Examples

Chapter 10 Class 12 Vector Algebra
Serial order wise

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Example 21 (Introduction) Show that the points A(β2π Μ + 3π Μ + 5π Μ), B(π Μ + 2π Μ + 3π Μ) and C(7π Μ β π Μ) are collinear. (1) Three points collinear i.e. AB + BC = AC (2) Three position vectors collinear i.e. |(π΄π΅) β | + |(π΅πΆ) β | = |(π΄πΆ) β | Example 21 Show that the points A(β2π Μ + 3π Μ + 5π Μ), B(π Μ + 2π Μ + 3π Μ) and C(7π Μ β π Μ) are collinear. Given A (β2π Μ + 3π Μ + 5π Μ) B (1π Μ + 2π Μ + 3π Μ) C (7π Μ + 0π Μ β 1π Μ) 3 points A, B, C are collinear if |(π¨π©) β | + |(π©πͺ) β | = |(π¨πͺ) β | Finding (π¨π©) β , (π©πͺ) β , (π¨πͺ) β (π¨π©) β = (1 β (-2)) π Μ + (2 β 3) π Μ + (3 β 5) π Μ = 3π Μ β 1π Μ β 2π Μ (π©πͺ) β = (7 β 1) π Μ + (0 β 2) π Μ + (-1β3) π Μ = 6π Μ β 2π Μ β 4π Μ (π¨πͺ) β = (7 β (-2)) π Μ + (0 β 3) π Μ + (-1 β 5) π Μ = 9π Μ β 3π Μ β 6π Μ Magnitude of |(π΄π΅) β | = β(3^2+(β1)^2+(β2)^2 ) |(π¨π©) β | = β(9+1+4) = βππ Magnitude of |(π΅πΆ) β | = β(6^2+(β2)^2+(β4)^2 ) |(π©πͺ) β | = β(36+4+16) = β56 = β(4 Γ 14) = 2βππ Magnitude of |(π΄πΆ) β | = β(9^2+(β3)^2+(β6)^2 ) |(π¨πͺ) β | = β(81+9+36) = β126 = β(9 Γ 14) = 3βππ Thus, |(π¨π©) β | + |(π©πͺ) β | = β14 + 2β14 = 3β14 = |(π¨πͺ) β | Thus, A, B and C are collinear.