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Chapter 11 Class 12 Three Dimensional Geometry
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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

Ex 11.3, 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3 Ex 11.3, 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.