Subscribe to our Youtube Channel - https://www.youtube.com/channel/UCZBx269Tl5Os5NHlSbVX4Kg

Slide42.JPG

Slide43.JPG
Slide44.JPG Slide45.JPG Slide46.JPG Slide47.JPG

  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.

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.