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  1. Chapter 10 Class 12 Vector Algebra (Term 2)
  2. Serial order wise

Transcript

Supplementary Exercise Q4 Show that (i) the vectors 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’ 4๐‘— ฬ‚ + 5๐‘˜ ฬ‚ are coplanar. 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’ 4๐‘— ฬ‚ + 5๐‘˜ ฬ‚ [๐‘Ž โƒ—,๐‘,๐‘ โƒ— ] = |โ– 8(2&โˆ’1&1@1&2&โˆ’3@3&โˆ’4&5)| = 2[(2ร—5)โˆ’(โˆ’4ร—โˆ’3)] โˆ’ (โˆ’1) [(1ร—5)โˆ’(3ร—โˆ’3) ] + 1[(1ร—โˆ’4)โˆ’(3ร—2) ] = 2 [10โˆ’12]+[5+9] + [โˆ’4โˆ’6] = โ€“4 + 14 โ€“ 10 = 0 โˆด [๐’‚ โƒ—,๐’ƒ,๐’„ โƒ— ] = 0 Therefore, ๐‘Ž โƒ—,๐‘ and ๐‘ โƒ— are coplanar. Supplementary Exercise Q4 Show that (ii) the vectors ๐‘Ž โƒ— = ๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + 3๐‘˜ ฬ‚, ๐‘ โƒ— = โˆ’2๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚, and ๐‘ โƒ— = ๐‘– ฬ‚ โˆ’ 3๐‘— ฬ‚ + 5๐‘˜ ฬ‚ are coplanar Three vectors ๐‘Ž โƒ—, ๐‘ โƒ—, ๐‘ โƒ— are coplanar if [ ๐’‚ โƒ—, ๐’ƒ โƒ—, ๐’„ โƒ— ] = 0 Given, ๐‘Ž โƒ— = ๐‘– ฬ‚ โˆ’ 2๐‘— ฬ‚ + 3๐‘˜ ฬ‚ ๐‘ โƒ— = โˆ’2๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚ ๐‘ โƒ— = ๐‘– ฬ‚ โˆ’ 3๐‘— ฬ‚ + 5๐‘˜ ฬ‚ [๐‘Ž โƒ—,๐‘,๐‘ โƒ— ] = |โ– 8(1&โˆ’2&3@โˆ’2&3&โˆ’4@1&โˆ’3&5)| = 1[(3ร—5)โˆ’(โˆ’3ร—4) ] โˆ’ (โˆ’2) [(โˆ’2ร—5)โˆ’(1ร—โˆ’4) ] + 3[(โˆ’2ร—โˆ’3)โˆ’(1ร—3) ] = 1 [15โˆ’12]+[2(โˆ’10โˆ’(โˆ’4)] + 3 [6โˆ’3] = 1(3) + 2 (โˆ’6) + 3 (3) = 3 โˆ’ 12 + 9 = 12 โˆ’ 12 = 0 โˆด [๐’‚ โƒ—,๐’ƒ,๐’„ โƒ— ] = 0 Therefore, ๐‘Ž โƒ—,๐‘ and ๐‘ โƒ— are coplanar.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.