Ex 12.2,  1 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 3

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Ex 12.2,  1 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 4

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

Transcript

Ex 12.2, 1 Find the distance between the following pairs of points: (ii) (–3, 7, 2) and (2, 4, –1) Let Point P be ( –3, 7, 2) Q be (2, 4, –1) PQ = √((x2−x1)2+(y2−y1)2+(z2 −z1)2) Here x1 = –3, y1 = 7, z1 = 2 x2 = 2, y2 = 4, z2 = – 1 PQ = √((2−(−3))2+(4−7)2+(−1 −2)2) = √((2+3)2+(−3)2+(−3)2) = √((5)2+9+9) = √(25+9+9) = √43 Thus, the required distance is √𝟒𝟑 units

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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.