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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
Ex 11.3, 2 Deleted for CBSE Board 2023 Exams
Ex 11.3, 3 (a) Deleted for CBSE Board 2023 Exams
Ex 11.3, 3 (b) Deleted for CBSE Board 2023 Exams
Ex 11.3, 3 (c) Important Deleted for CBSE Board 2023 Exams
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Ex 11.3, 4 (b) Deleted for CBSE Board 2023 Exams
Ex 11.3, 4 (c) Deleted for CBSE Board 2023 Exams
Ex 11.3, 4 (d) Important Deleted for CBSE Board 2023 Exams
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, 8 Deleted for CBSE Board 2023 Exams
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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
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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 22, 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 (3, β2, 1) So, π₯_1 = 3, π¦_1 = β2, π§_1 = 1 a And the equation of the plane is 2x β y + 2z + 3 = 0 2x β y + 2z = β3 β(2x β y + 2z) = 3 β2x + y β 2z = 3 Comparing with Ax + By + Cz = D, A = β2, B = 1, C = β2, D = 3 Now, Distance of the point form the plane = |((β2 Γ 3) + (1 Γ β2) + (β2 Γ 1) β 3)/β((β2)^2 + 1^2 + (2)^2 )| = |((β6) + (β2) + (β2) β 3)/β(4 + 1 + 4)| = |(β13)/β9| = |(β13)/3| = ππ/π