Question 3 - Section Formula in 3D Geometry - Chapter 11 Class 11 - Intro to Three Dimensional Geometry
Last updated at Dec. 16, 2024 by Teachoo
Section Formula in 3D Geometry
Section Formula in 3D Geometry
Last updated at Dec. 16, 2024 by Teachoo
Question 3 Find the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8). Let AB be the line segment joining points A (–2, 4, 7) & B (3, –5, 8) Let YZ Plane divide line AB at P (x, y, z) in the ratio k : 1 Co-ordinate of P that divide line segment joining point A (x1, y1, z1) & B((x2, y2, z2) in the ratio m : n is = ((𝑚𝑥2+ 𝑛𝑥1)/(𝑚 + 𝑛), (𝑚𝑦2+ 𝑛𝑦1)/(𝑚2+ 𝑛),(𝑚𝑧2+ 𝑛𝑧1)/(𝑚 + 𝑛)) Here, m = k , n = 1 x1 = – 2, y1 = 4, z1 = 7 x2 = 3, y2 = – 5, z2 = 8 Co- ordinate of P P (x, y , z) = ((𝑘 (3) + 1 (−2))/(𝑘 + 1), (𝑘(−5) + 1(4))/(𝑘 + 1), (𝑘(8) + 1(7))/(𝑘 + 1)) P (x, y , z) = ((3𝑘 − 2)/(k + 1),(−5𝑘 + 4)/(k + 1),(8𝑘 + 7)/(k + 1)) Since Point P (x, y, z) lie on the YZ plane its x – coordinate will be zero P(0, y , z) = ((3𝑘 − 2)/(k + 1),(−5𝑘 + 4)/(k + 1),(8𝑘 + 7)/(k + 1)) Comparing x – Co- ordinate 0 = (3𝑘 − 2)/(𝑘 + 1) (k + 1) (0) = 3k – 2 0 = 3k – 2 3k – 2 = 0 3k = 2 k = 2/3 𝑘/1 = 2/3 k : 1 = 2 : 3 Thus, YZ plane divides AB in the ratio 2 : 3