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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‘–ย ฬ‚ + 8๐‘—ย ฬ‚ + 12๐‘˜ย ฬ‚, 2๐‘–ย ฬ‚ + 4๐‘—ย ฬ‚ + 6๐‘˜ย ฬ‚, 3๐‘–ย ฬ‚ + 5๐‘—ย ฬ‚ + 4๐‘˜ย ฬ‚ and 5๐‘–ย ฬ‚ + 8๐‘—ย ฬ‚ + 5๐‘˜ย ฬ‚ are coplanarLet points be A = 4๐‘–ย ฬ‚ + 8๐‘—ย ฬ‚ + 12๐‘˜ย ฬ‚ B = 2๐‘–ย ฬ‚ + 4๐‘—ย ฬ‚ + 6๐‘˜ย ฬ‚ C = 3๐‘–ย ฬ‚ + 5๐‘—ย ฬ‚ + 4๐‘˜ย ฬ‚ D = 5๐‘–ย ฬ‚ + 8๐‘—ย ฬ‚ + 5๐‘˜ย ฬ‚ Four points A, B, C, D are coplanar if the three vectors (๐ด๐ต)ย โƒ— , (๐ด๐ถ)ย โƒ— and (๐ด๐ท)ย โƒ— are coplanar. i.e. [(๐‘จ๐‘ฉ)ย โƒ—, (๐‘จ๐‘ช)ย โƒ—, (๐‘จ๐‘ซ)ย โƒ— ] = 0 A (4๐‘–ย ฬ‚ + 8๐‘—ย ฬ‚ + 12๐‘˜ย ฬ‚) B (2๐‘–ย ฬ‚ + 4๐‘—ย ฬ‚ + 6๐‘˜ย ฬ‚) (๐‘จ๐‘ฉ)ย โƒ— = (2๐‘–ย ฬ‚ + 4๐‘—ย ฬ‚ + 6๐‘˜ย ฬ‚) โ€“ (4๐‘–ย ฬ‚ + 8๐‘—ย ฬ‚ + 12๐‘˜ย ฬ‚) = (2 โˆ’ 4) ๐‘–ย ฬ‚ + (4 โ€“ 8) ๐‘—ย ฬ‚ + (6 โ€“ 12)๐‘˜ย ฬ‚ = โ€“2๐’Šย ฬ‚ โˆ’ 4๐’‹ย ฬ‚ โˆ’ 6๐’Œย ฬ‚ A (4๐‘–ย ฬ‚ + 8๐‘—ย ฬ‚ + 12๐‘˜ย ฬ‚) C (3๐‘–ย ฬ‚ + 5๐‘—ย ฬ‚ + 4๐‘˜ย ฬ‚) (๐‘จ๐‘ช)ย โƒ— = (3๐‘–ย ฬ‚ + 5๐‘—ย ฬ‚ + 4๐‘˜ย ฬ‚) โ€“ (4๐‘–ย ฬ‚ + 8๐‘—ย ฬ‚ + 12๐‘˜ย ฬ‚) = (3 โˆ’ 4) ๐‘–ย ฬ‚ + (5 โ€“ 8) ๐‘—ย ฬ‚ + (4 โ€“ 12) ๐‘˜ย ฬ‚ = โ€“๐’Šย ฬ‚ โ€“ 3๐’‹ย ฬ‚ โ€“ 8๐’Œย ฬ‚ A (4๐‘–ย ฬ‚ + 8๐‘—ย ฬ‚ + 12๐‘˜ย ฬ‚) D (5๐‘–ย ฬ‚ + 8๐‘—ย ฬ‚ + 5๐‘˜ย ฬ‚) (๐‘จ๐‘ซ)ย โƒ— = (5๐‘–ย ฬ‚ + 8๐‘—ย ฬ‚ + 5๐‘˜ย ฬ‚) โ€“ (4๐‘–ย ฬ‚ + 8๐‘—ย ฬ‚ + 12๐‘˜ย ฬ‚) = (5 โˆ’ 4) ๐‘–ย ฬ‚ + (8 โ€“ 8) ๐‘—ย ฬ‚ + (5 โ€“ 12) ๐‘˜ย ฬ‚ = ๐’Šย ฬ‚ + 0๐’‹ย ฬ‚ โ€“ 7๐’Œย ฬ‚ Now, [(๐ด๐ต)ย โƒ—, (๐ด๐ถ)ย โƒ—, (๐ด๐ท)ย 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)ย โƒ—, (๐‘จ๐‘ช)ย โƒ—, (๐‘จ๐‘ซ)ย โƒ— ] = 0 Therefore, points A, B, C and D are coplanar.

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