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 , , .
Examples
Example 1
Example, 2 Important
Example, 3
Example, 4 Important
Example, 5 Important
Example, 6 Important
Example, 7
Example 8
Example, 9 Deleted for CBSE Board 2022 Exams
Example 10 Important Deleted for CBSE Board 2022 Exams
Example 11
Example 12 Important
Example 13 Important
Example 14 You are here
Example 15
Example 16 Important
Example 17
Example 18
Example 19 Important
Example 20 Important
Example 21 Important
Example 22 Deleted for CBSE Board 2022 Exams
Example 23 Important Deleted for CBSE Board 2022 Exams
Example 24
Example, 25 Important Deleted for CBSE Board 2022 Exams
Example 26
Example 27 Important
Example 28 Important
Example 29 Important
Example 30 Important
Chapter 11 Class 12 Three Dimensional Geometry (Term 2)
Serial order wise
About the Author
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.