Slide20.JPG

Slide21.JPG
Slide22.JPG

Subscribe to our Youtube Channel - https://you.tube/teachoo

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

Transcript

Ex 11.2, 9 Find the vector and the Cartesian equations of the line that passes through the points (3, โ€“ 2, โ€“ 5), (3, โ€“ 2, 6).Vector Equation Vector equation of a line passing through two points with position vectors ๐‘Ž โƒ— and ๐‘ โƒ— is ๐‘Ÿ โƒ— = ๐‘Ž โƒ— + ๐œ† (๐‘ โƒ— โˆ’ ๐‘Ž โƒ—) Given, the two points are So, ๐‘Ÿ โƒ— = (3๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ โˆ’ 5๐‘˜ ฬ‚) + ๐œ† ["(3" ๐‘– ฬ‚โˆ’"2" ๐‘— ฬ‚+"6" ๐‘˜ ฬ‚")" โˆ’"(3" ๐‘– ฬ‚โˆ’"2" ๐‘— ฬ‚ โˆ’"5" ๐‘˜ ฬ‚")" ] = 3๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ โˆ’ 5๐‘˜ ฬ‚ + ๐œ† ["(3" โˆ’3")" ๐‘– ฬ‚โˆ’"(2" โˆ’(โˆ’2))๐‘— ฬ‚+(6โˆ’(โˆ’5))๐‘˜ ฬ‚)] A (3, โˆ’ 2, โˆ’ 5) ๐‘Ž โƒ— = 3๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ โˆ’ 5๐‘˜ ฬ‚ B (3, โˆ’ 2, 6) ๐‘ โƒ— = 3๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + 6๐‘˜ ฬ‚ = 3๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ โˆ’ 5๐‘˜ ฬ‚ + ๐œ† [0๐‘– ฬ‚ + 0๐‘— ฬ‚ + 11๐‘˜ ฬ‚] = 3๐’Š ฬ‚ โˆ’ 2๐’‹ ฬ‚ โˆ’ 5๐’Œ ฬ‚ + ๐œ† (11๐’Œ ฬ‚) Therefore, the vector equation is ๐‘Ÿ โƒ— = 3๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ โˆ’ 5๐‘˜ ฬ‚ + ๐œ† (11๐‘˜ ฬ‚) Cartesian equation Cartesian equation of a line passing through two points A(x1, y1, z1) and B (x2, y2, z2) is (๐‘ฅ โˆ’ ๐‘ฅ1)/(๐‘ฅ2 โˆ’ ๐‘ฅ_1 ) = (๐‘ฆ โˆ’ ๐‘ฆ1)/(๐‘ฆ2 โˆ’ ๐‘ฆ1) = (๐‘ง โˆ’ ๐‘ง1)/(๐‘ง2 โˆ’ ๐‘ง1) Since the line passes through A (3, โˆ’2, โˆ’5) x1 = 3, y1 = โˆ’2, z1 = โˆ’ 5 And also passes through B (3, โˆ’2, 6) x2 = 3, y2 = โˆ’2, z2 = 6 Equation of line is (๐‘ฅ โˆ’ 3)/(3 โˆ’ 3) = (๐‘ฆ โˆ’ (โˆ’2))/( โˆ’2 โˆ’ (โˆ’2)) = (๐‘ง โˆ’ (โˆ’5))/(6 โˆ’ (โˆ’5)) (๐’™ โˆ’ ๐Ÿ‘)/๐ŸŽ = (๐’š + ๐Ÿ)/๐ŸŽ = (๐’› + ๐Ÿ“)/๐Ÿ๐Ÿ

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 9 years. He provides courses for Maths and Science at Teachoo.