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

Transcript

Misc 17 Find the equation of the plane which contains the line of intersection of the planes 𝑟 ⃗ . (𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂) – 4 = 0 , 𝑟 ⃗ . (2𝑖 ̂ + 𝑗 ̂ – 𝑘 ̂) + 5 = 0 and which is perpendicular to the plane 𝑟 ⃗ . (5𝑖 ̂ + 3𝑗 ̂ – 6𝑘 ̂) + 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 − d1) + 𝜆 (A2x + B2y + C2z – d2) = 0 Converting equation of planes to Cartesian form to find A1, B1, C1, d1 & A2, B2, C2, d2 𝒓 ⃗. (𝒊 ̂ + 2𝒋 ̂ + 3𝒌 ̂) − 4 = 0 𝑟 ⃗. (𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂) = 4 Putting 𝒓 ⃗ = x𝒊 ̂ + y𝒋 ̂ + z𝒌 ̂ (x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂).(𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂) = 4 (x × 1) + (y × 2) + (z × 3) = 4 1x + 2y + 3z = 4 Comparing with 𝐴_1 "x"+"B1y"+𝐶_1 "z = d1" 𝐴_1 = 1, 𝐵_1= 2 , 𝐶_1 = 3 , 𝑑_1 = 4 𝒓 ⃗. (2𝒊 ̂ + 𝒋 ̂ − 𝒌 ̂) + 5 = 0 𝑟 ⃗. (2𝑖 ̂ + 𝑗 ̂ − 𝑘 ̂) = − 5 −𝑟 ⃗. (2𝑖 ̂ + 𝑗 ̂ − 𝑘 ̂) = 5 𝑟 ⃗. ( −2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂) = 5 Putting 𝒓 ⃗ = x𝒊 ̂ + y𝒋 ̂ + z𝒌 ̂, (x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂).(-2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂) = 5 (x ×− 2) + (Y × − 1) + (z × 1) = 5 −2x − 1y + 1z = 5 Comparing with 𝐴_2 "x"+ "B2y"+ 𝐶_2 "z = d2" 𝐴_2 = −2, 𝐵_2= −1 , 𝐶_2 = 1 , 𝑑_2 = 5 Equation of plane is (A1x + B1y + C1z − d1) + 𝜆 (A2x + B2y + C2z = d2) = 0 Putting values (1x + 2y + 3z − 4) + 𝜆 ( − 2x − 1y + 1z − 5) = 0 (1 − 2𝜆) x + (2 − 𝜆)y + (3 + 𝜆) z + ( −4 − 5𝜆) = 0 Now, the plane is perpendicular to the plane 𝑟 ⃗.(5𝑖 ̂ + 3𝑗 ̂ − 6𝑘 ̂) + 8 = 0 So, normal to plane 𝑁 ⃗ will be perpendicular to normal 𝑛 ⃗ of 𝑟 ⃗.(5𝑖 ̂ + 3𝑗 ̂ − 6𝑘 ̂) + 8 = 0 Now, 𝑟 ⃗.(5𝑖 ̂ + 3𝑗 ̂ − 6𝑘 ̂) + 8 = 0 𝑟 ⃗ .(5𝑖 ̂ + 3𝑗 ̂ − 6𝑘 ̂) = –8 − 𝑟 ⃗ .(5𝑖 ̂ + 3𝑗 ̂ − 6𝑘 ̂) = 8 𝑟 ⃗ .( −5𝑖 ̂ − 3𝑗 ̂ + 6𝑘 ̂) = 8 Finding direction cosines of 𝑁 ⃗ & 𝑛 ⃗ Since, 𝑁 ⃗ is perpendicular to 𝑛 ⃗ 𝑎1 𝑎2 + b1 b2 + c1 c2 = 0 (1 − 2𝜆) × −5 + (2 − 𝜆) × −3 + (3 + 𝜆) × 6 = 0 Theory : Two lines with direction ratios 𝑎1, b1, c1 and 𝑎2, b2, c2 are perpendicular if 𝑎1 𝑎2 + b1b2 + c1 c2 = 0 𝑵 ⃗ = (1 − 2𝜆) 𝒊 ̂ + (2 − 𝜆) 𝒋 ̂ + (3 + 𝜆) 𝒌 ̂ Direction ratios = 1 − 2𝜆, 2 − 𝜆, 3 + 𝜆 ∴ 𝑎1 = 1 − 2𝜆, b1 = 2 − 𝜆, c1 = 3 + 𝜆 𝒏 ⃗ = − 5𝒊 ̂ – 3𝒋 ̂ + 6𝒌 ̂ Direction ratios = −5, −3, 6 ∴ 𝑎2 = − 5, b2 = −3, c2 = 6, − 5 + 10𝜆 − 6 + 3𝜆 + 18 + 6𝜆 = 0 19𝜆 + 7 = 0 ∴ 𝜆 = (−𝟕)/𝟏𝟗 Putting value of 𝜆 in (1), (1 − 2𝜆) x + (2 − 𝜆)y + (3 + 𝜆) z + ( −4 − 5𝜆) = 0 (1−2 ×(−7)/19) x + (2−(( −7)/19)) y + (3+(( − 7)/19)) z + ( −4−5×(−7)/19) = 0 (1 + 14/19) x + (2 + 7/19)y + (3 − 7/19)z + ( − 4 + 35/19) = 0 33/19 x + 45/19 y + 50/19 z − 41/19 = 0 1/19 (33x + 45y + 50z − 41) = 0 33x + 45y + 50z − 41 = 0

Miscellaneous 