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Ex 12.2, 2 - Show that points P (–2, 3, 5), Q (1, 2, 3) - Ex 12.2

  1. Chapter 12 Class 11 Introduction to Three Dimensional Geometry
  2. Serial order wise
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Ex 12.2,2 Show that the points P (–2, 3, 5), Q (1, 2, 3) and R (7, 0, –1) are collinear. If three points are collinear, then they lie on a line. Let first calculate distance between the 3 points i.e. PQ. QR and PR Calculating PQ P ( – 2, 3, 5) Q (1, 2, 3) Hence , PQ = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here , x1 = – 2, y1 = 3, z1 = 5 x2 = 1, y2 = 2, z2 = 3 PQ = ﷐﷮﷐1−−2﷯2+﷐2−3﷯2+﷐3−5﷯2﷯ = ﷐﷮﷐1+2﷯2+﷐2−3﷯2+﷐3−5﷯2﷯ = ﷐﷮32+﷐−1﷯2+(−2)2﷯ = ﷐﷮9+﷐−1﷯2+﷐−2﷯2﷯ = ﷐﷮9+1+4﷯ = ﷐﷮14﷯ Calculating QR Q ( 1, 2, 3) R (7, 0, –1) QR = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here x1 = 1, y1 = 2, z1 = 2 x2 = 7, y2 = 0, z2 = – 1 QR = ﷐﷮﷐7−1﷯2+﷐0−2﷯2+﷐−1−3﷯2﷯ = ﷐﷮﷐6﷯2+﷐−2﷯2+﷐−4﷯2﷯ = ﷐﷮36+4+16﷯ = ﷐﷮56﷯ = ﷐﷮14×2×2﷯ = 2﷐﷮14﷯ Calculating PR P ( –2, 3, 5) R (7, 0, –1) PR = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here, x1 = – 2, y1 = 3, z1 = 5 x2 = 7, y2 = 0, z2 = – 1 PR = ﷐﷮﷐7−(−2)﷯2+﷐0−3﷯2+﷐−1−5﷯2﷯ = ﷐﷮﷐7+2﷯2+﷐−3﷯2+﷐−6﷯2﷯ = ﷐﷮﷐9﷯2+9+36﷯ = ﷐﷮81+9+36﷯ = ﷐﷮126﷯ = ﷐﷮14×3×3﷯ = 3﷐﷮14﷯ Thus, PQ = ﷐﷮14﷯ , QR = 2﷐﷮14﷯ & PR = 3﷐﷮14﷯ So, PQ + QR = ﷐﷮14﷯ + 2﷐﷮14﷯ = 3﷐﷮14﷯ = PR Thus, PQ + QR = PR So, if we draw the points on a graph, with PQ + QR = PR We see that points P,Q,R lie on the same line. Thus, P, Q and R all collinear

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