Check sibling questions

Ex 10.3, 17 - Show vectors form vertices of right angled triangle

Ex 10.3, 17 - Chapter 10 Class 12 Vector Algebra - Part 2
Ex 10.3, 17 - Chapter 10 Class 12 Vector Algebra - Part 3
Ex 10.3, 17 - Chapter 10 Class 12 Vector Algebra - Part 4
Ex 10.3, 17 - Chapter 10 Class 12 Vector Algebra - Part 5
Ex 10.3, 17 - Chapter 10 Class 12 Vector Algebra - Part 6
Ex 10.3, 17 - Chapter 10 Class 12 Vector Algebra - Part 7

Solve all your doubts with Teachoo Black (new monthly pack available now!)


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.

Davneet Singh's photo - Co-founder, Teachoo

Made by

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.