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

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


Transcript

Ex 11.2, 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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.