Check sibling questions

Example 4 - Show that points P (-2, 3, 5), Q (1, 2, 3) and R (7, 0,-1)

Example 4 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 2
Example 4 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 3
Example 4 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 4
Example 4 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 5

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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 = √((π‘₯2βˆ’π‘₯1)2+(𝑦2βˆ’π‘¦1)2+(𝑧2 βˆ’π‘§1)2) PQ = √((1βˆ’(βˆ’2))2+(2βˆ’3)2+(3βˆ’5)2) = √((1+2)2+(2βˆ’3)2+(3βˆ’5)2) = √(32+(βˆ’1)2+(βˆ’2)2) = √(9+(βˆ’1)2+(βˆ’2)2) = √(9+1+4) = βˆšπŸπŸ’ Calculating QR Q ( 1, 2, 3) R (7, 0, –1) QR = √((x2βˆ’x1)2+(y2βˆ’y1)2+(z2 βˆ’z1)2) Here , x1 = – 2, y1 = 3, z1 = 5 x2 = 1, y2 = 2, z2 = 3 QR = √((7βˆ’1)2+(0βˆ’2)2+(βˆ’1βˆ’3)2) = √((6)2+(βˆ’2)2+(βˆ’4)2) = √(36+4+16) = √56 = √(14 Γ— 2 Γ— 2) = 2βˆšπŸπŸ’ Calculating PR P (–2, 3, 5), R (7, 0, –1) PR = √((x2βˆ’x1)2+(y2βˆ’y1)2+(z2 βˆ’z1)2) Here, x1 = –2, y1 = 3, z1 = 5 x2 = 7, y2 = 0, z2 = – 1 PR = √((7βˆ’(βˆ’2))2+(0βˆ’3)2+(βˆ’1βˆ’5)2) = √((7+2)2+(βˆ’3)2+(βˆ’6)2) = √((9)2+9+36) = √(81+9+36) = √126 = √(14 Γ— 3 Γ— 3) = πŸ‘βˆšπŸπŸ’ Thus, PQ = βˆšπŸπŸ’ , QR = 2βˆšπŸπŸ’ & PR = 3βˆšπŸπŸ’ 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.