Misc 7 - Find vector equation of line perpendicular to plane - Miscellaneous

Slide24.JPG

  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise
Ask Download

Transcript

Misc 7 Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane ๐‘Ÿ๏ทฏ.( ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ โˆ’ 5 ๐‘˜๏ทฏ) + 9 = 0 . The vector equation of a line passing through a point with position vector ๐‘Ž๏ทฏ and parallel to vector ๐‘๏ทฏ is ๐’“๏ทฏ = ๐’‚๏ทฏ + ๐œ† ๐’ƒ๏ทฏ Given, the line passes (1, 2, 3) So, ๐’‚๏ทฏ = 1 ๐’Š๏ทฏ + 2 ๐’‹๏ทฏ + 3 ๐’Œ๏ทฏ Finding normal of plane ๐‘Ÿ๏ทฏ.( ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ โˆ’ 5 ๐‘˜๏ทฏ) + 9 = 0 ๐‘Ÿ๏ทฏ. ( ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ โˆ’ 5 ๐‘˜๏ทฏ) = โˆ’ 9 โˆ’ ๐‘Ÿ๏ทฏ. ( ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ โˆ’ 5 ๐‘˜๏ทฏ) = 9 ๐‘Ÿ๏ทฏ. (โˆ’1 ๐‘–๏ทฏ โˆ’ 2 ๐‘—๏ทฏ + 5 ๐‘˜๏ทฏ) = 9 Comparing with ๐‘Ÿ๏ทฏ. ๐‘›๏ทฏ = d, ๐’๏ทฏ = โˆ’ ๐‘–๏ทฏ โˆ’ 2 ๐‘—๏ทฏ + 5 ๐‘˜๏ทฏ Since line is perpendicular to plane, the line will be parallel to the normal of the plane โˆด ๐’ƒ๏ทฏ = ๐’๏ทฏ = โˆ’1 ๐‘–๏ทฏ โˆ’ 2 ๐‘–๏ทฏ + 5 ๐‘˜๏ทฏ Hence, ๐‘Ÿ๏ทฏ = (1 ๐‘–๏ทฏ + 2 ๐‘–๏ทฏ + 3 ๐‘˜๏ทฏ) + ๐œ†(โˆ’1 ๐‘–๏ทฏ โˆ’ 2 ๐‘–๏ทฏ + 5 ๐‘˜๏ทฏ) ๐’“๏ทฏ = ( ๐’Š๏ทฏ + 2 ๐’‹๏ทฏ + 3 ๐’Œ๏ทฏ) โˆ’ ๐œ† ( ๐’Š๏ทฏ + 2 ๐’‹๏ทฏ โˆ’ 5 ๐’Œ๏ทฏ) โˆด Vector equation of line is ๐‘Ÿ๏ทฏ = ( ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ + 3 ๐‘˜๏ทฏ) โˆ’ ๐œ† ( ๐‘–๏ทฏ + 2 ๐‘—๏ทฏ โˆ’ 5 ๐‘˜๏ทฏ)

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
Jail