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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&1@1&2&โˆ’3@3&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&1@2&โˆ’๐œ†&1@5&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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo