Check sibling questions

Ex 11.2, 17 - Shortest distance r = (1-t)i + (t-2)j + (3-2t)k

Ex 11.2, 17 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Ex 11.2, 17 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3


Transcript

Ex 11.2, 17 Find the shortest distance between the lines whose vector equations are š‘Ÿ āƒ— = (1 āˆ’ t) š‘– Ģ‚ + (t āˆ’ 2) š‘— Ģ‚ + (3 āˆ’ 2t) š‘˜ Ģ‚ and š‘Ÿ āƒ— = (s + 1) š‘– Ģ‚ + (2s ā€“ 1) š‘— Ģ‚ ā€“ (2s + 1) š‘˜ Ģ‚ Shortest distance between lines with vector equations š‘Ÿ āƒ— = (š‘Ž1) āƒ— + šœ† (š‘1) āƒ— and š‘Ÿ āƒ— = (š‘Ž2) āƒ— + šœ‡(š‘2) āƒ— is |("(" (š’ƒšŸ) āƒ—Ć— (š’ƒšŸ) āƒ—")" ."(" (š’‚šŸ) āƒ— āˆ’ (š’‚šŸ) āƒ—")" )/|(š’ƒšŸ) āƒ— Ɨ (š’ƒšŸ) āƒ— | | š’“ āƒ— = (šŸ āˆ’ t) š’Š Ģ‚ + (š’•āˆ’šŸ)š’‹ Ģ‚ + (3 āˆ’ 2t) š’Œ Ģ‚ = 1š‘– Ģ‚ āˆ’ tš‘– Ģ‚ + tš‘— Ģ‚ āˆ’ 2š‘— Ģ‚ + 3š‘˜ Ģ‚ āˆ’ 2tš‘˜ Ģ‚ = (1š‘– Ģ‚ āˆ’ 2š‘— Ģ‚ + 3š‘˜ Ģ‚) + t(āˆ’1š‘– Ģ‚ + 1š‘— Ģ‚ āˆ’ 2š‘˜ Ģ‚) Comparing with š‘Ÿ āƒ— = (š‘Ž1) āƒ— + t (š‘1) āƒ—, (š‘Ž1) āƒ— = 1š‘– Ģ‚ ā€“ 2š‘— Ģ‚ + 3š‘˜ Ģ‚ & (š‘1) āƒ— = āˆ’ 1š‘– Ģ‚ + 1š‘— Ģ‚ āˆ’ 2š‘˜ Ģ‚ š’“ āƒ— = (š’” + 1) š’Š Ģ‚ + (šŸš’”" āˆ’ " šŸ)š’‹ Ģ‚ āˆ’ (2s + 1) š’Œ Ģ‚ = sš‘– Ģ‚ + 1š‘– Ģ‚ + 2sš‘— Ģ‚ āˆ’ 1š‘— Ģ‚ āˆ’ 2sš‘˜ Ģ‚ āˆ’ 1š‘˜ Ģ‚ = (1š‘– Ģ‚ āˆ’ 1š‘— Ģ‚ āˆ’ 1š‘˜ Ģ‚) + s(1š‘– Ģ‚ + 2š‘— Ģ‚ āˆ’ 2š‘˜ Ģ‚) Comparing with š‘Ÿ āƒ— = (š‘Ž2) āƒ— + s (š‘2) āƒ—, (š‘Ž2) āƒ— = 1š‘– Ģ‚ āˆ’ 1š‘— Ģ‚ āˆ’ 1š‘˜ Ģ‚ & (š‘2) āƒ— = 1š‘– Ģ‚ + 2š‘— Ģ‚ āˆ’ 2š‘˜ Ģ‚ Now, ((š’‚šŸ) āƒ— āˆ’ (š’‚_šŸ ) āƒ—) = (1š‘– Ģ‚ āˆ’ 1š‘— Ģ‚ āˆ’ 1š‘˜ Ģ‚) āˆ’ (1š‘– Ģ‚ āˆ’ 2š‘— + 3š‘˜ Ģ‚) = (1 āˆ’ 1) š‘– Ģ‚ + ( āˆ’ 1 + 2)š‘— Ģ‚ + ( āˆ’ 1 āˆ’ 3)š‘˜ Ģ‚ = 0š’Š Ģ‚ + 1š’‹ Ģ‚ āˆ’ 4š’Œ Ģ‚ ( (š’ƒ_šŸ ) āƒ—Ć— (š’ƒ_šŸ ) āƒ— ) = |ā– 8(š‘– Ģ‚&š‘— Ģ‚&š‘˜ Ģ‚@ āˆ’1&1& āˆ’2@1&2& āˆ’2)| = š‘– Ģ‚[(1Ɨāˆ’ 2)āˆ’(2Ɨāˆ’ 2)] āˆ’ š‘— Ģ‚[(āˆ’1Ɨāˆ’2)āˆ’(1Ɨāˆ’ 2)] + š‘˜ Ģ‚[(āˆ’ 1Ɨ2)āˆ’(1Ɨ1)] = š‘– Ģ‚[āˆ’2+4] āˆ’ š‘— Ģ‚[2+2] A + š‘˜ Ģ‚[āˆ’2āˆ’1] = 2š’Š Ģ‚ āˆ’ 4š’‹ Ģ‚ āˆ’ 3š’Œ Ģ‚ Magnitude of ((š‘1) āƒ—Ć—(š‘2) āƒ—) = āˆš(22+(āˆ’ 4)2+(āˆ’ 3)2) |(š’ƒšŸ) āƒ—Ć—(š’ƒšŸ) āƒ— | = āˆš(4+16+9) = āˆššŸšŸ— Also, ((š’ƒšŸ) āƒ— Ɨ (š’ƒšŸ) āƒ—) . ((š’‚šŸ) āƒ— ā€“ (š’‚šŸ) āƒ—) = (2š‘– Ģ‚ āˆ’ 4š‘— Ģ‚ āˆ’ 3š‘˜ Ģ‚) . (0š‘– Ģ‚ + 1š‘— Ģ‚ āˆ’ 4š‘˜ Ģ‚) = (2 Ɨ 0) + (āˆ’4 Ɨ 1) + (āˆ’3 Ɨ āˆ’4) = āˆ’0 + (āˆ’4) + 12 = 8 So, shortest distance = |(((š‘_1 ) āƒ— Ɨ (š‘_2 ) āƒ— ) . ((š‘Ž_2 ) āƒ— Ɨ (š‘Ž_1 ) āƒ— ).)/((š‘_1 ) āƒ— Ɨ (š‘_2 ) āƒ— )| = |8/āˆš29| = šŸ–/āˆššŸšŸ— Therefore, the shortest distance between the given two lines is 8/āˆš29 .

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.