Ex 12.3, 4 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry (Important Question)
Last updated at Jan. 31, 2020 by Teachoo
Chapter 12 Class 11 Introduction to Three Dimensional Geometry
Chapter 12 Class 11 Introduction to Three Dimensional Geometry
Last updated at Jan. 31, 2020 by Teachoo
Ex 12.3, 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