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  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise
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Ex 10.5, 5 (Supplementary NCERT) Show that the four points with position vectors 4๐‘– ฬ‚ + 8๐‘— ฬ‚ + 12๐‘˜ ฬ‚, 2๐‘– ฬ‚ + 4๐‘— ฬ‚ + 6๐‘˜ ฬ‚, 3๐‘– ฬ‚ + 5๐‘— ฬ‚ + 4๐‘˜ ฬ‚ and 5๐‘– ฬ‚ + 8๐‘— ฬ‚ + 5๐‘˜ ฬ‚ are coplanarLet points be A = 4๐‘– ฬ‚ + 8๐‘— ฬ‚ + 12๐‘˜ ฬ‚ B = 2๐‘– ฬ‚ + 4๐‘— ฬ‚ + 6๐‘˜ ฬ‚ C = 3๐‘– ฬ‚ + 5๐‘— ฬ‚ + 4๐‘˜ ฬ‚ D = 5๐‘– ฬ‚ + 8๐‘— ฬ‚ + 5๐‘˜ 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 (4๐‘– ฬ‚ + 8๐‘— ฬ‚ + 12๐‘˜ ฬ‚) B (2๐‘– ฬ‚ + 4๐‘— ฬ‚ + 6๐‘˜ ฬ‚) (๐‘จ๐‘ฉ) โƒ— = (2๐‘– ฬ‚ + 4๐‘— ฬ‚ + 6๐‘˜ ฬ‚) โ€“ (4๐‘– ฬ‚ + 8๐‘— ฬ‚ + 12๐‘˜ ฬ‚) = (2 โˆ’ 4) ๐‘– ฬ‚ + (4 โ€“ 8) ๐‘— ฬ‚ + (6 โ€“ 12)๐‘˜ ฬ‚ = โ€“2๐’Š ฬ‚ โˆ’ 4๐’‹ ฬ‚ โˆ’ 6๐’Œ ฬ‚ A (4๐‘– ฬ‚ + 8๐‘— ฬ‚ + 12๐‘˜ ฬ‚) C (3๐‘– ฬ‚ + 5๐‘— ฬ‚ + 4๐‘˜ ฬ‚) (๐‘จ๐‘ช) โƒ— = (3๐‘– ฬ‚ + 5๐‘— ฬ‚ + 4๐‘˜ ฬ‚) โ€“ (4๐‘– ฬ‚ + 8๐‘— ฬ‚ + 12๐‘˜ ฬ‚) = (3 โˆ’ 4) ๐‘– ฬ‚ + (5 โ€“ 8) ๐‘— ฬ‚ + (4 โ€“ 12) ๐‘˜ ฬ‚ = โ€“๐’Š ฬ‚ โ€“ 3๐’‹ ฬ‚ โ€“ 8๐’Œ ฬ‚ A (4๐‘– ฬ‚ + 8๐‘— ฬ‚ + 12๐‘˜ ฬ‚) D (5๐‘– ฬ‚ + 8๐‘— ฬ‚ + 5๐‘˜ ฬ‚) (๐‘จ๐‘ซ) โƒ— = (5๐‘– ฬ‚ + 8๐‘— ฬ‚ + 5๐‘˜ ฬ‚) โ€“ (4๐‘– ฬ‚ + 8๐‘— ฬ‚ + 12๐‘˜ ฬ‚) = (5 โˆ’ 4) ๐‘– ฬ‚ + (8 โ€“ 8) ๐‘— ฬ‚ + (5 โ€“ 12) ๐‘˜ ฬ‚ = ๐’Š ฬ‚ + 0๐’‹ ฬ‚ โ€“ 7๐’Œ ฬ‚ Now, [(๐ด๐ต) โƒ—, (๐ด๐ถ) โƒ—, (๐ด๐ท) โƒ— ] = |โ– 8(โˆ’2&โˆ’4&โˆ’6@โˆ’1&โˆ’3&โˆ’8@1&0&โˆ’7)| = โ€“2[(โˆ’3ร—โˆ’7)โˆ’(0ร—โˆ’8) ] โˆ’ (โˆ’4) [(โˆ’1ร—โˆ’7)โˆ’(1ร—โˆ’8)] + (โˆ’6)[(โˆ’1ร—0)โˆ’(1ร—โˆ’3) ] = โˆ’2[21โˆ’0]+4[7+8]โˆ’6[0+3] = โˆ’2[21]+4[15]โˆ’6[3] = โ€“ 42 + 60 โ€“ 18 = 60 โ€“ 60 = 0 โˆด [(๐‘จ๐‘ฉ) โƒ—, (๐‘จ๐‘ช) โƒ—, (๐‘จ๐‘ซ) โƒ— ] = 0 Therefore, points A, B, C and D 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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.