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


Transcript

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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.