Check sibling questions

Example 12 - Show that points A(2i - j + k), B(i - 3j + 5k), C (3i-4j-

Example 12 - Chapter 10 Class 12 Vector Algebra - Part 2
Example 12 - Chapter 10 Class 12 Vector Algebra - Part 3

Example 12 - Chapter 10 Class 12 Vector Algebra - Part 4 Example 12 - Chapter 10 Class 12 Vector Algebra - Part 5 Example 12 - Chapter 10 Class 12 Vector Algebra - Part 6 Example 12 - Chapter 10 Class 12 Vector Algebra - Part 7

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Example 12 (Method 1) Show that the points 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‚) are the vertices of a right angled triangle. Given 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, if their scalar product is zero. Finding (𝑨𝑩) βƒ— , (𝑩π‘ͺ) βƒ— , (𝑨π‘ͺ) βƒ— (𝑨𝑩) βƒ— = (𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚) βˆ’ (2𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + π‘˜ Μ‚) = (1 βˆ’ 2) 𝑖 Μ‚ + (βˆ’3 + 1) 𝑗 Μ‚ + (βˆ’5 βˆ’ 1) π‘˜ Μ‚ = βˆ’1π’Š Μ‚ βˆ’ 2𝒋 Μ‚ βˆ’ 6π’Œ Μ‚ (𝑩π‘ͺ) βƒ— = (3𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚) βˆ’ (𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚) = (3 βˆ’ 1) 𝑖 Μ‚ + (βˆ’4 + 3) 𝑗 Μ‚ + (βˆ’4 + 5) π‘˜ Μ‚ = 2π’Š Μ‚ βˆ’ 1𝒋 Μ‚ + 1π’Œ Μ‚ (π‘ͺ𝑨) βƒ— = (2𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + π‘˜ Μ‚) βˆ’ (3𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚) = (2 βˆ’ 3) 𝑖 Μ‚ + (βˆ’1 + 4) 𝑗 Μ‚ + (1 + 4) π‘˜ Μ‚ = βˆ’1π’Š Μ‚ + 3𝒋 Μ‚ + 5π’Œ Μ‚ Finding (𝑩π‘ͺ) βƒ—. (π‘ͺ𝑨) βƒ— (𝑩π‘ͺ) βƒ—. (π‘ͺ𝑨) 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 Example 12 (Method 2) Show that the points 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‚) are the vertices of a right angled triangle. Given 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" Finding (𝑨𝑩) βƒ— , (𝑩π‘ͺ) βƒ— , (𝑨π‘ͺ) βƒ— (𝑨𝑩) βƒ— = (𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚) βˆ’ (2𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + π‘˜ Μ‚) = (1 βˆ’ 2) 𝑖 Μ‚ + (βˆ’3 + 1) 𝑗 Μ‚ + (βˆ’5 βˆ’1) π‘˜ Μ‚ = βˆ’1π’Š Μ‚ βˆ’ 2𝒋 Μ‚ βˆ’ 6π’Œ Μ‚ (𝑩π‘ͺ) βƒ— = (3𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚) βˆ’ (𝑖 Μ‚ βˆ’ 3𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚) = (3 βˆ’ 1) 𝑖 Μ‚ + (βˆ’4 + 3) 𝑗 Μ‚ + (βˆ’4 + 5) π‘˜ Μ‚ = 2π’Š Μ‚ βˆ’ 1𝒋 Μ‚ + 1π’Œ Μ‚ (π‘ͺ𝑨) βƒ— = (2𝑖 Μ‚ βˆ’ 𝑗 Μ‚ + π‘˜ Μ‚) βˆ’ (3𝑖 Μ‚ βˆ’ 4𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚) = (2 βˆ’ 3) 𝑖 Μ‚ + (βˆ’1 + 4) 𝑗 Μ‚ + (1 + 4) π‘˜ Μ‚ = βˆ’1π’Š Μ‚ + 3𝒋 Μ‚ + 5π’Œ Μ‚ Now, "Magnitude of " (𝑨𝑩) βƒ—" = " √((βˆ’1)2+(βˆ’2)2+(βˆ’6)2) " " |(𝐴𝐡) βƒ— |" = " √(1+4+36) " = " βˆšπŸ’πŸ Magnitude of (𝑩π‘ͺ) βƒ— = √(22+(βˆ’1)2+1) |(𝐡𝐢) βƒ— | = √(4+1+1) = βˆšπŸ” Magnitude of (π‘ͺ𝑨) βƒ— = √((βˆ’1)2+32+52) |(𝐢𝐴) βƒ— | = √(1+9+25) = βˆšπŸ‘πŸ“ Now, |(𝑩π‘ͺ) βƒ— |^𝟐 + |(π‘ͺ𝑨) βƒ— |^𝟐 = (√6)2 + (√35)2 = 6 + 35 = 41 = (√41)2 = |(𝑨𝑩) βƒ— |^𝟐 Thus, |(𝑨𝑩) βƒ— |^𝟐 = |(𝑩π‘ͺ) βƒ— |^𝟐 + |(π‘ͺ𝑨) βƒ— |^𝟐 Hence, by Pythagoras Theorem, Ξ” ABC is a right angled triangle.

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.