Ex 10.3, 16 - Show that A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1) - Collinearity of 3 points or 3 position vectors

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  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise
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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﷯ = 1 𝑖﷯ + 4 𝑗﷯ − 4 𝑘﷯ 𝐵𝐶﷯ = (3 − 2) 𝑖﷯ + (10 − 6) 𝑗﷯ + (−1 − 3) 𝑘﷯ = 1 𝑖﷯ + 4 𝑗﷯ − 4 𝑘﷯ 𝐴𝐶﷯ = (3 − 1) 𝑖﷯ + (10 − 2) 𝑗﷯ + (−1 − 7) 𝑘﷯ = 2 𝑖﷯ + 8 𝑗﷯ − 8 𝑘﷯ 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.

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