Find the vector equation of the plane that passes through the point (1, 0, 0) and contains the line r = โ‹i ฬ‚ j ฬ‚

 

Find vector equation of plane that passes through the point (1,0,0)

Question 27 - CBSE Class 12 Sample Paper for 2021 Boards - Part 2
Question 27 - CBSE Class 12 Sample Paper for 2021 Boards - Part 3

  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Transcript

Equation of plane passing through point A whose position vector is ๐’‚ โƒ— & perpendicular to ๐’ โƒ— is (๐’“ โƒ— โˆ’ ๐’‚ โƒ—) . ๐’ โƒ— = 0 Given Plane passes through (1, 0, 0) So ๐’‚ โƒ— = 1๐‘– ฬ‚ + 0๐‘— ฬ‚ โˆ’ 0๐‘˜ ฬ‚ ๐’‚ โƒ— = ๐‘– ฬ‚ Thus, equation of plane is (๐‘Ÿ โƒ— โˆ’ ๐‘Ž โƒ—) . ๐‘› โƒ— = 0 (๐’“ โƒ— โˆ’ ๐’Š ฬ‚) . ๐’ โƒ— = 0 Now, The plane contains the line ๐‘Ÿ โƒ—= ๐œ†๐‘— ฬ‚ So, Line is perpendicular to normal of plane ๐’‹ ฬ‚ . ๐’ โƒ— = 0 And, Lines point (0, 0, 0) lies on plane as well Putting ๐‘Ÿ โƒ— = 0 โƒ— in equation of plane (0 โƒ— โˆ’ ๐‘– ฬ‚) . ๐‘› โƒ— = 0 โˆ’ ๐‘– ฬ‚ . ๐‘› โƒ— = 0 ๐’Š ฬ‚ . ๐’ โƒ— = 0 Since ๐’ โƒ— perpendicular to both ๐’Š ฬ‚ and ๐’‹ ฬ‚ Thus, ๐’ โƒ— = ๐’Œ ฬ‚ So, our equation of plane becomes (๐‘Ÿ โƒ— โˆ’ ๐‘– ฬ‚) . ๐‘› โƒ— = 0 (๐’“ โƒ— โˆ’ ๐’Š ฬ‚) . ๐’Œ ฬ‚ = 0 ๐‘Ÿ โƒ— . ๐‘˜ ฬ‚ โˆ’ ๐‘– ฬ‚ . ๐‘˜ ฬ‚ = 0 ๐’“ โƒ— . ๐’Œ ฬ‚ = 0

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
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.