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

Transcript

Ex 11.3, 14 In the following cases, find the distance of each of the given points from the corresponding given plane. The distance of the point (x1, y1, z1) from the plane Ax + By + Cz = D is |(๐‘จ๐’™_๐Ÿ + ใ€–๐‘ฉ๐’šใ€—_๐Ÿ +ใ€– ๐‘ช๐’›ใ€—_๐Ÿ โˆ’ ๐‘ซ)/โˆš(๐‘จ^๐Ÿ + ๐‘ฉ^๐Ÿ + ๐‘ช^๐Ÿ )| Given, the point is (0, 0, 0) So, ๐‘ฅ_1 = 0, ๐‘ฆ_1 = 0, ๐‘ง_1 = 0 and the equation of plane is 3x โˆ’ 4y + 12z = 3 Comparing with Ax + By + Cz = D, A = 3, B = โˆ’4, C = 12, D = 3 Now, Distance of point from the plane is = |((3 ร— 0) + (โˆ’4 ร— 0) + (12 ร— 0) โˆ’ 3)/( โˆš(3^2 + (โˆ’4)^2 + ใ€–12ใ€—^2 ))| = |(0 + 0 + 0 โˆ’ 3)/( โˆš(9 + 16 + 144))| = |3/( โˆš169)| = ๐Ÿ‘/๐Ÿ๐Ÿ‘ Ex 11.3, 14 In the following cases, find the distance of each of the given points from the corresponding given plane. The distance of the point (x1, y1, z1) from the plane Ax + By + Cz = D is |(๐‘จ๐’™_๐Ÿ + ใ€–๐‘ฉ๐’šใ€—_๐Ÿ +ใ€– ๐‘ช๐’›ใ€—_๐Ÿ โˆ’ ๐‘ซ)/โˆš(๐‘จ^๐Ÿ + ๐‘ฉ^๐Ÿ + ๐‘ช^๐Ÿ )| Given, the point is (3, โˆ’2, 1) So, ๐‘ฅ_1 = 3, ๐‘ฆ_1 = โˆ’2, ๐‘ง_1 = 1 a And the equation of the plane is 2x โˆ’ y + 2z + 3 = 0 2x โˆ’ y + 2z = โˆ’3 โˆ’(2x โˆ’ y + 2z) = 3 โˆ’2x + y โˆ’ 2z = 3 Comparing with Ax + By + Cz = D, A = โˆ’2, B = 1, C = โˆ’2, D = 3 Now, Distance of the point form the plane = |((โˆ’2 ร— 3) + (1 ร— โˆ’2) + (โˆ’2 ร— 1) โˆ’ 3)/โˆš((โˆ’2)^2 + 1^2 + (2)^2 )| = |((โˆ’6) + (โˆ’2) + (โˆ’2) โˆ’ 3)/โˆš(4 + 1 + 4)| = |(โˆ’13)/โˆš9| = |(โˆ’13)/3| = ๐Ÿ๐Ÿ‘/๐Ÿ‘ Ex 11.3, 14 In the following cases, find the distance of each of the given points from the corresponding given plane. The distance of the point (x1, y1, z1) from the plane Ax + By + Cz = D is |(๐‘จ๐’™_๐Ÿ + ใ€–๐‘ฉ๐’šใ€—_๐Ÿ +ใ€– ๐‘ช๐’›ใ€—_๐Ÿ โˆ’ ๐‘ซ)/โˆš(๐‘จ^๐Ÿ + ๐‘ฉ^๐Ÿ + ๐‘ช^๐Ÿ )| Given, the point is (2, 3, โˆ’5) So, ๐‘ฅ_1= 2, ๐‘ฆ_1= 3, ๐‘ง_1= โˆ’5 and the equation of the plane is 1x + 2y โˆ’ 2z = 9 Comparing with Ax + By + Cz = D, A = 1, B = 2, C = โˆ’2, D = 9 Now, Distance of the point from the plane = |((1 ร— 2) + (2 ร— 3) + (โˆ’2 ร— โˆ’5) โˆ’ 9)/โˆš(1^2 + 2^2 + ใ€–(โˆ’2)ใ€—^2 )| = |(2 + 6 + 10 โˆ’ 9)/โˆš(1 + 4 + 4)| = |(18 โˆ’ 9)/โˆš9| = 9/3 = 3 Ex 11.3, 14 In the following cases, find the distance of each of the given points from the corresponding given plane. The distance of the point (x1, y1, z1) from the plane Ax + By + Cz = D is |(๐‘จ๐’™_๐Ÿ + ใ€–๐‘ฉ๐’šใ€—_๐Ÿ +ใ€– ๐‘ช๐’›ใ€—_๐Ÿ โˆ’ ๐‘ซ)/โˆš(๐‘จ^๐Ÿ + ๐‘ฉ^๐Ÿ + ๐‘ช^๐Ÿ )| Given, the point is (โˆ’6, 0, 0) So, ๐‘ฅ_1 = โˆ’6, ๐‘ฆ_1 = 0, ๐‘ง_1 = 0 and the equation of plane is 2x โˆ’ 3y + 6z โˆ’ 2 = 0 2x โˆ’ 3y + 6z = 2 Comparing with Ax + By + Cz = D, A = 2, B = โˆ’3, C = 6 D = 3 Now, Distance of the point from the plane = |((2 ร— โˆ’6) + (โˆ’3 ร— 0) + (6 ร— 0)โˆ’ 2 )/โˆš(2^2+(โˆ’3)^2+6^2 )| = |(โˆ’12 + 0 + 0 โˆ’ 2)/โˆš(4 + 9 + 36)| = |(โˆ’14)/โˆš49| = |(โˆ’14)/7| = |โˆ’2| = 2

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.