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Misc 22 - Distance between planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z =

Misc 22 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Misc 22 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Misc 22 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4

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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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.