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

Transcript

Supplementary Example 5 Show that the four points with position vectors 6๐‘– ฬ‚ โˆ’ 7๐‘— ฬ‚, 16๐‘– ฬ‚ โˆ’ 19๐‘— ฬ‚ โˆ’ 4๐‘˜ ฬ‚, 3๐‘— ฬ‚ โˆ’ 6๐‘˜ ฬ‚ & 2๐‘– ฬ‚ + 5๐‘— ฬ‚ + 10๐‘˜ ฬ‚ are not co-planar Let points be A = 6๐‘– ฬ‚ โ€“ 7๐‘— ฬ‚ B = 16๐‘– ฬ‚ โ€“ 19๐‘— ฬ‚ โ€“ 4๐‘˜ ฬ‚ C = 3๐‘— ฬ‚ โ€“ 6๐‘˜ ฬ‚ D = 2๐‘– ฬ‚ + 5๐‘— ฬ‚ +10๐‘˜ ฬ‚ Four points A, B, C, D are coplanar if the three vectors (๐ด๐ต) โƒ— , (๐ด๐ถ) โƒ— and (๐ด๐ท) โƒ— are coplanar. i.e. [(๐‘จ๐‘ฉ) โƒ—, (๐‘จ๐‘ช) โƒ—, (๐‘จ๐‘ซ) โƒ— ] = 0 A (6๐’Š ฬ‚ โ€“ 7๐’‹ ฬ‚) B (16๐’Š ฬ‚ โ€“ 19๐’‹ ฬ‚ โ€“ 4๐’Œ ฬ‚) (๐‘จ๐‘ฉ) โƒ— = (16๐‘– ฬ‚ โ€“ 19๐‘— ฬ‚ โ€“ 4๐‘˜ ฬ‚) โ€“ (6๐‘– ฬ‚ โ€“ 7๐‘— ฬ‚ + 0๐‘˜ ฬ‚) = (16 โˆ’ 6) ๐‘– ฬ‚ + (โˆ’19 + 7) ๐‘— ฬ‚ โ€“ 4๐‘˜ ฬ‚ = 10๐’Š ฬ‚ โˆ’ 12๐’‹ ฬ‚ โˆ’ 4๐’Œ ฬ‚ A (6๐’Š ฬ‚ โ€“ 7๐’‹ ฬ‚) C (3๐’‹ ฬ‚ โ€“ 6๐’Œ ฬ‚) (๐‘จ๐‘ช) โƒ— = (0๐‘– ฬ‚ + 3๐‘— ฬ‚ โˆ’ 6๐‘˜ ฬ‚) โ€“ (6๐‘– ฬ‚ โ€“ 7๐‘— ฬ‚ + 0๐‘˜ ฬ‚) = (0 โˆ’ 6) ๐‘– ฬ‚ + (3 + 7) ๐‘— ฬ‚ + (โˆ’6 โ€“ 0) ๐‘˜ ฬ‚ = โ€“6๐’Š ฬ‚ + 10๐’‹ ฬ‚ โ€“ 6๐’Œ ฬ‚ A (6๐’Š ฬ‚ โ€“ 7๐’‹ ฬ‚) D (2๐’Š ฬ‚ + 5๐’‹ ฬ‚ +10๐’Œ ฬ‚) (๐‘จ๐‘ซ) โƒ— = (2๐‘– ฬ‚ + 5๐‘— ฬ‚ +10๐‘˜ ฬ‚) โ€“ (6๐‘– ฬ‚ โ€“ 7๐‘— ฬ‚ + 0๐‘˜ ฬ‚) = (2 โˆ’ 6) ๐‘– ฬ‚ + (5 + 7) ๐‘— ฬ‚ + (10 โ€“ 0) ๐‘˜ ฬ‚ = โ€“4๐’Š ฬ‚ + 12๐’‹ ฬ‚ + 10๐’Œ ฬ‚ Now, [(๐ด๐ต) โƒ—, (๐ด๐ถ) โƒ—, (๐ด๐ท) โƒ— ] = |โ– 8(10&โˆ’12&โˆ’4@โˆ’6&10&โˆ’6@โˆ’4&12&10)| = 10[(10ร—10)โˆ’(12ร—โˆ’6) ] โˆ’ (โˆ’12) [(โˆ’6ร—10)โˆ’(โˆ’4ร—โˆ’6)] + (โˆ’4)[(โˆ’6ร—12)โˆ’(โˆ’4ร—10) ] = 10[100+72]+12[โˆ’60โˆ’24]โˆ’4[โˆ’72+40] = 10[172]+12[โˆ’84]โˆ’4[โˆ’32] = 1720 โ€“ 1008 + 128 = 840 โˆด [(๐‘จ๐‘ฉ) โƒ—, (๐‘จ๐‘ช) โƒ—, (๐‘จ๐‘ซ) โƒ— ] โ‰  0 Therefore, points A, B, C and D are not 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.