Ex 11.3, 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 6

Ex 11.3, 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 7
Ex 11.3, 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 8
Ex 11.3, 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 9

Β 

  1. Chapter 11 Class 12 Three Dimensional Geometry (Term 2)
  2. Serial order wise

Transcript

Ex 11.3, 5 Find the vector and cartesian equations of the planes (b) That passes through the point (1, 4, 6) and the normal vector to the plane is 𝑖 Μ‚ – 2𝑗 Μ‚ + 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, 4, 6) So, 𝒂 βƒ— = 1𝑖 Μ‚ + 4𝑗 Μ‚ + 6π‘˜ Μ‚ Normal to plane = 𝑖 Μ‚ – 2𝑗 Μ‚ + π‘˜ Μ‚ 𝒏 βƒ— = 𝑖 Μ‚ – 2𝑗 Μ‚ + π‘˜ Μ‚ Vector equation of plane is (π‘Ÿ βƒ— βˆ’ π‘Ž βƒ—).𝑛 βƒ— = 0 [𝒓 βƒ—βˆ’(π’Š Μ‚+πŸ’π’‹ Μ‚+πŸ”π’Œ Μ‚)] . (π’Š Μ‚ βˆ’ 2𝒋 Μ‚ + π’Œ Μ‚) = 0 Now, Finding Cartesian form using two methods Cartesian form (Method 1) : Vector equation is [π‘Ÿ βƒ—βˆ’(𝑖 Μ‚+4𝑗 Μ‚+6π‘˜ Μ‚)] . (𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + π‘˜ Μ‚) = 0 Put 𝒓 βƒ— = xπ’Š Μ‚ + y𝒋 Μ‚ + zπ’Œ Μ‚ [(π‘₯𝑖 Μ‚+𝑦𝑗 Μ‚+π‘§π‘˜ Μ‚ )βˆ’(1𝑖 Μ‚+4𝑗 Μ‚+6π‘˜ Μ‚)]. (1𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + 1π‘˜ Μ‚) = 0 [(π‘₯βˆ’1) 𝑖 Μ‚ +(π‘¦βˆ’4) 𝑗 Μ‚+(π‘§βˆ’6)π‘˜ Μ‚ ]. (1𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + 1π‘˜ Μ‚) = 0 1(x βˆ’ 1) + (βˆ’2)(y βˆ’ 4) + 1 (z βˆ’ 6) = 0 x βˆ’ 1 βˆ’ 2(y βˆ’ 4) + z βˆ’ 6 = 0 x βˆ’ 2y + z + 1 = 0 ∴ Equation of plane in Cartesian form is x βˆ’ 2y + z + 1 = 0. 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, 4, 6) x1 = 1, y1 = 4, z1 = 6 And normal is 𝑖 Μ‚ – 2𝑗 Μ‚ + π‘˜ Μ‚ So, Direction ratios of line perpendicular to plane = 1, –2, 1 ∴ A = 1, B = –2, C = 1 Therefore, equation of line in Cartesian form is 1(x βˆ’ 1) – 2 (y βˆ’ 4) + 1(x βˆ’ 6) = 0 x βˆ’ 2y + z + 1 = 0

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.