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


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Supplementary Exercise Q5 Find the value of ๐œ† if the following vectors are co-planar (i) ๐‘Ž โƒ— = 2๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + ๐‘˜ ฬ‚, ๐‘ โƒ— = ๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚, and ๐‘ โƒ— = 3๐‘– ฬ‚ โˆ’ ๐œ†๐‘— ฬ‚ + 5๐‘˜ ฬ‚ Three vectors ๐‘Ž โƒ—, ๐‘ โƒ—, ๐‘ โƒ— are coplanar if [๐’‚ โƒ—" " ๐’ƒ โƒ—" " ๐’„ โƒ— ] = 0 Given, ๐‘Ž โƒ— = 2๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ + ๐‘˜ ฬ‚ ๐‘ โƒ— = ๐‘– ฬ‚ + 2๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚ ๐‘ โƒ— = 3๐‘– ฬ‚ โˆ’ ฮป ๐‘— ฬ‚ + 5๐‘˜ ฬ‚ [๐‘Ž โƒ—" " ๐‘ โƒ—" " ๐‘ โƒ— ] = |โ– 8(2&โˆ’1&[email protected]&2&โˆ’[email protected]&โˆ’๐œ†&5)| 0 = 2[(2ร—5)โˆ’(โˆ’ฮปร—โˆ’3) ] โˆ’ (โˆ’1) [(1ร—5)โˆ’(3ร—โˆ’3) ] + 1[(1ร—โˆ’ฮป)โˆ’(3ร—2) ] 0 = 2[10โˆ’3ฮป] + [5+9 ] + [โˆ’ฮปโˆ’6] 0 = 20 โ€“ 6ฮป + 14 โ€“ ฮป +โ€“ 6 0 = โ€“7ฮป + 28 7ฮป = 28 ฮป = 28/7 ๐€ = 4 Therefore, ๐‘Ž โƒ—,๐‘,๐‘ โƒ— are coplanar if ฮป = 4 Supplementary Exercise Q5 Find the value of ๐œ† if the following vectors are co-planar (ii) ๐‘Ž โƒ— = 2๐‘– ฬ‚ + ๐‘— ฬ‚ + ๐‘˜ ฬ‚, ๐‘ โƒ— = 2๐‘– ฬ‚ โˆ’ ๐œ†๐‘— ฬ‚ + ๐‘˜ ฬ‚, and ๐‘ โƒ— = 5๐‘– ฬ‚ โˆ’ ๐‘— ฬ‚ โˆ’ 3๐‘˜ ฬ‚ Three vectors ๐‘Ž โƒ—, ๐‘ โƒ—, ๐‘ โƒ— are coplanar if [๐’‚ โƒ—" " ๐’ƒ โƒ—" " ๐’„ โƒ— ] = 0 Given, ๐‘Ž โƒ— = 2๐‘– ฬ‚ + ๐‘— ฬ‚ + ๐‘˜ ฬ‚ ๐‘ โƒ— = 2๐‘– ฬ‚ โˆ’ ฮป๐‘— ฬ‚ + ๐‘˜ ฬ‚ ๐‘ โƒ— = 5๐‘– ฬ‚ + ๐‘— ฬ‚ โ€“ 3๐‘˜ ฬ‚ [๐‘Ž โƒ—" " ๐‘ โƒ—" " ๐‘ โƒ— ] = |โ– 8(2&1&[email protected]&โˆ’๐œ†&[email protected]&1&โˆ’3)| 0 = 2[(โˆ’ฮปร—โˆ’3)โˆ’(1ร—1)] โˆ’ 1[(2ร—โˆ’3)โˆ’(5ร—1)] + 1[(2ร—1)โˆ’(5ร—โˆ’ฮป) ] 0 = 2[3ฮปโˆ’1] โˆ’ 1[โˆ’6โˆ’5] + 1[2+5ฮป] 0 = 6ฮป โ€“ 2 + 11 + 2 + 5ฮป 0 = 11ฮป + 11 โ€“11ฮป = 11 ฮป = 11/(โˆ’11) ๐€ = โ€“1 Therefore, ๐‘Ž โƒ—,๐‘,๐‘ โƒ— are coplanar if ฮป = โ€“1

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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.