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Ex 11.3, 5 - Find vector and cartesian equation of planes (a) that

Ex 11.3, 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2

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Ex 11.3, 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 5

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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, –2) and the normal to the plane is 𝑖 Μ‚ + 𝑗 Μ‚ βˆ’ π‘˜ Μ‚.Vector equation of a plane passing through a point (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is [π‘Ÿ βƒ— βˆ’(π‘₯1𝑖 Μ‚+𝑦1𝑗 Μ‚+𝑧1π‘˜ Μ‚)]. (A𝑖 Μ‚ + B𝑗 Μ‚ + Cπ‘˜ Μ‚) = 0 or (π‘Ÿ βƒ— βˆ’ π‘Ž βƒ—).𝑛 βƒ— = 0 ("A" 𝑃) βƒ— is perpendicular to "n" βƒ— So, ("A" P) βƒ— . "n" βƒ— = 0 ("r" βƒ— βˆ’ "a" βƒ—)."n" βƒ— = 0 Ex 11.3, 5 Find the vector and Cartesian equations of the planes (a) that passes through the point (1, 0, – 2) and the normal to the plane is 𝑖 Μ‚ + 𝑗 Μ‚ βˆ’ π‘˜ Μ‚.Vector equation Equation of plane passing through point A whose position vector is 𝒂 βƒ— & perpendicular to 𝒏 βƒ— is (𝒓 βƒ— βˆ’ 𝒂 βƒ—) . 𝒏 βƒ— = 0 Given Plane passes through (1, 0, βˆ’2) So, 𝒂 βƒ— = 1𝑖 Μ‚ + 0𝑗 Μ‚ – 2π‘˜ Μ‚ Normal to plane = 𝑖 Μ‚ + 𝑗 Μ‚ – π‘˜ Μ‚ 𝒏 βƒ— = 𝑖 Μ‚ + 𝑗 Μ‚ βˆ’ π‘˜ Μ‚ Vector equation of plane is (𝒓 βƒ— βˆ’ 𝒂 βƒ—) . 𝒏 βƒ— = 0 [π‘Ÿ βƒ—βˆ’("1" 𝑖 Μ‚+"0" 𝑗 Μ‚βˆ’"2" π‘˜ Μ‚)].(𝑖 Μ‚ + 𝑗 Μ‚ βˆ’ π‘˜ Μ‚) = 0 [𝒓 βƒ—βˆ’(π’Š Μ‚βˆ’"2" π’Œ Μ‚)] . (π’Š Μ‚+𝒋 Μ‚βˆ’π’Œ Μ‚) = 0 Now, Finding Cartesian form using two methods Cartesian form (Method 1) Vector equation is [π‘Ÿ βƒ—βˆ’(𝑖 Μ‚βˆ’"2" π‘˜ Μ‚)] . (𝑖 Μ‚+𝑗 Μ‚βˆ’π‘˜ Μ‚) = 0 Put 𝒓 βƒ— = xπ’Š Μ‚ + y𝒋 Μ‚ + zπ’Œ Μ‚ [("x" 𝑖 Μ‚+"y" 𝑗 Μ‚+"z" π‘˜ Μ‚ )βˆ’("1" 𝑖 Μ‚+"0" 𝑗 Μ‚βˆ’"2" π‘˜ Μ‚)].(1𝑖 Μ‚ + 1𝑗 Μ‚ – 1π‘˜ Μ‚) = 0 [("x" βˆ’1) 𝑖 Μ‚+("y" βˆ’0) 𝐽 Μ‚+(π‘§βˆ’(βˆ’2))π‘˜ Μ‚ ].(1𝑖 Μ‚ + 1𝑗 Μ‚ – 1π‘˜ Μ‚) = 0 1(x βˆ’ 1) + 1(y βˆ’ 0) – 1 (z + 2) = 0 x βˆ’ 1 + y – z – 2 = 0 x + y – 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 βˆ’ x1) + B(y βˆ’ y1) + C (z βˆ’ z1) = 0 Since the plane passes through (1, 0, –2) x1 = 1, y1 = 0, z1 = 2 And normal is 𝑖 Μ‚ + 𝑗 Μ‚ – π‘˜ Μ‚ So, Direction ratios of line perpendicular to plane = 1, 1, –1 ∴ A = 1, B = 1, C = –1 Therefore, equation of line in Cartesian form is 1(x βˆ’ 1) + 1(y βˆ’ 0) βˆ’ 1(z βˆ’ (βˆ’2)) = 0 x + y – z = 3

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.