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Example 20 - Equation of plane passing through intersection

Example 20 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Example 20 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Example 20 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4


Transcript

Example 20 Find the vector equation of the plane passing through the intersection of the planes π‘Ÿ βƒ— . (𝑖 Μ‚ + 𝑗 Μ‚ + π‘˜ Μ‚) = 6 and π‘Ÿ βƒ— . (2𝑖 Μ‚ + 3𝑗 Μ‚ + 4π‘˜ Μ‚) = βˆ’ 5, and the point (1, 1, 1).The vector equation of a plane passing through the intersection of planes π‘Ÿ βƒ—. (𝑛1) βƒ— = d1 and π‘Ÿ βƒ—. (𝑛2) βƒ— = d2 and also through the point (x1, y1, z1) is 𝒓 βƒ—.((π’πŸ) βƒ— + πœ†(π’πŸ) βƒ—) = d1 + πœ†d2 Given, the plane passes through 𝒓 βƒ—.(π’Š Μ‚ + 𝒋 Μ‚ + π’Œ Μ‚) = 6 Comparing with π‘Ÿ βƒ—.(𝑛1) βƒ— = d1, (π’πŸ) βƒ— = π’Š Μ‚ + 𝒋 Μ‚ + π’Œ Μ‚ & d1 = 6 𝒓 βƒ—.(2π’Š Μ‚ + 3𝒋 Μ‚ + 4π’Œ Μ‚) = βˆ’5 β€“π‘Ÿ βƒ—.(2𝑖 Μ‚ + 3𝑗 Μ‚ + 4π‘˜ Μ‚) = 5 π‘Ÿ βƒ— .(βˆ’ 2𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚) = 5 Comparing with π‘Ÿ βƒ—.(𝑛2) βƒ— = d2 (π’πŸ) βƒ— = βˆ’ 2π’Š Μ‚ βˆ’ 3𝒋 Μ‚ βˆ’ 4π’Œ Μ‚ & d2 = 5 Equation of plane is π‘Ÿ βƒ—. [(𝑖 Μ‚+𝑗 Μ‚+π‘˜ Μ‚ )+"πœ†" (βˆ’2𝑖 Μ‚βˆ’3𝑗 Μ‚βˆ’4π‘˜ Μ‚)] = 6 + πœ†5 𝒓 βƒ—. [(π’Š Μ‚" " +𝒋 Μ‚" " +π’Œ Μ‚ )βˆ’"πœ†" (πŸπ’Š Μ‚+πŸ‘π’‹ Μ‚+πŸ’π’Œ Μ‚)] = 6 + 5πœ† Now to find πœ† , put 𝒓 βƒ— = xπ’Š Μ‚ + y𝒋 Μ‚ + zπ’Œ Μ‚ (x𝑖 Μ‚ + y𝑗 Μ‚ + zπ‘˜ Μ‚). [(𝑖 Μ‚+𝑗 Μ‚+π‘˜ Μ‚ )βˆ’"πœ†" (2𝑖 Μ‚+3𝑗 Μ‚+4π‘˜ Μ‚)] = 5πœ† + 6 (x𝑖 Μ‚ + y𝑗 Μ‚ + zπ‘˜ Μ‚).(𝑖 Μ‚+𝑗 Μ‚+π‘˜ Μ‚ ) βˆ’ πœ† (x𝑖 Μ‚ + y𝑗 Μ‚ + zπ‘˜ Μ‚).(2𝑖 Μ‚+3𝑗 Μ‚+4π‘˜ Μ‚) = 5πœ† + 6 (x Γ— 1) + (y Γ— 1) + (z Γ— 1) βˆ’ πœ†[(π‘₯Γ—2)+(𝑦×3)+(𝑧×4)] = 5πœ† + 6 x + y + z βˆ’ πœ†[2π‘₯+3𝑦+4𝑧] = 5πœ† + 6 x + y + z βˆ’ 2πœ†π‘₯ βˆ’ 3πœ†y βˆ’ 4πœ†z = 5πœ† + 6 (1 βˆ’ 2πœ†)x + (1 βˆ’ 3πœ†)y + (1 βˆ’ 4πœ†) z = 5πœ† + 6 Since the plane passes through (1, 1, 1), Putting (1, 1, 1) in (2) (1 βˆ’ 2πœ†)x + (1 βˆ’ 3πœ†)y + (1 βˆ’ 4πœ†) z = 5πœ† + 6 (1 βˆ’2πœ†) Γ— 1 + (1 βˆ’ 3πœ†) Γ— 1 + (1 βˆ’ 4πœ†) Γ— 1 = 5πœ† + 6 1 βˆ’2πœ† + 1 βˆ’ 3πœ† + 1 βˆ’ 4πœ†= 5πœ† + 6 3 βˆ’ 9πœ† = 5πœ† + 6 βˆ’14πœ† = 3 ∴ πœ† = (βˆ’πŸ‘)/πŸπŸ’ Putting value of πœ† in (1), π‘Ÿ βƒ—. [(𝑖 Μ‚" " +" " 𝑗 Μ‚" " +" " π‘˜ Μ‚ )βˆ’(( βˆ’3)/14)(2𝑖 Μ‚+3𝑗 Μ‚+"4" π‘˜ Μ‚)]= 6 + 5 Γ— ( βˆ’3)/14 π‘Ÿ βƒ—. [(𝑖 Μ‚+𝑗 Μ‚+" " π‘˜ Μ‚ )+3/14(2𝑖 Μ‚+3𝑗 Μ‚+"4" π‘˜ Μ‚)]= 6 βˆ’ 15/14 π‘Ÿ βƒ—. [𝑖 Μ‚+𝑗 Μ‚" " +π‘˜ Μ‚+ 6/14 𝑖 Μ‚+9/14 𝑗 Μ‚+12/14 π‘˜ Μ‚ ]= 69/14 π‘Ÿ βƒ—. [(1+6/14) 𝑖 Μ‚ +(1+9/14) 𝑗 Μ‚+(1+12/14) π‘˜ Μ‚ ]= 69/14 π‘Ÿ βƒ—. [20/14 𝑖 Μ‚ + 23/14 𝑗 Μ‚ + 26/14 π‘˜ Μ‚ ]= 69/14 π‘Ÿ βƒ—. [1/14(20𝑖 Μ‚+23𝑗 Μ‚+26π‘˜ Μ‚)]= 69/14 1/14 π‘Ÿ βƒ—. (20𝑖 Μ‚ + 23𝑗 Μ‚ + 26π‘˜ Μ‚) = 69/14 π‘Ÿ βƒ—. (20𝑖 Μ‚ + 23𝑗 Μ‚ + 26π‘˜ Μ‚) = 69 Therefore, the vector equation of the required plane is 𝒓 βƒ—.(πŸπŸŽπ’Š Μ‚ + πŸπŸ‘π’‹ Μ‚ + πŸπŸ”π’Œ Μ‚) = πŸ”πŸ—

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Davneet Singh

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.