Misc 17 - Plane which contains line of intersection of planes

Misc 17 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Misc 17 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3 Misc 17 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4 Misc 17 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5

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

Transcript

Misc 17 Find the equation of the plane which contains the line of intersection of the planes π‘Ÿ βƒ— . (𝑖 Μ‚ + 2𝑗 Μ‚ + 3π‘˜ Μ‚) – 4 = 0 , π‘Ÿ βƒ— . (2𝑖 Μ‚ + 𝑗 Μ‚ – π‘˜ Μ‚) + 5 = 0 and which is perpendicular to the plane π‘Ÿ βƒ— . (5𝑖 Μ‚ + 3𝑗 Μ‚ – 6π‘˜ Μ‚) + 8 = 0 .Equation of a plane passing through the intersection of the places A1x + B1y + C1z = d1 and A2x + B2y + C2z = d2 is (A1x + B1y + C1z βˆ’ d1) + πœ† (A2x + B2y + C2z – d2) = 0 Converting equation of planes to Cartesian form to find A1, B1, C1, d1 & A2, B2, C2, d2 𝒓 βƒ—. (π’Š Μ‚ + 2𝒋 Μ‚ + 3π’Œ Μ‚) βˆ’ 4 = 0 π‘Ÿ βƒ—. (𝑖 Μ‚ + 2𝑗 Μ‚ + 3π‘˜ Μ‚) = 4 Putting 𝒓 βƒ— = xπ’Š Μ‚ + y𝒋 Μ‚ + zπ’Œ Μ‚ (x𝑖 Μ‚ + y𝑗 Μ‚ + zπ‘˜ Μ‚).(𝑖 Μ‚ + 2𝑗 Μ‚ + 3π‘˜ Μ‚) = 4 (x Γ— 1) + (y Γ— 2) + (z Γ— 3) = 4 1x + 2y + 3z = 4 Comparing with 𝐴_1 "x"+"B1y"+𝐢_1 "z = d1" 𝐴_1 = 1, 𝐡_1= 2 , 𝐢_1 = 3 , 𝑑_1 = 4 𝒓 βƒ—. (2π’Š Μ‚ + 𝒋 Μ‚ βˆ’ π’Œ Μ‚) + 5 = 0 π‘Ÿ βƒ—. (2𝑖 Μ‚ + 𝑗 Μ‚ βˆ’ π‘˜ Μ‚) = βˆ’ 5 βˆ’π‘Ÿ βƒ—. (2𝑖 Μ‚ + 𝑗 Μ‚ βˆ’ π‘˜ Μ‚) = 5 π‘Ÿ βƒ—. ( βˆ’2𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + π‘˜ Μ‚) = 5 Putting 𝒓 βƒ— = xπ’Š Μ‚ + y𝒋 Μ‚ + zπ’Œ Μ‚, (x𝑖 Μ‚ + y𝑗 Μ‚ + zπ‘˜ Μ‚).(-2𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + π‘˜ Μ‚) = 5 (x Γ—βˆ’ 2) + (Y Γ— βˆ’ 1) + (z Γ— 1) = 5 βˆ’2x βˆ’ 1y + 1z = 5 Comparing with 𝐴_2 "x"+ "B2y"+ 𝐢_2 "z = d2" 𝐴_2 = βˆ’2, 𝐡_2= βˆ’1 , 𝐢_2 = 1 , 𝑑_2 = 5 Equation of plane is (A1x + B1y + C1z βˆ’ d1) + πœ† (A2x + B2y + C2z = d2) = 0 Putting values (1x + 2y + 3z βˆ’ 4) + πœ† ( βˆ’ 2x βˆ’ 1y + 1z βˆ’ 5) = 0 (1 βˆ’ 2πœ†) x + (2 βˆ’ πœ†)y + (3 + πœ†) z + ( βˆ’4 βˆ’ 5πœ†) = 0 Now, the plane is perpendicular to the plane π‘Ÿ βƒ—.(5𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ 6π‘˜ Μ‚) + 8 = 0 So, normal to plane 𝑁 βƒ— will be perpendicular to normal 𝑛 βƒ— of π‘Ÿ βƒ—.(5𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ 6π‘˜ Μ‚) + 8 = 0 Now, π‘Ÿ βƒ—.(5𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ 6π‘˜ Μ‚) + 8 = 0 π‘Ÿ βƒ— .(5𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ 6π‘˜ Μ‚) = –8 βˆ’ π‘Ÿ βƒ— .(5𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ 6π‘˜ Μ‚) = 8 π‘Ÿ βƒ— .( βˆ’5𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ + 6π‘˜ Μ‚) = 8 Finding direction cosines of 𝑁 βƒ— & 𝑛 βƒ— Since, 𝑁 βƒ— is perpendicular to 𝑛 βƒ— π‘Ž1 π‘Ž2 + b1 b2 + c1 c2 = 0 (1 βˆ’ 2πœ†) Γ— βˆ’5 + (2 βˆ’ πœ†) Γ— βˆ’3 + (3 + πœ†) Γ— 6 = 0 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‚ + (2 βˆ’ πœ†) 𝒋 Μ‚ + (3 + πœ†) π’Œ Μ‚ Direction ratios = 1 βˆ’ 2πœ†, 2 βˆ’ πœ†, 3 + πœ† ∴ π‘Ž1 = 1 βˆ’ 2πœ†, b1 = 2 βˆ’ πœ†, c1 = 3 + πœ† 𝒏 βƒ— = βˆ’ 5π’Š Μ‚ – 3𝒋 Μ‚ + 6π’Œ Μ‚ Direction ratios = βˆ’5, βˆ’3, 6 ∴ π‘Ž2 = βˆ’ 5, b2 = βˆ’3, c2 = 6, βˆ’ 5 + 10πœ† βˆ’ 6 + 3πœ† + 18 + 6πœ† = 0 19πœ† + 7 = 0 ∴ πœ† = (βˆ’πŸ•)/πŸπŸ— Putting value of πœ† in (1), (1 βˆ’ 2πœ†) x + (2 βˆ’ πœ†)y + (3 + πœ†) z + ( βˆ’4 βˆ’ 5πœ†) = 0 (1βˆ’2 Γ—(βˆ’7)/19) x + (2βˆ’(( βˆ’7)/19)) y + (3+(( βˆ’ 7)/19)) z + ( βˆ’4βˆ’5Γ—(βˆ’7)/19) = 0 (1 + 14/19) x + (2 + 7/19)y + (3 βˆ’ 7/19)z + ( βˆ’ 4 + 35/19) = 0 33/19 x + 45/19 y + 50/19 z βˆ’ 41/19 = 0 1/19 (33x + 45y + 50z βˆ’ 41) = 0 33x + 45y + 50z βˆ’ 41 = 0

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.