Last updated at Feb. 1, 2020 by Teachoo
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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/โ29 units Distance between two parallel planes Ax + By + Cz = ๐_1 and Ax + By + Cz = ๐_2 is |(๐ _๐ โ ๐ _๐)/(โ(๐จ^๐ + ๐ฉ^๐ + ๐ช^๐ ) )| 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 โ 6)/โ(2^2 + 3^2 + 4^2 )| = |(โ2)/โ(4 + 9 + 16)| = ๐/โ๐๐ 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/โ29 units Distance of a point (๐ฅ_1, ๐ฆ_1, ๐ง_1) from the plane Ax + By + Cz = D is |(๐จ๐_๐ + ๐ฉ๐_๐ + ๐ช๐_๐โ ๐ซ)/โ(๐จ^๐ + ๐ฉ^๐ + ๐ช^๐ )| Let us take a point P (๐ฅ_1, ๐ฆ_1, ๐ง_1) on the plane 2x + 3y + 4z = 4 2๐ฅ_1 + 3๐ฆ_1 + 4๐ง_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 (๐ฅ_1, ๐ฆ_1, ๐ง_1) from the plane 4x + 6y + 8z = 12 = |(4๐ฅ_1+ ใ6๐ฆใ_1 + 8๐ง_1โ 12)/โ(4^2 + 6^2 + 8^2 )| = |(2(๐๐_๐ + ใ๐๐ใ_๐ + ๐๐_๐ )โ 12)/โ(16 + 36 + 64)| = |(2 ร ๐ โ 12)/โ116| = |(8 โ 12)/โ(4 ร 29)| = |(โ4)/(2โ29)| = ๐/โ๐๐ Hence, (D) is the correct option
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