Example 14 - Chapter 11 Class 12 Three Dimensional Geometry (Term 2)
Last updated at May 29, 2018 by Teachoo
Transcript
Example 14 Find the direction cosines of the unit vector perpendicular to the plane .(6 3 2 ) + 1 = 0 passing through the origin. Vector equation of a plane at a distance d from the origin and unit vector to normal from origin is . = d Unit vector of = = 1 ( ) Given, equation of plane is .(6 3 2 ) + 1 = 0 .(6 3 2 ) = 1 Multiplying with 1 on both sides, .(6 3 2 ) = 1 1 . ( 6 + 3 + 2 ) = 1 So; = 6 + 3 + 2 Magnitude of = 6 2+32+22 = 36+9+4 = 49 = 7 Now, = 1 ( ) = 1 7 ( 6 + 3 + 2 ) = + + Direction cosines of unit vector perpendicular to the given plane i.e. in are , , .
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Chapter 11 Class 12 Three Dimensional Geometry (Term 2)
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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.