Example 11 - Chapter 11 Class 12 Three Dimensional Geometry
Last updated at May 29, 2018 by Teachoo
Transcript
Example 11 Find the shortest distance between the lines l1 and l2 whose vector equations are = + + (2 + ) and = 2 + + (3 5 + 2 ) Shortest distance between lines with vector equations = 1 + 1 and = 2 + 2 is 1 2 . 2 1 1 2 Given, Now = (2 + 1 1 ) (1 + 1 + 0 ) = (2 1) + (1 1) + ( 1 0) = 1 + 0 1 = 2 1 1 3 5 2 = 1 2 ( 5 1) 2 2 ( 3 1) + 2 5 (3 1) = 2+5 4 3 + 10+3 = (3) (1) + ( 7) = 3 7 Magnitude of ( 1 2 ) = 32+ 1 2+ 7 2 = 9+1+49 = Also, ( ) .( ) = (3 7 ) . (1 + 0 1 ) = (3 1) + ( 1 0) + ( 7 1) = 3 + 0 + 7 = 10 Shortest distance = 1 2 . 2 1 1 2 = 10 59 = Therefore the shortest distance between the given lines is 10 59 .
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Chapter 11 Class 12 Three Dimensional Geometry
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.