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

Transcript

Example 13 Find the vector equation of the plane which is at a distance of 6/โˆš29 from the origin and its normal vector from the origin is 2๐‘– ฬ‚ โˆ’ 3๐‘— ฬ‚ + 4๐‘˜ ฬ‚.Vector equation of a plane at a distance โ€˜dโ€™ from the origin and unit vector to normal from origin ๐‘› ฬ‚ is ๐’“ โƒ—.๐’ ฬ‚ = d Unit vector of ๐‘› โƒ— = ๐‘› ฬ‚ = 1/|๐‘› โƒ— | (๐‘› โƒ—) Now, distance from origin = d = 6/โˆš29 ๐‘› โƒ— = 2๐‘– ฬ‚ โˆ’ 3๐‘— ฬ‚ + 4๐‘˜ ฬ‚ Magnitude of ๐‘› ฬ‚ = โˆš(22+(โˆ’3)^2+4^2 ) |๐‘› โƒ— | = โˆš(4+9+16) = โˆš29 Now, ๐‘› ฬ‚ = 1/|๐‘› โƒ— | " (" ๐‘› โƒ—")" = 1/โˆš29 " (2" ๐‘– ฬ‚โˆ’"3" ๐‘— ฬ‚+"4" ๐‘˜ ฬ‚")" = 2/โˆš29 ๐‘– ฬ‚ โˆ’ 3/โˆš29 ๐‘— ฬ‚ + 4/โˆš29 ๐‘˜ ฬ‚ Vector equation of plane is ๐‘Ÿ โƒ—.๐‘› ฬ‚ = d ๐’“ โƒ— . (๐Ÿ/โˆš๐Ÿ๐Ÿ— ๐’Š ฬ‚โˆ’๐Ÿ‘/โˆš๐Ÿ๐Ÿ— " " ๐’‹ ฬ‚+ ๐Ÿ’/โˆš๐Ÿ๐Ÿ— " " ๐’Œ ฬ‚ ) = ๐Ÿ”/โˆš๐Ÿ๐Ÿ— Cartesian equation Equation of a plane in Cartesian form which is at a distance โ€˜dโ€™ from the origin and has a normal vector ๐‘› โƒ— = ๐‘Ž๐‘– ฬ‚ + b๐‘— ฬ‚ + c๐‘˜ ฬ‚ is lx + my + nz = d where l, m, n are direction cosines of ๐‘› โƒ— l = ๐‘Ž/โˆš(๐‘Ž^2 + ๐‘^2 + ๐‘^2 ) , m = ๐‘/โˆš(๐‘Ž^2 + ๐‘^2 + ๐‘^2 ) , n = ๐‘/โˆš(๐‘Ž^2 + ๐‘^2 + ๐‘^2 ) Distance form origin = d = 6/โˆš29 ๐‘› โƒ— = 2๐‘– ฬ‚ โˆ’ 3๐‘— ฬ‚ + 4๐‘˜ ฬ‚ Comparing with ๐‘› โƒ— = ๐‘Ž๐‘– ฬ‚ + ๐‘๐‘— ฬ‚ + ๐‘๐‘˜ ฬ‚, ๐‘Ž = 2, b = โˆ’3, c = 4 Also, โˆš(๐‘Ž2+๐‘2+๐‘2) = โˆš(22+(โˆ’3)2+42) = โˆš(4+9+16) = โˆš29 So, direction cosines are l = 2/โˆš29 , m = ( โˆ’3)/โˆš29 , n = 4/โˆš29 โˆด Cartesian equations of plane is lx + my + nz = d 2/โˆš29 x + (( โˆ’3)/โˆš29)y + 4/โˆš29 z = 6/โˆš29 2x โˆ’ 3y + 4z = 6 Therefore, the equation of plane in Cartesian form is 2x โˆ’ 3y + 4z = 6

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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.