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  1. Chapter 12 Class 11 Introduction to Three Dimensional Geometry
  2. Serial order wise

Transcript

Example 4 Show that the points P (โ€“2, 3, 5), Q (1, 2, 3) and R (7, 0, โ€“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 ( โ€“ 2, 3, 5) Q (1, 2, 3) Hence , PQ = โˆš((๐‘ฅ2โˆ’๐‘ฅ1)2+(๐‘ฆ2โˆ’๐‘ฆ1)2+(๐‘ง2 โˆ’๐‘ง1)2) PQ = โˆš((1โˆ’(โˆ’2))2+(2โˆ’3)2+(3โˆ’5)2) = โˆš((1+2)2+(2โˆ’3)2+(3โˆ’5)2) = โˆš(32+(โˆ’1)2+(โˆ’2)2) = โˆš(9+(โˆ’1)2+(โˆ’2)2) = โˆš(9+1+4) = โˆš๐Ÿ๐Ÿ’ Calculating QR Q ( 1, 2, 3) R (7, 0, โ€“1) QR = โˆš((x2โˆ’x1)2+(y2โˆ’y1)2+(z2 โˆ’z1)2) Here , x1 = โ€“ 2, y1 = 3, z1 = 5 x2 = 1, y2 = 2, z2 = 3 QR = โˆš((7โˆ’1)2+(0โˆ’2)2+(โˆ’1โˆ’3)2) = โˆš((6)2+(โˆ’2)2+(โˆ’4)2) = โˆš(36+4+16) = โˆš56 = โˆš(14 ร— 2 ร— 2) = 2โˆš๐Ÿ๐Ÿ’ Calculating PR P (โ€“2, 3, 5), R (7, 0, โ€“1) PR = โˆš((x2โˆ’x1)2+(y2โˆ’y1)2+(z2 โˆ’z1)2) Here, x1 = โ€“2, y1 = 3, z1 = 5 x2 = 7, y2 = 0, z2 = โ€“ 1 PR = โˆš((7โˆ’(โˆ’2))2+(0โˆ’3)2+(โˆ’1โˆ’5)2) = โˆš((7+2)2+(โˆ’3)2+(โˆ’6)2) = โˆš((9)2+9+36) = โˆš(81+9+36) = โˆš126 = โˆš(14 ร— 3 ร— 3) = ๐Ÿ‘โˆš๐Ÿ๐Ÿ’ Thus, PQ = โˆš๐Ÿ๐Ÿ’ , QR = 2โˆš๐Ÿ๐Ÿ’ & PR = 3โˆš๐Ÿ๐Ÿ’ So, PQ + QR = โˆš14 + 2โˆš14 = 3โˆš14 = 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.