Collinearity of 3 points or 3 position vectors

Chapter 10 Class 12 Vector Algebra
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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. 