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

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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’™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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.