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Last updated at April 16, 2024 by Teachoo

Misc 8 (Introduction) Show that the points A(1, β 2, β 8), B (5, 0, β 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. |(π΄π΅) β | + |(π΅πΆ) β | = |(π΄πΆ) β | Misc 8 Show that the points A(1, β 2, β 8), B (5, 0, β 2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC. vGiven A (1, -2, β8) B (5, 0,β2) C (11, 3, 7) 3 points A, B, C are collinear if |(π¨π©) β | + |(π©πͺ) β | = |(π¨πͺ) β | Finding (π¨π©) β , (π©πͺ) β , (π¨πͺ) β (π¨π©) β = (5 β 1) π Μ + (0 β (β2)) π Μ + (β2β(β8)) π Μ = 4π Μ + 2π Μ + 6π Μ (π©πͺ) β = (11 β 5) π Μ + (3 β 0) π Μ + (7β(β2)) π Μ = 6π Μ + 3π Μ + 9π Μ (π¨πͺ) β = (11 β 1) π Μ + (3 β (β2)) π Μ + (7β(β8)) π Μ = 10π Μ + 5π Μ + 15π Μ Magnitude of (π΄π΅) β = β(42+22+62) |(π¨π©) β | = β(16+4+36) = β56 = β(4Γ14 ) = 2βππ Magnitude of (π΅πΆ) β = β(62+32+92) |(π©πͺ) β |= β(36+9+81) = β126 = β(9Γ14 ) = 3βππ Magnitude of (π΄πΆ) β = β(102+52+152) |(π¨πͺ) β |= β(100+25+225)= β350 = β(25 Γ 14 ) = 5βππ Thus, |(π¨π©) β | + |(π©πͺ) β | = 2β(14 ) + 3β(14 ) = 5β(14 ) = |(π¨πͺ) β | 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, (πΆπ¨) β = 1π Μ β 2π Μ β 8π Μ (πΆπ©) β = 5π Μ + 0π Μ β 2π Μ and (πΆπͺ) β = 11π Μ + 3π Μ + 7π Μ Position vector of π© = (π(πΆπͺ) β + π.(πΆπ¨) β)/(π + π) 5π Μ + 0π Μ β 2π Μ = (π(11π Μ + 3π Μ + 7π Μ ) + 1(1π Μ β 2π Μ β 8π Μ ))/(π + 1) 5π Μ + 0π Μ β 2π Μ = (11ππ Μ + 3ππ Μ + 7π π Μ + π Μ β 2π Μ β 8π Μ)/(π + 1) 5π Μ + 0π Μ β 2π Μ = ((πππ + π) π Μ + (ππ β π) π Μ + (ππ β π) π Μ)/(π + π) Since the two vectors are equal, corresponding components are also equal. So, (πππ + π)/(π + π) = 5 11k + 1 = 5k + 5 11k β 5k = 5 β 1 6k = 4 k = 4/6 = 2/3 Thus, B divides AC in the ratio 2 : 3