Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Miscellaneous
Misc 2 Important
Misc 3 Important
Misc 4 Important
Misc 5 Important
Question 1 Important Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Important Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 8 Important Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 10 Important Deleted for CBSE Board 2024 Exams
Question 11 Important Deleted for CBSE Board 2024 Exams You are here
Question 12 Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 14 Important Deleted for CBSE Board 2024 Exams
Question 15 Deleted for CBSE Board 2024 Exams
Question 16 Important Deleted for CBSE Board 2024 Exams
Question 17 (MCQ) Important Deleted for CBSE Board 2024 Exams
Question 18 (MCQ) Important Deleted for CBSE Board 2024 Exams
Miscellaneous
Last updated at May 29, 2023 by Teachoo
Question 11 Find the equation of the plane passing through the line of intersection of the planes 𝑟 ⃗ . (𝑖 ̂ + 𝑗 ̂ + 𝑘 ̂) =1 and 𝑟 ⃗ . (2𝑖 ̂ + 3𝑗 ̂ – 𝑘 ̂) + 4 = 0 and parallel to x-axis. Equation of a plane passing through the intersection of two planes 𝐴_1x + B1y + 𝐶_1z = d1 and 𝐴_2x + B2y + 𝐶_2z = d2 is (𝑨_𝟏 "x " +" B1y" + 𝑪_𝟏 "z – d1 " ) + 𝜆 (𝑨_𝟐 "x" +"B2y" +𝑪_𝟐 "z – d2 " ) = 0. Converting equation of planes to Cartesian form to find A1, B1, C1, d1 & A2, B2, C2, d2 𝒓 ⃗. (𝒊 ̂ + 𝒋 ̂ + 𝒌 ̂) = 1 Putting 𝒓 ⃗ = x𝒊 ̂ + y𝒋 ̂ + z𝒌 ̂, (x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂).(𝑖 ̂ + 𝑗 ̂ + 𝑘 ̂) = 1 (x × 1) + (Y × 1) + (z × 1) = 1 1x + 1y + 1z = 1 Comparing with 𝐴_1 "x "+"B1y "+" " 𝐶_1 "z = d1" 𝐴_1 = 1 , 𝐵_1 = 1 , 𝐶_1 = 1 , 𝑑_1 = 1 𝒓 ⃗. (2𝒊 ̂ + 3𝒋 ̂ − 𝒌 ̂) + 4 = 0 𝑟 ⃗. (2𝑖 ̂ + 3𝑗 ̂ − 𝑘 ̂) = −4 –𝑟 ⃗. (2𝑖 ̂ + 3𝑗 ̂ − 𝑘 ̂) = 4 𝑟 ⃗. ( −2𝑖 ̂ −3𝑗 ̂ + 𝑘 ̂) = 4 Putting 𝒓 ⃗ = x𝒊 ̂ + y𝒋 ̂ + z𝒌 ̂, (x𝑖 ̂ + y𝑗 ̂ + z𝑘 ̂).(-2𝑖 ̂ − 3𝑗 ̂ + 1𝑘 ̂) = 4 (x × −2) + (y × − 3) + (z × 1) = 4 − 2x − 3y + 1z = 4 Comparing with 𝐴_2 "x "+"B2y "+" " 𝐶_2 "z = d2" 𝐴_2 = –2 , 𝐵_2 = –3 , 𝐶_2 = 1 , 𝑑_2 = 4 Equation of plane is (𝐴_1 "x " +" B1y" + 𝐶_1 "z – d1 " ) + 𝜆 (𝐴_2 "x" +"B2y" +𝐶_2 "z – d2 " ) = 0 (1x + 1y + 1z − 1) + 𝜆 ( − 2x − 3y + 1z − 4) = 0 (1 − 2𝜆) x + (1 − 3𝜆)y + (1 + 𝜆) z + (− 1− 4𝜆) = 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 𝑎1, b1, c1 and 𝑎2, b2, c2 are perpendicular if 𝑎1 𝑎2 + b1b2 + c1 c2 = 0 𝑵 ⃗ = (1 − 2𝜆) 𝒊 ̂ + (1 − 3𝜆) 𝒋 ̂ + (1 + 𝜆) 𝒌 ̂ Direction ratios = (1 −2𝜆), (1 − 3𝜆), (1 + 𝜆)𝑘 ̂ ∴ 𝑎1 = 1 − 2𝜆, b1 = 1 − 3𝜆, c1 = 1 + 𝜆 (𝑶𝑿) ⃗ = 1𝒊 ̂ + 0𝒋 ̂ + 0𝒌 ̂ Direction ratios = 1, 0, 0 ∴ 𝑎2 = 1, b2 = 0, c2 = 0, So, 𝑎1 𝑎2 + b1 b2 + c1 c2 = 0 (1 − 2𝜆) × 1 + (1 − 3𝜆) × 0 + (1 + 𝜆) × 0 = 0 (1 − 2𝜆) + 0 + 0 = 0 1 = 2𝜆 ∴ 𝜆 = 𝟏/𝟐 Putting value of 𝜆 in (1) (1− 2. 1/2)x + (1− 3. 1/2) y + (1+ 1/2) + (−1 − 4.1/2) = 0 (1 − 1)x + (1 − 3/2) y + (1+1/2) z + (−1 − 2) = 0 0x − 1/2 y + 3/2 z − 3 = 0 0x − 1/2 y + 3/2 z = 3 − y + 3z = 6 0 = y – 3z + 6 y – 3z + 6 = 0 Therefore, the equation of the plane is y − 3z + 6 = 0