# Ex 12.3, 5

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 12.3, 5 (Method 1) Find the coordinates of the points which trisect the line segment joining the points P (4, 2, –6) and Q (10, –16, 6). Let Point A (a, b, c) & point B (p, q, r) trisect the line segment PQ i.e. PA = AB = BC Point A divides PQ in the ratio of 1 : 2 We know that , Coordinate of point that divides the line segment joining A(x1, y1, z1) & B(x2, y2, z2) internally in the ratio m: n is P(x, y, z) = 𝑚 𝑥2+ 𝑛 𝑥1𝑚+𝑛,𝑚 𝑦2+ 𝑛 𝑦1𝑚+𝑛,𝑚 𝑧2+ 𝑛 𝑧1𝑚+𝑛 Here, m = 1 , n = 2 x1 = 4 , y1 = 2 , z1 = – 6 x2 = 10 , y2 = – 16 , z2 = 6 Coordinate of A are (a, b, c) = 10 1 + 4 21 + 2,−16 1 + 2 21 + 2,6 1 + − 6 21 + 2 (a, b, c) = 10 + 83,− 16 + 43,6 − 123 (a, b, c) = ( 6, – 4, –2) Hence coordinate of A = (6, – 4, –2) Similarly Point B (p, q, r) divide line PQ in the ratio 2 : 1 We know that coordinate of point that divide line segment joining (x1 y1 z1) , ( x2 , y2 z2) in the ration m : n are = 𝑚𝑥2+𝑛𝑥1𝑚+𝑛,𝑚𝑦2+𝑛𝑦1𝑚+𝑛,𝑚𝑧2+𝑛𝑧1𝑚+𝑛 Here, m = 2 , n = 1 x1 = 4 , y1 = 2 , z1 = – 6 x2 = 10 , y2 = – 16 , z2 = 6 Coordinate of B are (p, q, r) = 10 2 + 4 12 + 1,−16 2 + 2 12 + 1,6 2 + − 6 12 + 1 (p, q, r) = 20 + 42 + 1,− 32 + 22 + 1,12 − 6 2 + 1 (p, q, r) = 243,− 303,63 (p, q, r) = ( 8, – 10, 2) Hence coordinate of Point B = (6, – 10, 2) Ex 12.3, 5 (Method 2) Find the coordinates of the points which trisect the line segment joining the points P (4, 2, –6) and Q (10, –16, 6). Let Point A (a, b, c) & point B (p, q, r) trisect the line segment PQ i.e. PA = AB = BC Point A divides PQ in the ratio of 1 : 2 We know that , Coordinate of point that divides the line segment joining A(x1, y1, z1) & B(x2, y2, z2) internally in the ratio m: n is P(x, y, z) = 𝑚 𝑥2+ 𝑛 𝑥1𝑚+𝑛,𝑚 𝑦2+ 𝑛 𝑦1𝑚+𝑛,𝑚 𝑧2+ 𝑛 𝑧1𝑚+𝑛 Here, m = 1 , n = 2 x1 = 4 , y1 = 2 , z1 = – 6 x2 = 10 , y2 = – 16 , z2 = 6 Coordinate of A are (a, b, c) = 10 1 + 4 21 + 2,−16 1 + 2 21 + 2,6 1 + − 6 21 + 2 (a, b, c) = 10 + 83,− 16 + 43,6 − 123 (a, b, c) = ( 6, – 4, –2) Hence coordinate of A = (6, – 4, –2) Now, Point B (p, q, r) is the mid-point of line AQ Coordinate of point B which is mid point of line AQ A = (6, – 4, – 2) Q = (10, – 16, 6) Here, x1 = 6 , y1 = – 4 , z1 = – 2 x2 = 10 , y2 = – 16 , z2 = 6 Coordinate of B are = 6 + 102,− 4 + ( − 16)2,− 2 + 62 = 162,− 202,42 = ( 8, – 10, 2) Thus point A (6, – 4, 2) & B (8, – 10, 2) trisect the line segment PQ.

Chapter 12 Class 11 Introduction to Three Dimensional Geometry

Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.