Question 2 - Section Formula in 3D Geometry - Chapter 11 Class 11 - Intro to Three Dimensional Geometry
Last updated at April 16, 2024 by Teachoo
Section Formula in 3D Geometry
Section Formula in 3D Geometry
Last updated at April 16, 2024 by Teachoo
Question 2 Given that P (3, 2, –4), Q (5, 4, –6) and R (9, 8, –10) are collinear. Find the ratio in which Q divides PR. Given that Point P (3, 2, –4), Q (5, 4, –6) & R (9, 8, –10) are collinear Q must divide line segment PR in some ratio externally & 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)) Let point Q (5, 4, –6) divide line segment P (3, 2, – 4), R (9, 8, – 10) in the ratio k : 1 Here, x1 = 3, y1 = 2, z1 = – 4 x2 = 9, y2 = 8, z2 = – 10 & m = k , n = 1 Putting values Q (5, 4, – 6) = ((k(9) + 3)/(k+1),(k(8) + 2)/(k+1),(k(−10) − 4)/(k+1)) (5, 4, – 6) = ((9𝑘 + 3)/(k + 1),(8𝑘 + 2)/(k + 1),(− 10𝑘 −4)/(k+1)) Comparing x – coordinate of Q 5 = (9k + 3)/(k + 1) 5 (k + 1) = 9k + 3 5k + 5 = 9k + 3 5k – 9k = 3 – 5 – 4k = – 2 k = (−2)/(−4) k = 1/2 So, k : 1 = 1 : 2 Thus, point Q divides PR in the Ratio 1 : 2