Check sibling questions

Slide50.JPG

Slide51.JPG
Slide52.JPG

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 11.2, 15 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 .

Ask a doubt
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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.