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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.

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

About the Author

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