Misc 8 - Show A, B, C are collinear, find ratio where B - Collinearity of 3 points or 3 position vectors

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

Transcript

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. 𝐴đĩīˇ¯īˇ¯ + đĩđļīˇ¯īˇ¯ = 𝐴đļīˇ¯īˇ¯ 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. 𝑨𝑩īˇ¯īˇ¯ + 𝑩đ‘Ēīˇ¯īˇ¯ = 𝑨đ‘Ēīˇ¯īˇ¯ A (1, -2, −8) B (5, 0,−2) C (11, 3, 7) 𝐴đĩīˇ¯ = (5 − 1) 𝑖īˇ¯ + (0 − (−2)) 𝑗īˇ¯ + (−2−(−8)) 𝑘īˇ¯ = 4 𝑖īˇ¯ + 2 𝑗īˇ¯ + 6 𝑘īˇ¯ đĩđļīˇ¯ = (11 − 5) 𝑖īˇ¯ + (3 − 0) 𝑗īˇ¯ + (7−(−2)) 𝑘īˇ¯ = 6 𝑖īˇ¯ + 3 𝑗īˇ¯ + 9 𝑘īˇ¯ 𝐴đļīˇ¯ = (11 − 1) 𝑖īˇ¯ + (3 − (−2)) 𝑗īˇ¯ + (7−(−8)) 𝑘īˇ¯ = 10 𝑖īˇ¯ + 5 𝑗īˇ¯ + 15 𝑘īˇ¯ Magnitude of 𝐴đĩīˇ¯ = īˇŽ42+22+62īˇ¯ 𝐴đĩīˇ¯ = īˇŽ16+4+36īˇ¯ = īˇŽ56īˇ¯ = īˇŽ4×14 īˇ¯ = 2 īˇŽ14īˇ¯ Magnitude of đĩđļīˇ¯ = īˇŽ62+32+92īˇ¯ đĩđļīˇ¯īˇ¯= īˇŽ36+9+81īˇ¯ = īˇŽ126īˇ¯ = īˇŽ9×14 īˇ¯ = 3 īˇŽ14īˇ¯ Magnitude of 𝐴đļīˇ¯ = īˇŽ102+52+152īˇ¯ 𝐴đļīˇ¯īˇ¯= īˇŽ100+25+225īˇ¯= īˇŽ350īˇ¯ = īˇŽ25×14 īˇ¯ = 5 īˇŽ14īˇ¯ 𝐴đĩīˇ¯īˇ¯ + đĩđļīˇ¯īˇ¯ = 2 īˇŽ14 īˇ¯ + 3 īˇŽ14 īˇ¯ = 5 īˇŽ14 īˇ¯ = 𝐴đļīˇ¯īˇ¯ ∴ A, B and C are collinear. Finding the ratio in which B divides AC Let B divide AC in the ratio k : 1 𝑂𝐴īˇ¯ = 1 𝑖īˇ¯ − 2 𝑗īˇ¯ − 8 𝑘īˇ¯ 𝑂đĩīˇ¯ = 5 𝑖īˇ¯ + 0 𝑗īˇ¯ − 2 𝑘īˇ¯ and 𝑂đļīˇ¯ = 11 𝑖īˇ¯ + 3 𝑗īˇ¯ + 7 𝑘īˇ¯ Position vector of đĩ = 𝑘 𝑂đļīˇ¯+1. 𝑂𝐴īˇ¯īˇŽđ‘˜+1īˇ¯ 𝑂đĩīˇ¯ = 𝑘 11 𝑖īˇ¯ + 3 𝑗īˇ¯ + 7 𝑘īˇ¯īˇ¯ + 1 1 𝑖īˇ¯ −2 𝑗īˇ¯ − 8 𝑘īˇ¯īˇ¯īˇŽđ‘˜ + 1īˇ¯ 5 𝑖īˇ¯ + 0 𝑗īˇ¯ − 2 𝑘īˇ¯ = 11𝑘 𝑖īˇ¯ + 3𝑘 𝑗īˇ¯ + 7𝑘 𝑘īˇ¯ + 𝑖īˇ¯ − 2 𝑗īˇ¯ − 8 𝑘īˇ¯īˇŽđ‘˜ + 1īˇ¯ 5 𝑖īˇ¯ + 0 𝑗īˇ¯ − 2 𝑘īˇ¯ = 11𝑘 + 1īˇ¯ 𝑖īˇ¯ + 3𝑘 − 2īˇ¯ 𝑗īˇ¯ + (7𝑘− 8) 𝑘īˇ¯īˇŽđ‘˜ + 1īˇ¯ ∴ 5 𝑖īˇ¯ + 0 𝑗īˇ¯ − 2 𝑘īˇ¯ = (11𝑘 + 1) īˇŽ(𝑘 + 1)īˇ¯ 𝑖īˇ¯ + (3𝑘 − 2) īˇŽ(𝑘 + 1)īˇ¯ 𝑗īˇ¯ + (7𝑘 − 8) īˇŽ(𝑘 + 1)īˇ¯ 𝑘īˇ¯ Since the two vectors are equal, corresponding components are also equal. So, 11𝑘+1īˇŽđ‘˜+1īˇ¯ = 5 11k + 1 = 5k + 5 11k – 5k = 5 − 1 6k = 4 k = 4īˇŽ6īˇ¯ = 2īˇŽ3īˇ¯ Thus, B divides AC in the ratio 2īˇŽ3īˇ¯ : 1 or 2 : 3

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