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Example 4 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) 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 = – 2, y1 = 3, z1 = 5 x2 = 1, y2 = 2, z2 = 3 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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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.