9)| =|6/โˆš34| = 6/โˆš34 = 6/โˆš34 ร— โˆš34/โˆš34 = (6โˆš34)/34 = (๐Ÿ‘โˆš๐Ÿ‘๐Ÿ’)/๐Ÿ๐Ÿ•

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.