Miscellaneous
Misc 2 Important
Misc 3 Important
Misc 4 Important
Misc 5 Important
Question 1 Important Deleted for CBSE Board 2025 Exams
Question 2 Deleted for CBSE Board 2025 Exams
Question 3 Important Deleted for CBSE Board 2025 Exams
Question 4 Deleted for CBSE Board 2025 Exams
Question 5 Important Deleted for CBSE Board 2025 Exams
Question 6 Deleted for CBSE Board 2025 Exams
Question 7 Important Deleted for CBSE Board 2025 Exams
Question 8 Important Deleted for CBSE Board 2025 Exams
Question 9 Important Deleted for CBSE Board 2025 Exams
Question 10 Important Deleted for CBSE Board 2025 Exams
Question 11 Important Deleted for CBSE Board 2025 Exams
Question 12 Deleted for CBSE Board 2025 Exams
Question 13 Important Deleted for CBSE Board 2025 Exams
Question 14 Important Deleted for CBSE Board 2025 Exams
Question 15 Deleted for CBSE Board 2025 Exams
Question 16 Important Deleted for CBSE Board 2025 Exams
Question 17 (MCQ) Important Deleted for CBSE Board 2025 Exams You are here
Question 18 (MCQ) Important Deleted for CBSE Board 2025 Exams
Miscellaneous
Last updated at April 16, 2024 by Teachoo
Misc 22 (Method 1) Distance between two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is (A) 2 units (B) 4 units (C) 8 units (D) 2/√29 units Distance between two parallel planes Ax + By + Cz = 𝑑_1 and Ax + By + Cz = 𝑑_2 is |(𝒅_𝟏 − 𝒅_𝟐)/(√(𝑨^𝟐 + 𝑩^𝟐 + 𝑪^𝟐 ) )| 2x + 3y + 4z = 4 Comparing with Ax + By + Cz = d1 A = 2, B = 3, C = 4, d1 = 4 4x + 6y + 8z = 12 2 (2x + 3y + 4z) = 12 Dividing by 2 2x + 3y + 4z = 6 Comparing with Ax + By + Cz = d2 A = 2, B = 3, C = 4 , d2 = 6 So, Distance between the two planes = |(4 − 6)/√(2^2 + 3^2 + 4^2 )| = |(−2)/√(4 + 9 + 16)| = 𝟐/√𝟐𝟗 Hence, (D) is the correct option Misc 22 (Method 2) Distance between the two planes : 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is (A) 2 units (B) 4 units (C) 8 units (D)2/√29 units Distance of a point (𝑥_1, 𝑦_1, 𝑧_1) from the plane Ax + By + Cz = D is |(𝑨𝒙_𝟏 + 𝑩𝒚_𝟏 + 𝑪𝒛_𝟏− 𝑫)/√(𝑨^𝟐 + 𝑩^𝟐 + 𝑪^𝟐 )| Let us take a point P (𝑥_1, 𝑦_1, 𝑧_1) on the plane 2x + 3y + 4z = 4 2𝑥_1 + 3𝑦_1 + 4𝑧_1 = 4 Now, to find the distance of point P form plane 4x + 6y + 8z = 12, Comparing with Ax + By + Cz = D, A = 4, B = 6, C = 8, D = 12 Distance of P (𝑥_1, 𝑦_1, 𝑧_1) from the plane 4x + 6y + 8z = 12 = |(4𝑥_1+ 〖6𝑦〗_1 + 8𝑧_1− 12)/√(4^2 + 6^2 + 8^2 )| = |(2(𝟐𝒙_𝟏 + 〖𝟑𝒚〗_𝟏 + 𝟒𝒛_𝟏 )− 12)/√(16 + 36 + 64)| = |(2 × 𝟒 − 12)/√116| = |(8 − 12)/√(4 × 29)| = |(−4)/(2√29)| = 𝟐/√𝟐𝟗 Hence, (D) is the correct option