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Last updated at Aug. 11, 2021 by
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Misc 23 (Method 1) The planes: 2x โ y + 4z = 5 and 5x โ 2.5y + 10z = 6 are (A) Perpendicular (B) Parallel (C) intersect y-axis (D) passes through (0,0, 5/4) Angle between two planes A1x + B1y + C1z = d1 and A2x + B2y + C2z = d2 is given by cos ฮธ = (๐จ_๐ ๐จ_๐ + ๐ฉ_๐ ๐ฉ_๐ + ๐ช_๐ ๐ช_๐)/(โ(ใ๐จ_๐ใ^๐ + ใ๐ฉ_๐ใ^๐ + ใ๐ช_๐ใ^๐ ) โ(ใ๐จ_๐ใ^๐ + ใ๐ฉ_๐ใ^๐ + ใ๐ช_๐ใ^๐ )) Given the two planes are 2x โ 1y + 4z = 5 Comparing with A1x + B1y + C1z = d1 A1 = 2 , B1 = โ1 , C1 = 4 , ๐_1= 5 5x โ 2.5y + 10z = 6 Multiplying by 2 on both sides, 10x โ 5y + 20z = 12 Comparing with A2x + B2y + C2z = d2 A2 = 10 , B2 = โ5 , C2 = 20 , ๐_2= 12 So, cos ๐ = |((2 ร 10) + (โ1 ร โ5) + (4 ร 20))/(โ(2^2 + (ใโ1)ใ^2 + 4^2 ) โ(ใ10ใ^(2 )+ (ใโ5)ใ^2 + ใ20ใ^2 ))| = |(20 + 5 + 80)/(โ(4 + 1 + 16) โ(100 + 25 + 400))| = |105/(โ21 โ525)| = |105/(โ21 ร โ(25 ร 21))| = |105/(โ21 ร 5 โ21)| = |105/(21 ร 5)| = 1 So, cos ฮธ = 1 โด ฮธ = 0ยฐ Since angle between the planes is 0ยฐ, Therefore, the planes are parallel. So, Option (B) is correct Misc 23 (Method 2) The planes: 2x โ y + 4z = 5 and 5x โ 2.5y + 10z = 6 are (A) Perpendicular (B) Parallel (C) intersect y-axis (D) passes through (0,0, 5/4) 2x โ 1y + 4z = 5 Comparing with A1x + B1y + C1z = d1 Direction ratios of normal = 2, โ1, 4 A1 = 2 , B1 = โ1 , C1 = 4 5x โ 2.5y + 10z = 6 Multiplying by 2 on both sides, 10x โ 5y + 20z = 12 Comparing with A2x + B2y + C2z = d2 Direction ratios of normal = 10, โ5, 20 A2 = 10 , B2 = โ5 , C2 = 20 Two lines are parallel if their direction ratios are proportional. ๐ด_1/๐ด_2 = 2/10 = 1/5 , ๐ต_1/๐ต_2 = (โ1)/(โ5) = 1/5 , ๐ถ_1/๐ถ_2 = 4/20 = 1/5 a Since, ๐จ_๐/๐จ_๐ = ๐ฉ_๐/๐ฉ_๐ = ๐ช_๐/๐ช_๐ = ๐/๐ Therefore, the normal vectors of the two planes are parallel. So, the two planes are parallel. So, option (B) is correct
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