Misc 15 - Plane intersection of planes, parallel to x-axis  - Equation of plane - Passing Through Intersection Of Planes

Slide57.JPG
Slide58.JPG Slide59.JPG Slide60.JPG

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

Transcript

Misc 15 Find the equation of the plane passing through the line of intersection of the planes π‘ŸΒ βƒ— . (𝑖 ̂ + 𝑗 ̂ + π‘˜Β Μ‚) =1 and π‘ŸΒ βƒ— . (2𝑖 ̂ + 3𝑗 ̂ – π‘˜Β Μ‚) + 4 = 0 and parallel to x-axis. Equation of a plane passing through the intersection of two planes 𝐴_1x + B1y + 𝐢_1z = d1 and 𝐴_2x + B2y + 𝐢_2z = d2 is (𝑨_𝟏 "x " +" B1y" + π‘ͺ_𝟏 "z – d1 " ) + πœ† (𝑨_𝟐 "x" +"B2y" +π‘ͺ_𝟐 "z – d2 " ) = 0. Converting equation of planes to Cartesian form to find A1, B1, C1, d1 & A2, B2, C2, d2 𝒓 ⃗. (π’ŠΒ Μ‚ + 𝒋 ̂ + π’ŒΒ Μ‚) = 1 Putting 𝒓 ⃗ = xπ’ŠΒ Μ‚ + y𝒋 ̂ + zπ’ŒΒ Μ‚, (x𝑖 ̂ + y𝑗 ̂ + zπ‘˜Β Μ‚).(𝑖 ̂ + 𝑗 ̂ + π‘˜Β Μ‚) = 1 (x Γ— 1) + (Y Γ— 1) + (z Γ— 1) = 1 1x + 1y + 1z = 1 Comparing with 𝐴_1 "x "+"B1y "+" " 𝐢_1 "z = d1" 𝐴_1 = 1 , 𝐡_1 = 1 , 𝐢_1 = 1 , 𝑑_1 = 1 𝒓 ⃗. (2π’ŠΒ Μ‚ + 3𝒋 ̂ βˆ’ π’ŒΒ Μ‚) + 4 = 0 π‘ŸΒ βƒ—. (2𝑖 ̂ + 3𝑗 ̂ βˆ’ π‘˜Β Μ‚) = βˆ’4 β€“π‘ŸΒ βƒ—. (2𝑖 ̂ + 3𝑗 ̂ βˆ’ π‘˜Β Μ‚) = 4 π‘ŸΒ βƒ—. ( βˆ’2𝑖 ̂ βˆ’3𝑗 ̂ + π‘˜Β Μ‚) = 4 Putting 𝒓 ⃗ = xπ’ŠΒ Μ‚ + y𝒋 ̂ + zπ’ŒΒ Μ‚, (x𝑖 ̂ + y𝑗 ̂ + zπ‘˜Β Μ‚).(-2𝑖 ̂ βˆ’ 3𝑗 ̂ + 1π‘˜Β Μ‚) = 4 (x Γ— βˆ’2) + (y Γ— βˆ’ 3) + (z Γ— 1) = 4 βˆ’ 2x βˆ’ 3y + 1z = 4 Comparing with 𝐴_2 "x "+"B2y "+" " 𝐢_2 "z = d2" 𝐴_2 = –2 , 𝐡_2 = –3 , 𝐢_2 = 1 , 𝑑_2 = 4 Equation of plane is (𝐴_1 "x " +" B1y" + 𝐢_1 "z – d1 " ) + πœ† (𝐴_2 "x" +"B2y" +𝐢_2 "z – d2 " ) = 0 (1x + 1y + 1z βˆ’ 1) + πœ† ( βˆ’ 2x βˆ’ 3y + 1z βˆ’ 4) = 0 (1 βˆ’ 2πœ†) x + (1 βˆ’ 3πœ†)y + (1 + πœ†) z + (βˆ’ 1βˆ’ 4πœ†) = 0 Also, the plane is parallel to x – axis. So, Normal vector 𝑁 ⃗ to the plane is perpendicular to x – axis. Finding direction ratios of normal and x-axis Theory : Two lines with direction ratios π‘Ž1, b1, c1 and π‘Ž2, b2, c2 are perpendicular if π‘Ž1 π‘Ž2 + b1b2 + c1 c2 = 0 𝑡 ⃗ = (1 βˆ’ 2πœ†) π’ŠΒ Μ‚ + (1 βˆ’ 3πœ†) 𝒋 ̂ + (1 + πœ†) π’ŒΒ Μ‚ Direction ratios = (1 βˆ’2πœ†), (1 βˆ’ 3πœ†), (1 + πœ†)π‘˜Β Μ‚ ∴ π‘Ž1 = 1 βˆ’ 2πœ†, b1 = 1 βˆ’ 3πœ†, c1 = 1 + πœ† (𝑢𝑿)Β βƒ— = 1π’ŠΒ Μ‚ + 0𝒋 ̂ + 0π’ŒΒ Μ‚ Direction ratios = 1, 0, 0 ∴ π‘Ž2 = 1, b2 = 0, c2 = 0, So, π‘Ž1 π‘Ž2 + b1 b2 + c1 c2 = 0 (1 βˆ’ 2πœ†) Γ— 1 + (1 βˆ’ 3πœ†) Γ— 0 + (1 + πœ†) Γ— 0 = 0 (1 βˆ’ 2πœ†) + 0 + 0 = 0 1 = 2πœ† ∴ πœ† = 𝟏/𝟐 Putting value of πœ† in (1) (1βˆ’ 2. 1/2)x + (1βˆ’ 3. 1/2) y + (1 + 1/2) + (βˆ’ 1 βˆ’ 4.1/2) = 0 (1 βˆ’ 1)x + (1 βˆ’ 3/2) y + (1 + 1/2) z + (βˆ’1 βˆ’ 2) = 0 0x βˆ’ 1/2 y + 3/2 z βˆ’ 3 = 0 0x βˆ’ 1/2 y + 3/2 z = 3 βˆ’ y + 3z = 6 0 = y – 3z + 6 y – 3z + 6 = 0 Therefore, the equation of the plane is y βˆ’ 3z + 6 = 0

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
Jail