Misc 14 - If points (1, 1, p) and (-3, 0, 1) be equidistant from plane

Misc 14 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Misc 14 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3
Misc 14 - Chapter 11 Class 12 Three Dimensional Geometry - Part 4

Go Ad-free

Transcript

Question 10 If the points (1, 1 , p) and (ā€“3 , 0, 1) be equidistant from the plane š‘Ÿ āƒ—. (3š‘– Ģ‚ + 4š‘— Ģ‚ āˆ’ 12š‘˜ Ģ‚) + 13 = 0, then find the value of p. The distance of a point with position vector š‘Ž āƒ— from the plane š‘Ÿ āƒ—.š‘› āƒ— = d is |(š’‚ āƒ—.š’ āƒ— āˆ’ š’…)/|š’ āƒ— | | Given, the points are The equation of plane is š‘Ÿ āƒ—. (3š‘– Ģ‚ + 4š‘— Ģ‚ āˆ’ 12š‘˜ Ģ‚) + 13 = 0 š‘Ÿ āƒ—.(3š‘– Ģ‚ + 4š‘— Ģ‚ āˆ’ 12š‘˜ Ģ‚) = āˆ’13 (1, 1, p) So, (š‘Ž_1 ) āƒ— = 1š‘– Ģ‚ + 1š‘— Ģ‚ + pš‘˜ Ģ‚ (āˆ’3, 0, 1) So, (š‘Ž_2 ) āƒ— = āˆ’3š‘– Ģ‚ + 0š‘— Ģ‚ + 1š‘˜ Ģ‚ ā€“š‘Ÿ āƒ—.(3š‘– Ģ‚ + 4š‘— Ģ‚ āˆ’ 12š‘˜ Ģ‚) = 13 š’“ āƒ—.(ā€“3š’Š Ģ‚ ā€“ 4š’‹ Ģ‚ + 12š’Œ Ģ‚) = 13 Comparing with š‘Ÿ āƒ—.š‘› āƒ— = d, š‘› āƒ— = āˆ’3š‘– Ģ‚ āˆ’ 4š‘— Ģ‚ + 12š‘˜ Ģ‚ d = 13 Magnitude of š‘› āƒ— = āˆš((āˆ’3)^2+(āˆ’4)^2+怖12怗^2 ) |š‘› āƒ— | = āˆš(9+16+144) = āˆš169 = 13 Distance of point (š’‚šŸ) āƒ— from plane |((š‘Ž1) āƒ—"." š‘› āƒ—" " āˆ’ š‘‘)/|š‘› āƒ— | | = |((1š‘– Ģ‚ + 1š‘— Ģ‚ + š‘š‘˜ Ģ‚ ).(āˆ’3š‘– Ģ‚āˆ’4š‘— Ģ‚+12š‘˜ Ģ‚ )āˆ’13)/13| = |((1Ɨāˆ’3)+(1Ɨāˆ’4) +(š‘Ć—12)āˆ’13)/13| = |(āˆ’3āˆ’4+12š‘āˆ’13)/13| = |(12š‘ āˆ’ 20)/13| Distance of point (š’‚šŸ) āƒ— from plane |((š‘Ž2) āƒ—"." š‘› āƒ— āˆ’ š‘‘)/|š‘› āƒ— | | = |((āˆ’3š‘– Ģ‚ +0š‘— Ģ‚ +1š‘˜ Ģ‚ ).(āˆ’3š‘– Ģ‚āˆ’4š‘— Ģ‚+12š‘˜ Ģ‚ )āˆ’13)/13| = |((āˆ’3Ɨāˆ’3)+(0Ɨāˆ’4) +(1Ɨ12)āˆ’13)/13| = |(9 + 0 +12āˆ’13)/13| = |8/13| = 8/13 Since the plane is equidistance from both the points, |(šŸšŸš’‘ āˆ’ šŸšŸŽ)/šŸšŸ‘| = šŸ–/šŸšŸ‘ |12š‘āˆ’20| = 8 (12p ā€“ 20) = Ā± 8 12p āˆ’ 20 = 8 12p = 8 + 20 12p = 28 p = 28/12 p = 7/3 12p āˆ’ 20 = āˆ’8 12p = āˆ’8 + 20 12p = 12 p = 12/12 p = 1 Answer does not match at end. If mistake, please comment

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo