Collinearity of 3 points or 3 position vectors

Chapter 10 Class 12 Vector Algebra
Concept wise

### Transcript

Example 21 (Introduction) Show that the points A(−2𝑖 ̂ + 3𝑗 ̂ + 5𝑘 ̂), B(𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂) and C(7𝑖 ̂ − 𝑘 ̂) are collinear. (1) Three points collinear i.e. AB + BC = AC (2) Three position vectors collinear i.e. |(𝐴𝐵) ⃗ | + |(𝐵𝐶) ⃗ | = |(𝐴𝐶) ⃗ | Example 21 Show that the points A(−2𝑖 ̂ + 3𝑗 ̂ + 5𝑘 ̂), B(𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂) and C(7𝑖 ̂ − 𝑘 ̂) are collinear. Given A (−2𝑖 ̂ + 3𝑗 ̂ + 5𝑘 ̂) B (1𝑖 ̂ + 2𝑗 ̂ + 3𝑘 ̂) C (7𝑖 ̂ + 0𝑗 ̂ − 1𝑘 ̂) 3 points A, B, C are collinear if |(𝑨𝑩) ⃗ | + |(𝑩𝑪) ⃗ | = |(𝑨𝑪) ⃗ | Finding (𝑨𝑩) ⃗ , (𝑩𝑪) ⃗ , (𝑨𝑪) ⃗ (𝑨𝑩) ⃗ = (1 – (-2)) 𝑖 ̂ + (2 − 3) 𝑗 ̂ + (3 − 5) 𝑘 ̂ = 3𝒊 ̂ – 1𝒋 ̂ – 2𝒌 ̂ (𝑩𝑪) ⃗ = (7 − 1) 𝑖 ̂ + (0 − 2) 𝑗 ̂ + (-1−3) 𝑘 ̂ = 6𝒊 ̂ – 2𝒋 ̂ – 4𝒌 ̂ (𝑨𝑪) ⃗ = (7 − (-2)) 𝑖 ̂ + (0 − 3) 𝑗 ̂ + (-1 − 5) 𝑘 ̂ = 9𝒊 ̂ – 3𝒋 ̂ – 6𝒌 ̂ Magnitude of |(𝐴𝐵) ⃗ | = √(3^2+(−1)^2+(−2)^2 ) |(𝑨𝑩) ⃗ | = √(9+1+4) = √𝟏𝟒 Magnitude of |(𝐵𝐶) ⃗ | = √(6^2+(−2)^2+(−4)^2 ) |(𝑩𝑪) ⃗ | = √(36+4+16) = √56 = √(4 × 14) = 2√𝟏𝟒 Magnitude of |(𝐴𝐶) ⃗ | = √(9^2+(−3)^2+(−6)^2 ) |(𝑨𝑪) ⃗ | = √(81+9+36) = √126 = √(9 × 14) = 3√𝟏𝟒 Thus, |(𝑨𝑩) ⃗ | + |(𝑩𝑪) ⃗ | = √14 + 2√14 = 3√14 = |(𝑨𝑪) ⃗ | Thus, A, B and C are collinear.

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