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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 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.