Slide64.JPG

Slide65.JPG
Slide66.JPG Slide67.JPG Slide68.JPG Slide69.JPG Slide70.JPG Slide71.JPG

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

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

Transcript

Misc 18 (Method 1) Find the distance of the point (โ€“1, โ€“5, โ€“10) from the point of intersection of the line ๐‘Ÿ โƒ— = 2๐‘– ฬ‚ โ€“ ๐‘— ฬ‚ + 2๐‘˜ ฬ‚ + ๐œ† (3๐‘– ฬ‚ + 4๐‘— ฬ‚ + 2๐‘˜ ฬ‚) and the plane ๐‘Ÿ โƒ— . (๐‘– ฬ‚ โ€“ ๐‘— ฬ‚ + ๐‘˜ ฬ‚) = 5 .Given, the equation of line is ๐‘Ÿ โƒ— = (2๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + 2๐‘˜ ฬ‚) + ๐œ† (3๐‘– ฬ‚ + 4๐‘— ฬ‚ + 2๐‘˜ ฬ‚) and the equation of the plane is ๐‘Ÿ โƒ—.(๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + ๐‘˜ ฬ‚) = 5 To find point of intersection of line and plane, Putting value of ๐’“ โƒ— from equation of line into equation of plane. ["(2" ๐‘– ฬ‚" โˆ’ " ๐‘— ฬ‚" + 2" ๐‘˜ ฬ‚") + ๐œ† (3" ๐‘– ฬ‚" + 4" ๐‘— ฬ‚" + 2" ๐‘˜ ฬ‚")" ] . (๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + ๐‘˜ ฬ‚) = 5 ["(2" ๐‘– ฬ‚" โˆ’ 1" ๐‘— ฬ‚" + 2" ๐‘˜ ฬ‚+3"๐œ†" ๐‘– ฬ‚" + 4๐œ†" ๐‘— ฬ‚+2"๐œ†" ๐‘˜ ฬ‚")" ] . (1๐‘– ฬ‚ โˆ’ 1๐‘— ฬ‚ + 1๐‘˜ ฬ‚) = 5 ["(2 + 3๐œ†) " ๐‘– ฬ‚" + (" โˆ’1" + 4๐œ†) " ๐‘— ฬ‚+(2+2"๐œ†" )๐‘˜ ฬ‚ ] . (1๐‘– ฬ‚ โˆ’ 1๐‘— ฬ‚ + 1๐‘˜ ฬ‚) = 5 (2 + 3๐œ†) ร— 1 + (โˆ’1 + 4๐œ†) ร— (โˆ’1) + (2 + 2๐œ†) ร— 1 = 5 2 + 3๐œ† + 1 โˆ’ 4๐œ† + 2 + 2๐œ† = 5 ๐œ† + 5 = 5 ๐œ† = 5 โˆ’ 5 ๐œ† = 0 So, the equation of line is ๐‘Ÿ โƒ— = (2๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + 2๐‘˜ ฬ‚) + ๐œ† (3๐‘– ฬ‚ + 4๐‘— ฬ‚ + 2๐‘˜ ฬ‚) ๐’“ โƒ— = 2๐’Š ฬ‚ โˆ’ ๐’‹ ฬ‚ + 2๐’Œ ฬ‚ Let the point of intersection be (x, y, z) So, ๐‘Ÿ โƒ— = x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚ x๐‘– ฬ‚ + y๐‘— ฬ‚ + z๐‘˜ ฬ‚ = 2๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + 2๐‘˜ ฬ‚ Hence, x = 2 , y = โˆ’1, z = 2 Therefore, the point of intersection is (2, โˆ’1, 2) Now, the distance between two points (๐‘ฅ_1, ๐‘ฆ_1, ๐‘ง_1) and (๐‘ฅ_2, ๐‘ฆ_2, ๐‘ง_2) is โˆš((๐‘ฅ_2โˆ’๐‘ฅ_1 )^2 ใ€–+ (๐‘ฆ_2โˆ’๐‘ฆ_1 )ใ€—^2+ (๐‘ง_2โˆ’๐‘ง_1 )^2 ) Distance between (2, โˆ’1, 2) and (โˆ’1, โˆ’5, โˆ’10) = โˆš((โˆ’1โˆ’2)^2 ใ€–+ (โˆ’5+1)ใ€—^2+ (โˆ’10โˆ’2)^2 ) = โˆš((โˆ’3)^2 ใ€–+ (โˆ’4)ใ€—^2+ (โˆ’12)^2 ) = โˆš(9+16+144) = โˆš169 = 13. Misc 18 (Method 2) Find the distance of the point (โ€“1, โ€“5, โ€“10) from the point of intersection of the line ๐‘Ÿ โƒ— = 2๐‘– ฬ‚ โ€“ ๐‘— ฬ‚ + 2๐‘˜ ฬ‚ + ๐œ† (3๐‘– ฬ‚ + 4๐‘— ฬ‚ + 2๐‘˜ ฬ‚) and the plane ๐‘Ÿ โƒ— . (๐‘– ฬ‚ โ€“ ๐‘— ฬ‚ + ๐‘˜ ฬ‚) = 5 .Given, the equation of line is ๐‘Ÿ โƒ— = (2๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + 2๐‘˜ ฬ‚) + ๐œ† (3๐‘– ฬ‚ + 4๐‘— ฬ‚ + 2๐‘˜ ฬ‚) Comparing with ๐’“ โƒ— = ๐’‚ โƒ— + ๐œ†๐’ƒ โƒ— , ๐’‚ โƒ— = 2๐’Š ฬ‚ โˆ’ ๐’‹ ฬ‚ + 2๐’Œ ฬ‚ Comparing with ๐‘Ž โƒ— = ๐‘ฅ_1 ๐‘– ฬ‚ + ๐‘ฆ_1 ๐‘— ฬ‚ + ๐‘ง_1 ๐‘˜ ฬ‚, โˆด ๐‘ฅ_1= 2, ๐‘ฆ_1= โˆ’1, ๐‘ง_1= 2, ๐’ƒ โƒ— = 3๐’Š ฬ‚ + 4๐’‹ ฬ‚ + 2๐’Œ ฬ‚ Comparing with ๐‘ โƒ— = ๐‘Ž๐‘– ฬ‚ + ๐‘๐‘— ฬ‚ + ๐‘๐‘˜ ฬ‚, โˆด ๐‘Ž = 3, ๐‘ = 4, ๐‘ = 2, Equation of line in Cartesian form is (๐‘ฅ โˆ’ ๐‘ฅ_1)/๐‘Ž = (๐‘ฆ โˆ’ ๐‘ฆ_1)/๐‘ = (๐‘ง โˆ’ ๐‘ง_1)/๐‘ (๐‘ฅ โˆ’ 2)/3 = (๐‘ฆ โˆ’ (โˆ’1))/4 = (๐‘ง โˆ’ 2)/2 (๐’™ โˆ’ ๐Ÿ)/๐Ÿ‘ = (๐’š + ๐Ÿ)/๐Ÿ’ = (๐’› โˆ’ ๐Ÿ)/๐Ÿ = k So, Also, the equation of plane is ๐‘Ÿ โƒ—.(๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + ๐‘˜ ฬ‚) = 5 Comparing with ๐‘Ÿ โƒ—.๐‘› โƒ— = d, ๐‘› โƒ— = 1๐‘– ฬ‚ โˆ’ 1๐‘— ฬ‚ + 1๐‘˜ ฬ‚ & d = 5 Comparing ๐‘› โƒ— with A๐‘– ฬ‚ + B๐‘— ฬ‚ + C๐‘˜ ฬ‚, A = 1, B = โˆ’1, C = 1 Equation of plane in Cartesian form is Ax + By + Cz = d 1x โˆ’ 1y + 1z = 5 x โˆ’ y + z = 5 Let the point of intersection of line and plane be (x, y, z) Putting values of x, y, z in equation of plane, (3k + 2) โˆ’ (4k โˆ’ 1) + (2k + 2) = 5 3k + 2 โˆ’ 4k + 1 + 2k + 2 = 5 k + 5 = 5 โˆด k = 0 So, x = 3k + 2 = 3 ร— 0 + 2 = 2 y = 4k โˆ’ 1 = 4 ร— 0 โˆ’ 1 = โˆ’1 z = 2k + 2 = 2 ร— 0 + 2 = 2 Therefore, the point of intersection is (2, โˆ’1, 2). Distance between two points (๐‘ฅ_1, ๐‘ฆ_1, ๐‘ง_1) & (๐‘ฅ_2, ๐‘ฆ_2, ๐‘ง_2) = โˆš((๐‘ฅ_2โˆ’๐‘ฅ_1 )^2 (๐‘ฆ_2โˆ’๐‘ฆ_1 )^2+(๐‘ง_2โˆ’๐‘ง_1 )^2 ) โˆด Distance between (2, โˆ’1, 2) and (โˆ’1, โˆ’5, โˆ’10) = โˆš((โˆ’1โˆ’2)^2+(โˆ’5+1)^2+(โˆ’10โˆ’2)^2 ) = โˆš((โˆ’3)^2+(โˆ’4)^2+(โˆ’12)^2 ) = โˆš(9+16+144) = โˆš169 = 13

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.