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Last updated at May 29, 2018 by Teachoo
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Ex 11.3, 3 Find the Cartesian equation of the following planes: (a) ๐๏ทฏ . ( ๐๏ทฏ + ๐๏ทฏ โ ๐๏ทฏ) = 2 Putting ๐๏ทฏ = x ๐๏ทฏ + y ๐๏ทฏ + z ๐๏ทฏ in equation ๐๏ทฏ.( ๐๏ทฏ + ๐๏ทฏ โ ๐๏ทฏ) = 2 (x ๐๏ทฏ + y ๐๏ทฏ + z ๐๏ทฏ). ( ๐๏ทฏ + ๐๏ทฏ โ ๐๏ทฏ) = 2 (x ร 1) + (y ร 1) + (z ร โ1) = 2 x + y โ z = 2 is the cartesian equation of the given plane. Ex 11.3, 3 Find the Cartesian equation of the following planes: (b) ๐๏ทฏ . (2 ๐๏ทฏ + 3 ๐๏ทฏ โ 4 ๐๏ทฏ) = 1 Putting ๐๏ทฏ = x ๐๏ทฏ + y ๐๏ทฏ + z ๐๏ทฏ in equation ๐๏ทฏ.(2 ๐๏ทฏ + 3 ๐๏ทฏ โ 4 ๐๏ทฏ) = 1 (x ๐๏ทฏ + y ๐๏ทฏ + z ๐๏ทฏ). (2 ๐๏ทฏ + 3 ๐๏ทฏ โ 4 ๐๏ทฏ) = 1 (x ร 2) + (y ร 3) + (z รโ 4) = 1 ๐๐ + 3y โ 4z = 1 Which is the Cartesian equation of the plane. Ex 11.3, 3 Find the Cartesian equation of the following planes: (c) ๐๏ทฏ . [(s โ 2t) ๐๏ทฏ + (3 โ t) ๐๏ทฏ + (2s + t) ๐๏ทฏ] = 15 Putting ๐๏ทฏ = x ๐๏ทฏ + y ๐๏ทฏ + z ๐๏ทฏ in equation ๐๏ทฏ . [(s โ 2t) ๐๏ทฏ + (3 โ t) ๐๏ทฏ + (2s + t) ๐๏ทฏ] = 15 (x ๐๏ทฏ + y ๐๏ทฏ + z ๐๏ทฏ) . [(s โ 2t) ๐๏ทฏ + (3 โ t) ๐๏ทฏ + (2s + t) ๐๏ทฏ] = 15 x (s โ 2t) + y(3 โ t) + z(2s + t) = 15 (s โ 2t) x + (3 โ t) y + (2s + t) z = 15 which is the equation of the plane in Cartesian form.
Ex 11.3
Ex 11.3, 2
Ex 11.3, 3 You are here
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)
Ex 11.3, 14 (d)
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