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Example 13 - Find vector equation of plane which is at distance

Example 13 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Example 13 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Example 13 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4

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Example 13 Find the vector equation of the plane which is at a distance of 6/√29 from the origin and its normal vector from the origin is 2𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ + 4π‘˜ Μ‚.Vector equation of a plane at a distance β€˜d’ from the origin and unit vector to normal from origin 𝑛 Μ‚ is 𝒓 βƒ—.𝒏 Μ‚ = d Unit vector of 𝑛 βƒ— = 𝑛 Μ‚ = 1/|𝑛 βƒ— | (𝑛 βƒ—) Now, distance from origin = d = 6/√29 𝑛 βƒ— = 2𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ + 4π‘˜ Μ‚ Magnitude of 𝑛 Μ‚ = √(22+(βˆ’3)^2+4^2 ) |𝑛 βƒ— | = √(4+9+16) = √29 Now, 𝑛 Μ‚ = 1/|𝑛 βƒ— | " (" 𝑛 βƒ—")" = 1/√29 " (2" 𝑖 Μ‚βˆ’"3" 𝑗 Μ‚+"4" π‘˜ Μ‚")" = 2/√29 𝑖 Μ‚ βˆ’ 3/√29 𝑗 Μ‚ + 4/√29 π‘˜ Μ‚ Vector equation of plane is π‘Ÿ βƒ—.𝑛 Μ‚ = d 𝒓 βƒ— . (𝟐/βˆšπŸπŸ— π’Š Μ‚βˆ’πŸ‘/βˆšπŸπŸ— " " 𝒋 Μ‚+ πŸ’/βˆšπŸπŸ— " " π’Œ Μ‚ ) = πŸ”/βˆšπŸπŸ— Cartesian equation Equation of a plane in Cartesian form which is at a distance β€˜d’ from the origin and has a normal vector 𝑛 βƒ— = π‘Žπ‘– Μ‚ + b𝑗 Μ‚ + cπ‘˜ Μ‚ is lx + my + nz = d where l, m, n are direction cosines of 𝑛 βƒ— l = π‘Ž/√(π‘Ž^2 + 𝑏^2 + 𝑐^2 ) , m = 𝑏/√(π‘Ž^2 + 𝑏^2 + 𝑐^2 ) , n = 𝑐/√(π‘Ž^2 + 𝑏^2 + 𝑐^2 ) Distance form origin = d = 6/√29 𝑛 βƒ— = 2𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ + 4π‘˜ Μ‚ Comparing with 𝑛 βƒ— = π‘Žπ‘– Μ‚ + 𝑏𝑗 Μ‚ + π‘π‘˜ Μ‚, π‘Ž = 2, b = βˆ’3, c = 4 Also, √(π‘Ž2+𝑏2+𝑐2) = √(22+(βˆ’3)2+42) = √(4+9+16) = √29 So, direction cosines are l = 2/√29 , m = ( βˆ’3)/√29 , n = 4/√29 ∴ Cartesian equations of plane is lx + my + nz = d 2/√29 x + (( βˆ’3)/√29)y + 4/√29 z = 6/√29 2x βˆ’ 3y + 4z = 6 Therefore, the equation of plane in Cartesian form is 2x βˆ’ 3y + 4z = 6

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.