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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


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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

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