Misc 8 - Show A, B, C are collinear, find ratio where B - Miscellaneou

Misc 8 - Chapter 10 Class 12 Vector Algebra - Part 2

Misc 8 - Chapter 10 Class 12 Vector Algebra - Part 3 Misc 8 - Chapter 10 Class 12 Vector Algebra - Part 4 Misc 8 - Chapter 10 Class 12 Vector Algebra - Part 5 Misc 8 - Chapter 10 Class 12 Vector Algebra - Part 6

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

Transcript

Misc 8 (Introduction) Show that the points A(1, – 2, – 8), B (5, 0, – 2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC. (1) Three points collinear i.e. AB + BC = AC (2) Three position vectors collinear i.e. |(𝐴𝐡) βƒ— | + |(𝐡𝐢) βƒ— | = |(𝐴𝐢) βƒ— | Misc 8 Show that the points A(1, – 2, – 8), B (5, 0, – 2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC. vGiven A (1, -2, βˆ’8) B (5, 0,βˆ’2) C (11, 3, 7) 3 points A, B, C are collinear if |(𝑨𝑩) βƒ— | + |(𝑩π‘ͺ) βƒ— | = |(𝑨π‘ͺ) βƒ— | Finding (𝑨𝑩) βƒ— , (𝑩π‘ͺ) βƒ— , (𝑨π‘ͺ) βƒ— (𝑨𝑩) βƒ— = (5 βˆ’ 1) 𝑖 Μ‚ + (0 βˆ’ (βˆ’2)) 𝑗 Μ‚ + (βˆ’2βˆ’(βˆ’8)) π‘˜ Μ‚ = 4𝑖 Μ‚ + 2𝑗 Μ‚ + 6π‘˜ Μ‚ (𝑩π‘ͺ) βƒ— = (11 βˆ’ 5) 𝑖 Μ‚ + (3 βˆ’ 0) 𝑗 Μ‚ + (7βˆ’(βˆ’2)) π‘˜ Μ‚ = 6π’Š Μ‚ + 3𝒋 Μ‚ + 9π’Œ Μ‚ (𝑨π‘ͺ) βƒ— = (11 βˆ’ 1) 𝑖 Μ‚ + (3 βˆ’ (βˆ’2)) 𝑗 Μ‚ + (7βˆ’(βˆ’8)) π‘˜ Μ‚ = 10π’Š Μ‚ + 5𝒋 Μ‚ + 15π’Œ Μ‚ Magnitude of (𝐴𝐡) βƒ— = √(42+22+62) |(𝑨𝑩) βƒ— | = √(16+4+36) = √56 = √(4Γ—14 ) = 2βˆšπŸπŸ’ Magnitude of (𝐡𝐢) βƒ— = √(62+32+92) |(𝑩π‘ͺ) βƒ— |= √(36+9+81) = √126 = √(9Γ—14 ) = 3βˆšπŸπŸ’ Magnitude of (𝐴𝐢) βƒ— = √(102+52+152) |(𝑨π‘ͺ) βƒ— |= √(100+25+225)= √350 = √(25 Γ— 14 ) = 5βˆšπŸπŸ’ Thus, |(𝑨𝑩) βƒ— | + |(𝑩π‘ͺ) βƒ— | = 2√(14 ) + 3√(14 ) = 5√(14 ) = |(𝑨π‘ͺ) βƒ— | Thus, A, B and C are collinear. Finding the ratio in which B divides AC Let B divide AC in the ratio k : 1 Here, (𝑢𝑨) βƒ— = 1𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ βˆ’ 8π‘˜ Μ‚ (𝑢𝑩) βƒ— = 5𝑖 Μ‚ + 0𝑗 Μ‚ βˆ’ 2π‘˜ Μ‚ and (𝑢π‘ͺ) βƒ— = 11𝑖 Μ‚ + 3𝑗 Μ‚ + 7π‘˜ Μ‚ Position vector of 𝑩 = (π’Œ(𝑢π‘ͺ) βƒ— + 𝟏.(𝑢𝑨) βƒ—)/(π’Œ + 𝟏) 5𝑖 Μ‚ + 0𝑗 Μ‚ βˆ’ 2π‘˜ Μ‚ = (π‘˜(11𝑖 Μ‚ + 3𝑗 Μ‚ + 7π‘˜ Μ‚ ) + 1(1𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ βˆ’ 8π‘˜ Μ‚ ))/(π‘˜ + 1) 5𝑖 Μ‚ + 0𝑗 Μ‚ βˆ’ 2π‘˜ Μ‚ = (11π‘˜π‘– Μ‚ + 3π‘˜π‘— Μ‚ + 7π‘˜ π‘˜ Μ‚ + 𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ βˆ’ 8π‘˜ Μ‚)/(π‘˜ + 1) 5π’Š Μ‚ + 0𝒋 Μ‚ βˆ’ 2π’Œ Μ‚ = ((πŸπŸπ’Œ + 𝟏) π’Š Μ‚ + (πŸ‘π’Œ βˆ’ 𝟐) 𝒋 Μ‚ + (πŸ•π’Œ βˆ’ πŸ–) π’Œ Μ‚)/(π’Œ + 𝟏) Since the two vectors are equal, corresponding components are also equal. So, (πŸπŸπ’Œ + 𝟏)/(π’Œ + 𝟏) = 5 11k + 1 = 5k + 5 11k – 5k = 5 βˆ’ 1 6k = 4 k = 4/6 = 2/3 Thus, B divides AC in the ratio 2 : 3

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.