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  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise

Transcript

Ex 10.3, 17 (Method 1) Show that the vectors 2𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + π‘˜ Μ‚, 𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚ , 3𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚ form the vertices of a right angled triangle. Let A(2𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + π‘˜ Μ‚), B(𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚) C(3𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚) We know that two vectors are perpendicular to each other, i.e have an angle of 90Β° between them , if their scalar product is zero. (𝐴𝐡) βƒ— = (𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚) βˆ’ (2𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + π‘˜ Μ‚) = 1𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚ βˆ’ 2𝑖 Μ‚ + 1𝑗 Μ‚ βˆ’ 1π‘˜ Μ‚ = (1 βˆ’ 2) 𝑖 Μ‚ + (βˆ’3 + 1) 𝑗 Μ‚ + (βˆ’5 βˆ’1) π‘˜ Μ‚ = βˆ’1𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ βˆ’ 6π‘˜ Μ‚ (𝐡𝐢) βƒ— = (3𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚) βˆ’ (𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚) = 3𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚ βˆ’ 1𝑖 Μ‚ + 3𝑗 Μ‚ + 5π‘˜ Μ‚ = (3 βˆ’ 1) 𝑖 Μ‚ + (βˆ’4 + 3) 𝑗 Μ‚ + (βˆ’4 + 5) π‘˜ Μ‚ = 2𝑖 Μ‚ βˆ’ 1𝑗 Μ‚ + 1π‘˜ Μ‚ (𝐢𝐴) βƒ— = (2𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + π‘˜ Μ‚) βˆ’ (3𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚) = 2𝑖 Μ‚ βˆ’ 1𝑗 Μ‚ + 1π‘˜ Μ‚ βˆ’ 3𝑖 Μ‚ + 4𝑗 Μ‚ + 4π‘˜ Μ‚ = (2 βˆ’ 3) 𝑖 Μ‚ + (βˆ’1 + 4) 𝑗 Μ‚ + (1 + 4) π‘˜ Μ‚ = βˆ’1𝑖 Μ‚ + 3𝑗 Μ‚ + 5π‘˜ Μ‚ Now, (𝐡𝐢) βƒ—. (𝐢𝐴) βƒ— = (2𝑖 Μ‚ βˆ’ 1𝑗 Μ‚ + 1π‘˜ Μ‚) . (-1𝑖 Μ‚ + 3𝑗 Μ‚ + 5π‘˜ Μ‚) = (2 Γ— –1) + (βˆ’1 Γ— 3) + (1 Γ— 5) = (βˆ’2) + (βˆ’3) + 5 = βˆ’5 + 5 = 0 Since, (𝑩π‘ͺ) βƒ—. (π‘ͺ𝑨) βƒ— = 0 Therefore, (𝐡𝐢) βƒ— is perpendicular to (𝐢𝐴) βƒ— . Hence Ξ” ABC is a right angled triangle Ex 10.3, 17 (Method 2) Show that the vectors 2𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + π‘˜ Μ‚, 𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚ , 3𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚ form the vertices of a right angled triangle. Let A(2𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + π‘˜ Μ‚), B(𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚) C(3𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚) Considering βˆ†ABC as a right angled triangle, By Pythagoras theorem, AB2 = BC2 + CA2 or |("AB" ) βƒ— |"2" = |("BC" ) βƒ— |"2" + |("CA" ) βƒ— |"2" (𝐴𝐡) βƒ— = (𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚) βˆ’ (2𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + π‘˜ Μ‚) = 1𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚ βˆ’ 2𝑖 Μ‚ + 1𝑗 Μ‚ βˆ’ 1π‘˜ Μ‚ = (1 βˆ’ 2) 𝑖 Μ‚ + (βˆ’3 + 1) 𝑗 Μ‚ + (βˆ’5 βˆ’1) π‘˜ Μ‚ = βˆ’1𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ βˆ’ 6π‘˜ Μ‚ (𝐡𝐢) βƒ— = (3𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚) βˆ’ (𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚) = 3𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚ βˆ’ 1𝑖 Μ‚ + 3𝑗 Μ‚ + 5π‘˜ Μ‚ = (3 βˆ’ 1) 𝑖 Μ‚ + (βˆ’4 + 3) 𝑗 Μ‚ + (βˆ’4 + 5) π‘˜ Μ‚ = 2𝑖 Μ‚ βˆ’ 1𝑗 Μ‚ + 1π‘˜ Μ‚ (𝐢𝐴) βƒ— = (2𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + π‘˜ Μ‚) βˆ’ (3𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚) = 2𝑖 Μ‚ βˆ’ 1𝑗 Μ‚ + 1π‘˜ Μ‚ βˆ’ 3𝑖 Μ‚ + 4𝑗 Μ‚ + 4π‘˜ Μ‚ = (2 βˆ’ 3) 𝑖 Μ‚ + (βˆ’1 + 4) 𝑗 Μ‚ + (1 + 4) π‘˜ Μ‚ = βˆ’1𝑖 Μ‚ + 3𝑗 Μ‚ + 5π‘˜ Μ‚ Now, "Magnitude of " (𝐴𝐡) βƒ—" = " √((βˆ’1)2+(βˆ’2)2+(βˆ’6)2) " " |(𝐴𝐡) βƒ— |" = " √(1+4+36) " = " √41 Magnitude of (𝐡𝐢) βƒ— = √(22+(βˆ’1)2+1) |(𝐡𝐢) βƒ— | = √(4+1+1) = √6 Magnitude of (𝐢𝐴) βƒ— = √((βˆ’1)2+32+52) |(𝐢𝐴) βƒ— | = √(1+9+25) = √35 Now, |(𝐡𝐢) βƒ— |2 + |(𝐢𝐴) βƒ— |2 = (√6)2 + (√35)2 = 6 + 35 = 41 = (√41)2 = |(𝐴𝐡) βƒ— |2 Thus, |(𝐴𝐡) βƒ— |^2 = |(𝐡𝐢) βƒ— |^2 + |(𝐢𝐴) βƒ— |^2 So, ABC is a right angled triangle.

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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.