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Last updated at Aug. 18, 2020 by Teachoo
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Ex 11.3, 14 In the following cases, find the distance of each of the given points from the corresponding given plane. The distance of the point (x1, y1, z1) from the plane Ax + By + Cz = D is |(๐จ๐_๐ + ใ๐ฉ๐ใ_๐ +ใ ๐ช๐ใ_๐ โ ๐ซ)/โ(๐จ^๐ + ๐ฉ^๐ + ๐ช^๐ )| Given, the point is (2, 3, โ5) So, ๐ฅ_1= 2, ๐ฆ_1= 3, ๐ง_1= โ5 and the equation of the plane is 1x + 2y โ 2z = 9 Comparing with Ax + By + Cz = D, A = 1, B = 2, C = โ2, D = 9 Now, Distance of the point from the plane = |((1 ร 2) + (2 ร 3) + (โ2 ร โ5) โ 9)/โ(1^2 + 2^2 + ใ(โ2)ใ^2 )| = |(2 + 6 + 10 โ 9)/โ(1 + 4 + 4)| = |(18 โ 9)/โ9| = 9/3 = 3
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
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) You are here
Ex 11.3, 14 (d)
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