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Ex 11.3

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Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at Aug. 24, 2021 by Teachoo

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Ex 11.3, 5 (Introduction) Find the vector and cartesian equations of the planes (a) that passes through the point (1, 0, โ2) and the normal to the plane is ๐ ฬ + ๐ ฬ โ ๐ ฬ.Vector equation of a plane passing through a point (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is [๐ โ โ(๐ฅ1๐ ฬ+๐ฆ1๐ ฬ+๐ง1๐ ฬ)]. (A๐ ฬ + B๐ ฬ + C๐ ฬ) = 0 or (๐ โ โ ๐ โ).๐ โ = 0 ("A" ๐) โ is perpendicular to "n" โ So, ("A" P) โ . "n" โ = 0 ("r" โ โ "a" โ)."n" โ = 0 Ex 11.3, 5 Find the vector and Cartesian equations of the planes (a) that passes through the point (1, 0, โ 2) and the normal to the plane is ๐ ฬ + ๐ ฬ โ ๐ ฬ.Vector equation Equation of plane passing through point A whose position vector is ๐ โ & perpendicular to ๐ โ is (๐ โ โ ๐ โ) . ๐ โ = 0 Given Plane passes through (1, 0, โ2) So, ๐ โ = 1๐ ฬ + 0๐ ฬ โ 2๐ ฬ Normal to plane = ๐ ฬ + ๐ ฬ โ ๐ ฬ ๐ โ = ๐ ฬ + ๐ ฬ โ ๐ ฬ Vector equation of plane is (๐ โ โ ๐ โ) . ๐ โ = 0 [๐ โโ("1" ๐ ฬ+"0" ๐ ฬโ"2" ๐ ฬ)].(๐ ฬ + ๐ ฬ โ ๐ ฬ) = 0 [๐ โโ(๐ ฬโ"2" ๐ ฬ)] . (๐ ฬ+๐ ฬโ๐ ฬ) = 0 Now, Finding Cartesian form using two methods Cartesian form (Method 1) Vector equation is [๐ โโ(๐ ฬโ"2" ๐ ฬ)] . (๐ ฬ+๐ ฬโ๐ ฬ) = 0 Put ๐ โ = x๐ ฬ + y๐ ฬ + z๐ ฬ [("x" ๐ ฬ+"y" ๐ ฬ+"z" ๐ ฬ )โ("1" ๐ ฬ+"0" ๐ ฬโ"2" ๐ ฬ)].(1๐ ฬ + 1๐ ฬ โ 1๐ ฬ) = 0 [("x" โ1) ๐ ฬ+("y" โ0) ๐ฝ ฬ+(๐งโ(โ2))๐ ฬ ].(1๐ ฬ + 1๐ ฬ โ 1๐ ฬ) = 0 1(x โ 1) + 1(y โ 0) โ 1 (z + 2) = 0 x โ 1 + y โ z โ 2 = 0 x + y โ z = 3 Cartesian form (Method 2) Equation of plane passing through (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is A(x โ x1) + B(y โ y1) + C (z โ z1) = 0 Since the plane passes through (1, 0, โ2) x1 = 1, y1 = 0, z1 = 2 And normal is ๐ ฬ + ๐ ฬ โ ๐ ฬ So, Direction ratios of line perpendicular to plane = 1, 1, โ1 โด A = 1, B = 1, C = โ1 Therefore, equation of line in Cartesian form is 1(x โ 1) + 1(y โ 0) โ 1(z โ (โ2)) = 0 x + y โ z = 3