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Last updated at Feb. 1, 2020 by Teachoo
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Ex 11.3, 12 Find the angle between the planes whose vector equations are ๐ โ . (2๐ ฬ + 2๐ ฬ โ 3๐ ฬ) = 5 and ๐ โ . (3๐ ฬ โ 3๐ ฬ + 5๐ ฬ) = 3 .Angle between two planes ๐ โ . (๐_1 ) โ = d1 and ๐ โ.(๐2) โ = d2 is given by cos ๐ = |((๐๐) โ. (๐๐) โ)/|(๐๐) โ ||(๐๐) โ | | ๐ โ.(2๐ ฬ + 2๐ ฬ โ 3๐ ฬ) = 5 Comparing with ๐ โ.(๐1) โ = (๐1) โ, (๐1) โ = 2๐ ฬ + 2๐ ฬโ3๐ ฬ Magnitude of (๐1) โ = โ(2^2+2^2+ใ(โ3)ใ^2 ) |(๐1) โ | = โ(4+4+9) = โ17 So, cos ฮธ = |((2๐ ฬ + 2๐ ฬ โ 3๐ ฬ ) . (3๐ ฬ โ 3๐ ฬ + 5๐ ฬ ))/(โ17 ร โ43)| = |((2 ร 3) + (2 ร โ3) + (โ3 ร 5))/โ(17 ร 43)| = |(6 โ 6 โ 15)/โ731| = |(โ15)/โ731| = 15/โ731 So, cos ฮธ = 15/โ731 โด ฮธ = cosโ1(๐๐/โ๐๐๐) Therefore, the angle between the planes is cosโ1(15/โ731).
Ex 11.3
Ex 11.3, 2
Ex 11.3, 3
Ex 11.3, 4 Important
Ex 11.3, 5 Important
Ex 11.3, 6 Important
Ex 11.3, 7
Ex 11.3, 8
Ex 11.3, 9
Ex 11.3, 10 Important
Ex 11.3, 11 Important
Ex 11.3, 12 Important Not in Syllabus - CBSE Exams 2021 You are here
Ex 11.3, 13 (a) Important
Ex 11.3, 13 (b)
Ex 11.3, 13 (c)
Ex 11.3, 13 (d)
Ex 11.3, 13 (e)
Ex 11.3, 14 (a) Important
Ex 11.3, 14 (b)
Ex 11.3, 14 (c)
Ex 11.3, 14 (d)
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