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Examples
Last updated at Jan. 31, 2020 by Teachoo
Example 4 Show that the points P (β2, 3, 5), Q (1, 2, 3) and R (7, 0, β1) 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 ( β 2, 3, 5) Q (1, 2, 3) Hence , PQ = β((π₯2βπ₯1)2+(π¦2βπ¦1)2+(π§2 βπ§1)2) PQ = β((1β(β2))2+(2β3)2+(3β5)2) = β((1+2)2+(2β3)2+(3β5)2) = β(32+(β1)2+(β2)2) = β(9+(β1)2+(β2)2) = β(9+1+4) = βππ Calculating QR Q ( 1, 2, 3) R (7, 0, β1) QR = β((x2βx1)2+(y2βy1)2+(z2 βz1)2) Here , x1 = β 2, y1 = 3, z1 = 5 x2 = 1, y2 = 2, z2 = 3 QR = β((7β1)2+(0β2)2+(β1β3)2) = β((6)2+(β2)2+(β4)2) = β(36+4+16) = β56 = β(14 Γ 2 Γ 2) = 2βππ Calculating PR P (β2, 3, 5), R (7, 0, β1) PR = β((x2βx1)2+(y2βy1)2+(z2 βz1)2) Here, x1 = β2, y1 = 3, z1 = 5 x2 = 7, y2 = 0, z2 = β 1 PR = β((7β(β2))2+(0β3)2+(β1β5)2) = β((7+2)2+(β3)2+(β6)2) = β((9)2+9+36) = β(81+9+36) = β126 = β(14 Γ 3 Γ 3) = πβππ Thus, PQ = βππ , QR = 2βππ & PR = 3βππ So, PQ + QR = β14 + 2β14 = 3β14 = 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