Last updated at Dec. 8, 2016 by Teachoo

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Misc 8(Introduction) Show that the points A(1, – 2, – 8), B (5, 0, – 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 vectors. i.e. 𝐴𝐵 + 𝐵𝐶 = 𝐴𝐶 Misc 8 Show that the points A(1, – 2, – 8), B (5, 0, – 2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC. 3 points A, B, C are collinear if i.e. 𝑨𝑩 + 𝑩𝑪 = 𝑨𝑪 A (1, -2, −8) B (5, 0,−2) C (11, 3, 7) 𝐴𝐵 = (5 − 1) 𝑖 + (0 − (−2)) 𝑗 + (−2−(−8)) 𝑘 = 4 𝑖 + 2 𝑗 + 6 𝑘 𝐵𝐶 = (11 − 5) 𝑖 + (3 − 0) 𝑗 + (7−(−2)) 𝑘 = 6 𝑖 + 3 𝑗 + 9 𝑘 𝐴𝐶 = (11 − 1) 𝑖 + (3 − (−2)) 𝑗 + (7−(−8)) 𝑘 = 10 𝑖 + 5 𝑗 + 15 𝑘 Magnitude of 𝐴𝐵 = 42+22+62 𝐴𝐵 = 16+4+36 = 56 = 4×14 = 2 14 Magnitude of 𝐵𝐶 = 62+32+92 𝐵𝐶= 36+9+81 = 126 = 9×14 = 3 14 Magnitude of 𝐴𝐶 = 102+52+152 𝐴𝐶= 100+25+225= 350 = 25×14 = 5 14 𝐴𝐵 + 𝐵𝐶 = 2 14 + 3 14 = 5 14 = 𝐴𝐶 ∴ A, B and C are collinear. Finding the ratio in which B divides AC Let B divide AC in the ratio k : 1 𝑂𝐴 = 1 𝑖 − 2 𝑗 − 8 𝑘 𝑂𝐵 = 5 𝑖 + 0 𝑗 − 2 𝑘 and 𝑂𝐶 = 11 𝑖 + 3 𝑗 + 7 𝑘 Position vector of 𝐵 = 𝑘 𝑂𝐶+1. 𝑂𝐴𝑘+1 𝑂𝐵 = 𝑘 11 𝑖 + 3 𝑗 + 7 𝑘 + 1 1 𝑖 −2 𝑗 − 8 𝑘𝑘 + 1 5 𝑖 + 0 𝑗 − 2 𝑘 = 11𝑘 𝑖 + 3𝑘 𝑗 + 7𝑘 𝑘 + 𝑖 − 2 𝑗 − 8 𝑘𝑘 + 1 5 𝑖 + 0 𝑗 − 2 𝑘 = 11𝑘 + 1 𝑖 + 3𝑘 − 2 𝑗 + (7𝑘− 8) 𝑘𝑘 + 1 ∴ 5 𝑖 + 0 𝑗 − 2 𝑘 = (11𝑘 + 1) (𝑘 + 1) 𝑖 + (3𝑘 − 2) (𝑘 + 1) 𝑗 + (7𝑘 − 8) (𝑘 + 1) 𝑘 Since the two vectors are equal, corresponding components are also equal. So, 11𝑘+1𝑘+1 = 5 11k + 1 = 5k + 5 11k – 5k = 5 − 1 6k = 4 k = 46 = 23 Thus, B divides AC in the ratio 23 : 1 or 2 : 3

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .