Ex 12.3, 4 - Using section formula, show that points A (2, -3, 4), B

Ex 12.3,  4 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 2
Ex 12.3,  4 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 3

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Question 4 Using section formula, show that the points A (2, –3, 4), B (–1, 2, 1) and C (0, 1/3,2) are collinear. Given Points A (2, –3, 4) , B (–1, 2, 1) & C (0, 1/3 ,2) Point A, B & C are collinear if , point C divides AB in some ratio externally or internally. We know that Co-ordinate of point P(x, y ,z) that divides line segment joining (x1, y1, z1) & (x2, y2, z2) in ration m : n is (x, y ,z) = ((mx2 + nx1)/(m + n),(my2 + ny1)/(m + n), (〖𝑚𝑧〗_2 + 〖𝑛𝑧〗_1)/(𝑚 + 𝑛)) Here, let point C (0, 1/3 ,2) divide A(2, –3, 4), B(–1, 2, 1) in the ratio k : 1 Here, m = k , n = 1 x1 = 2, y1 = 3, z1 = 4 x2 = – 1, y2 = 2, z2 = 1 Putting values ("0, " 1/3 ", 2" ) =((k(−1) + 1(2))/(k + 1),(k(2) + 1(3))/(k + 1),(k(1) + 1(4))/(k + 1)) ("0, " 1/3 ", 2" ) =((−k + 2)/(k + 1),(2k + 3)/(k + 1),(k + 4)/(k + 1)) Comparing x – coordinate 0 = (−k + 2)/(k + 1) 0 (k + 1) = – k + 2 0 = – k + 2 – k + 2 = 0 – k = – 2 k = 2 So, k : 1 = 2 : 1 Hence point C divide line segment AB in the ratio 2 : 1 Hence points A, B & C are collinear

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.