Example 10 - Find ratio in which line joining (4, 8, 10)

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Example,  10 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 2

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Example,  10 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 3 Example,  10 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 4

  1. Chapter 12 Class 11 - Intro to Three Dimensional Geometry (Term 2)
  2. Concept wise

Transcript

Example 10 Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, –8) is divided by the YZ-plane. Let AB be the line segment joining points A (4, 8, 10) & (6, 10, –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)/(𝑚 + 𝑛),(𝑚𝑧2 + 𝑛𝑧1)/(𝑚 + 𝑛)) Here, m = k , n = 1 x1 = 4, y1 = 8, z1 = 10 x2 = 6 , y2 = 10, z2 = –8 Co- ordinate of P P (x, y , z) = ((𝑘 (6) + 1 (4))/(𝑘 + 1), (𝑘(10) + 1(8))/(𝑘 + 1), (𝑘(−8) + 1(10))/(𝑘 + 1)) P (x, y , z) = ((6𝑘 + 4)/(k + 1),(10k + 8)/(k + 1),(−8𝑘 + 10)/(k + 1)) Since Point P (x, y, z) lies on the YZ plane its x – coordinate will be zero P (0, y , z) = ((6𝑘 + 4)/(k + 1),(10k + 8)/(k + 1),(−8𝑘 + 10)/(k + 1)) Comparing x – Coordinate 0 = (6𝑘 + 4)/(𝑘 + 1) (k + 1) (0) = 6k + 4 0 = 6k + 4 6k + 4 = 0 6k = – 4 k = (−4)/6 k = (−2)/3 𝑘/1 = (−2)/3 Thus, k : 1 = – 2 : 3 Since value of k is negative Hence YZ Plane divide AB externally in the ratio 2 : 3

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.