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Misc 15 - Plane intersection of planes, parallel to x-axis

Misc 15 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Misc 15 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Misc 15 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4
Misc 15 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5


Transcript

Misc 15 Find the equation of the plane passing through the line of intersection of the planes π‘Ÿ βƒ— . (𝑖 Μ‚ + 𝑗 Μ‚ + π‘˜ Μ‚) =1 and π‘Ÿ βƒ— . (2𝑖 Μ‚ + 3𝑗 Μ‚ – π‘˜ 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 βƒ—. (π’Š Μ‚ + 𝒋 Μ‚ + π’Œ Μ‚) = 1 Putting 𝒓 βƒ— = xπ’Š Μ‚ + y𝒋 Μ‚ + zπ’Œ Μ‚, (x𝑖 Μ‚ + y𝑗 Μ‚ + zπ‘˜ Μ‚).(𝑖 Μ‚ + 𝑗 Μ‚ + π‘˜ 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 βƒ—. (2π’Š Μ‚ + 3𝒋 Μ‚ βˆ’ π’Œ Μ‚) + 4 = 0 π‘Ÿ βƒ—. (2𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ π‘˜ Μ‚) = βˆ’4 β€“π‘Ÿ βƒ—. (2𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ π‘˜ Μ‚) = 4 π‘Ÿ βƒ—. ( βˆ’2𝑖 Μ‚ βˆ’3𝑗 Μ‚ + π‘˜ Μ‚) = 4 Putting 𝒓 βƒ— = xπ’Š Μ‚ + y𝒋 Μ‚ + zπ’Œ Μ‚, (x𝑖 Μ‚ + y𝑗 Μ‚ + zπ‘˜ Μ‚).(-2𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ + 1π‘˜ 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 βƒ— 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 βƒ— = (1 βˆ’ 2πœ†) π’Š Μ‚ + (1 βˆ’ 3πœ†) 𝒋 Μ‚ + (1 + πœ†) π’Œ Μ‚ Direction ratios = (1 βˆ’2πœ†), (1 βˆ’ 3πœ†), (1 + πœ†)π‘˜ Μ‚ ∴ π‘Ž1 = 1 βˆ’ 2πœ†, b1 = 1 βˆ’ 3πœ†, c1 = 1 + πœ† (𝑢𝑿) βƒ— = 1π’Š Μ‚ + 0𝒋 Μ‚ + 0π’Œ 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

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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.