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

Transcript

Misc 14 If the points (1, 1 , p) and (โ€“3 , 0, 1) be equidistant from the plane ๐‘Ÿ โƒ—. (3๐‘– ฬ‚ + 4๐‘— ฬ‚ โˆ’ 12๐‘˜ ฬ‚) + 13 = 0, then find the value of p. The distance of a point with position vector ๐‘Ž โƒ— from the plane ๐‘Ÿ โƒ—.๐‘› โƒ— = d is |(๐’‚ โƒ—.๐’ โƒ— โˆ’ ๐’…)/|๐’ โƒ— | | Given, the points are The equation of plane is ๐‘Ÿ โƒ—. (3๐‘– ฬ‚ + 4๐‘— ฬ‚ โˆ’ 12๐‘˜ ฬ‚) + 13 = 0 ๐‘Ÿ โƒ—.(3๐‘– ฬ‚ + 4๐‘— ฬ‚ โˆ’ 12๐‘˜ ฬ‚) = โˆ’13 (1, 1, p) So, (๐‘Ž_1 ) โƒ— = 1๐‘– ฬ‚ + 1๐‘— ฬ‚ + p๐‘˜ ฬ‚ (โˆ’3, 0, 1) So, (๐‘Ž_2 ) โƒ— = โˆ’3๐‘– ฬ‚ + 0๐‘— ฬ‚ + 1๐‘˜ ฬ‚ โ€“๐‘Ÿ โƒ—.(3๐‘– ฬ‚ + 4๐‘— ฬ‚ โˆ’ 12๐‘˜ ฬ‚) = 13 ๐’“ โƒ—.(โ€“3๐’Š ฬ‚ โ€“ 4๐’‹ ฬ‚ + 12๐’Œ ฬ‚) = 13 Comparing with ๐‘Ÿ โƒ—.๐‘› โƒ— = d, ๐‘› โƒ— = โˆ’3๐‘– ฬ‚ โˆ’ 4๐‘— ฬ‚ + 12๐‘˜ ฬ‚ d = 13 Magnitude of ๐‘› โƒ— = โˆš((โˆ’3)^2+(โˆ’4)^2+ใ€–12ใ€—^2 ) |๐‘› โƒ— | = โˆš(9+16+144) = โˆš169 = 13 Distance of point (๐’‚๐Ÿ) โƒ— from plane |((๐‘Ž1) โƒ—"." ๐‘› โƒ—" " โˆ’ ๐‘‘)/|๐‘› โƒ— | | = |((1๐‘– ฬ‚ + 1๐‘— ฬ‚ + ๐‘๐‘˜ ฬ‚ ).(โˆ’3๐‘– ฬ‚โˆ’4๐‘— ฬ‚+12๐‘˜ ฬ‚ )โˆ’13)/13| = |((1ร—โˆ’3)+(1ร—โˆ’4) +(๐‘ร—12)โˆ’13)/13| = |(โˆ’3โˆ’4+12๐‘โˆ’13)/13| = |(12๐‘ โˆ’ 20)/13| Distance of point (๐’‚๐Ÿ) โƒ— from plane |((๐‘Ž2) โƒ—"." ๐‘› โƒ— โˆ’ ๐‘‘)/|๐‘› โƒ— | | = |((โˆ’3๐‘– ฬ‚ +0๐‘— ฬ‚ +1๐‘˜ ฬ‚ ).(โˆ’3๐‘– ฬ‚โˆ’4๐‘— ฬ‚+12๐‘˜ ฬ‚ )โˆ’13)/13| = |((โˆ’3ร—โˆ’3)+(0ร—โˆ’4) +(1ร—12)โˆ’13)/13| = |(9 + 0 +12โˆ’13)/13| = |8/13| = 8/13 Since the plane is equidistance from both the points, |(๐Ÿ๐Ÿ๐’‘ โˆ’ ๐Ÿ๐ŸŽ)/๐Ÿ๐Ÿ‘| = ๐Ÿ–/๐Ÿ๐Ÿ‘ |12๐‘โˆ’20| = 8 (12p โ€“ 20) = ยฑ 8 12p โˆ’ 20 = 8 12p = 8 + 20 12p = 28 p = 28/12 p = 7/3 12p โˆ’ 20 = โˆ’8 12p = โˆ’8 + 20 12p = 12 p = 12/12 p = 1 Answer does not match at end. If mistake, please comment

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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.