Check sibling questions


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Ex 10.5, 4 (Supplementary NCERT) Let ๐‘Ž โƒ— = ๐‘– ฬ‚ + ๐‘— ฬ‚ + ๐‘˜ ฬ‚, ๐‘ โƒ— = ๐‘– ฬ‚ and ๐‘ โƒ— = c1๐‘– ฬ‚ + c2๐‘— ฬ‚ + c3๐‘˜ ฬ‚ are coplanar (a) If c1 = 1 and c2 = 2, find c3 which makes ๐‘Ž โƒ—, ๐‘ โƒ—, ๐‘ โƒ— coplanar Given c1 = 1 and c2 = 2 So, our vectors become ๐’‚ โƒ— = ๐’Š ฬ‚ + ๐’‹ ฬ‚ + ๐’Œ ฬ‚ ๐’ƒ โƒ— = ๐‘– ฬ‚ ๐’„ โƒ— = c1๐‘– ฬ‚ + c2๐‘— ฬ‚ + c3๐‘˜ ฬ‚ Three vectors ๐‘Ž โƒ—, ๐‘ โƒ—, ๐‘ โƒ— are coplanar if [๐’‚ โƒ—" " ๐’ƒ โƒ—" " ๐’„ โƒ— ] = 0 Ex 10.5, 4 (Supplementary NCERT) Let ๐‘Ž โƒ— = ๐‘– ฬ‚ + ๐‘— ฬ‚ + ๐‘˜ ฬ‚, ๐‘ โƒ— = ๐‘– ฬ‚ and ๐‘ โƒ— = c1๐‘– ฬ‚ + c2๐‘— ฬ‚ + c3๐‘˜ ฬ‚ are coplanar (b) If c2 = โ€“1 and c3 = 1, show that no value of c1 can make ๐‘Ž โƒ—, ๐‘ โƒ—, ๐‘ โƒ— coplanar Given c2 = โ€“1 and c3 = 1 So, our vectors ๐’‚ โƒ— = ๐‘– ฬ‚ + ๐‘— ฬ‚ + ๐‘˜ ฬ‚ ๐’ƒ โƒ— = ๐‘– ฬ‚ ๐’„ โƒ— = c1๐‘– ฬ‚ + c2๐‘— ฬ‚ + c3๐‘˜ ฬ‚ Three vectors ๐‘Ž โƒ—, ๐‘ โƒ—, ๐‘ โƒ— are coplanar if [๐’‚ โƒ—" " ๐’ƒ โƒ—" " ๐’„ โƒ— ] = 0 Finding [๐’‚ โƒ—" " ๐’ƒ โƒ—" " ๐’„ โƒ— ] [๐‘Ž โƒ—" " ๐‘ โƒ—" " ๐‘ โƒ— ] = |โ– 8(1&1&1@1&0&0@๐‘_1&โˆ’1&1)| = 1[(0ร—1)โˆ’(0ร—โˆ’1) ] โˆ’ 1[(1ร—1)โˆ’(๐‘_1ร—0) ] + 1[(1ร—โˆ’1)โˆ’(๐‘_1ร—0) ] = 1 [0โˆ’0]โˆ’1[1โˆ’0]+1[โˆ’1โˆ’0] = 0 โ€“ 1 โ€“ 1 = โ€“2 Thus, [๐’‚ โƒ—" " ๐’ƒ โƒ—" " ๐’„ โƒ— ] โ‰  0 for any value of c1 So, we can write that ๐‘Ž โƒ—,๐‘ โƒ—,๐‘ โƒ— are not coplanar for any value of c1

  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