Miscellaneous
Misc 2
Misc 3 Deleted for CBSE Board 2022 Exams
Misc 4 Important
Misc 5 Important
Misc 6 Important
Misc 7
Misc 8 Important
Misc 9 Important
Misc 10
Misc 11 Important
Misc 12 Important
Misc 13 Important
Misc 14 Important
Misc 15 Important
Misc 16
Misc 17 Important
Misc 18 Important
Misc 19
Misc 20 Important
Misc 21 Important
Misc 22 (MCQ) Important You are here
Misc 23 (MCQ) Important
Miscellaneous
Last updated at Aug. 11, 2021 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