Ex 12.3, 4 - Using section formula, A (2,-3,4), B (-1,2,1) - Ex 12.3

  1. Chapter 12 Class 11 Introduction to Three Dimensional Geometry
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Ex 12.3, 4 Using section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and C ﷐0,﷐1﷮3﷯,2﷯ are collinear. Given Points A (2, –3, 4) , B (–1, 2, 1) & C ﷐0, ﷐1﷮3﷯ ,2﷯ Point A, B & C are collinear if , point C divides AB in some ratio externally or internally. We know that Co-ordinate of point P(x, y ,z) that divides line segment joining (x1, y1, z1) & (x2, y2, z2) in ration m : n is (x, y ,z) = ﷐﷐mx2 + nx1﷮m + n﷯,﷐my2+ my1﷮m + n﷯﷯ Here, let point C(14, 0, – 2) divide A(– 4, 6, 10) , B(2, 4, 6) in the ratio k : 1 Here, m = k , n = 1 x1 = 2, y1 = 3, z1 = 4 x2 = – 1, y2 = 2, z2 = 1 Putting values ﷐0, ﷐1﷮3﷯, 2﷯ =﷐﷐k﷐−1﷯ + 1(2)﷮k + 1﷯,﷐k﷐2﷯ + 1(3)﷮k + 1﷯,﷐k﷐1﷯ + 1(4)﷮k + 1﷯﷯ ﷐0, ﷐1﷮3﷯, 2﷯ =﷐﷐− k + 2﷮k + 1﷯,﷐2k + 3﷮k + 1﷯,﷐k + 4﷮k + 1﷯﷯ Comparing x – coordinate 0 = ﷐−k + 2﷮k + 1﷯ 0 (k + 1) = – k + 2 0 = – k + 2 – k + 2 = 0 – k = – 2 k = 2 So, k : 1 = 2 : 1 Hence point C divide line segment AB in the ratio 2 : 1 Hence points A, B & C are collinear

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