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1. Chapter 10 Class 12 Vector Algebra
2. Serial order wise
3. Ex 10.3

Transcript

Ex 10.3, 16 (Introduction) Show that the points A (1, 2, 7), B (2, 6, 3) and C (3, 10, –1) are collinear. (1) Three points collinear i.e. AB + BC = AC (2) Three vectors collinear i.e. |(𝐴𝐵) ⃗ | + |(𝐵𝐶) ⃗ | = |(𝐴𝐶) ⃗ | Ex 10.3, 16 Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear. 3 points A, B, C are collinear if i.e. |(𝑨𝑩) ⃗ | + |(𝑩𝑪) ⃗ | = |(𝑨𝑪) ⃗ | A (1, 2, 7) B (2, 6, 3) C (3, 10, –1) (𝐴𝐵) ⃗ = (2 − 1) 𝑖 ̂ + (6 − 2) 𝑗 ̂ + (3 − 7) 𝑘 ̂ = 1𝑖 ̂ + 4𝑗 ̂ − 4𝑘 ̂ (𝐵𝐶) ⃗ = (3 − 2) 𝑖 ̂ + (10 − 6) 𝑗 ̂ + (−1 − 3) 𝑘 ̂ = 1𝑖 ̂ + 4𝑗 ̂ − 4𝑘 ̂ (𝐴𝐶) ⃗ = (3 − 1) 𝑖 ̂ + (10 − 2) 𝑗 ̂ + (−1 − 7) 𝑘 ̂ = 2𝑖 ̂ + 8𝑗 ̂ − 8𝑘 ̂ Magnitude of (𝐴𝐵) ⃗ = √(12+42+(−4)2) |(𝐴𝐵) ⃗ | = √(1+16+16) = √33 Magnitude of (𝐵𝐶) ⃗ = √(12+42+(−4)2) |(𝐵𝐶) ⃗ | = √(1+16+16) = √33 Magnitude of (𝐴𝐶) ⃗ = √(22+82+(−8)2) |(𝐵𝐶) ⃗ | = √(4+64+64) = √132 = √(4×33 )= 2√(33 ) Thus, |(𝐴𝐵) ⃗ | + |(𝐵𝐶) ⃗ | = √(33 ) + √(33 ) = 2√(33 ) = |(𝐴𝐶) ⃗ | Thus, A, B and C are collinear.

Ex 10.3 