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Ex 11.3
Ex 11.3, 1 (b) Deleted for CBSE Board 2023 Exams
Ex 11.3, 1 (c) Important Deleted for CBSE Board 2023 Exams
Ex 11.3, 1 (d) Important Deleted for CBSE Board 2023 Exams
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Ex 11.3, 3 (a) Deleted for CBSE Board 2023 Exams
Ex 11.3, 3 (b) Deleted for CBSE Board 2023 Exams
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Ex 11.3, 4 (a) Important Deleted for CBSE Board 2023 Exams
Ex 11.3, 4 (b) Deleted for CBSE Board 2023 Exams
Ex 11.3, 4 (c) Deleted for CBSE Board 2023 Exams
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Ex 11.3, 5 (a) Important Deleted for CBSE Board 2023 Exams
Ex 11.3, 5 (b) Deleted for CBSE Board 2023 Exams
Ex 11.3, 6 (a) Important Deleted for CBSE Board 2023 Exams
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Ex 11.3, 7 Deleted for CBSE Board 2023 Exams
Ex 11.3, 8 Deleted for CBSE Board 2023 Exams
Ex 11.3, 9 Deleted for CBSE Board 2023 Exams
Ex 11.3, 10 Important Deleted for CBSE Board 2023 Exams
Ex 11.3, 11 Important Deleted for CBSE Board 2023 Exams
Ex 11.3, 12 Important Deleted for CBSE Board 2023 Exams
Ex 11.3, 13 (a) Important Deleted for CBSE Board 2023 Exams
Ex 11.3, 13 (b) Important Deleted for CBSE Board 2023 Exams
Ex 11.3, 13 (c) Deleted for CBSE Board 2023 Exams
Ex 11.3, 13 (d) Deleted for CBSE Board 2023 Exams
Ex 11.3, 13 (e) Deleted for CBSE Board 2023 Exams
Ex 11.3, 14 (a) Important Deleted for CBSE Board 2023 Exams You are here
Ex 11.3, 14 (b) Deleted for CBSE Board 2023 Exams
Ex 11.3, 14 (c) Deleted for CBSE Board 2023 Exams
Ex 11.3, 14 (d) Important Deleted for CBSE Board 2023 Exams
Last updated at March 30, 2023 by Teachoo
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 |(π¨π_π + γπ©πγ_π +γ πͺπγ_π β π«)/β(π¨^π + π©^π + πͺ^π )| Given, the point is (0, 0, 0) So, π₯_1 = 0, π¦_1 = 0, π§_1 = 0 and the equation of plane is 3x β 4y + 12z = 3 Comparing with Ax + By + Cz = D, A = 3, B = β4, C = 12, D = 3 Now, Distance of point from the plane is = |((3 Γ 0) + (β4 Γ 0) + (12 Γ 0) β 3)/( β(3^2 + (β4)^2 + γ12γ^2 ))| = |(0 + 0 + 0 β 3)/( β(9 + 16 + 144))| = |3/( β169)| = π/ππ