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  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise
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Ex 10.5, 6 (Supplementary NCERT) Find x such that A (3, 2, 1), B = (4, x, 5), C(4, 2, โ€“2) and D (6, 5, โ€“1) are coplanar 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 (3, 2, 1), B (4, x, 5) (๐‘จ๐‘ฉ)ย โƒ— = (4 โ€“ 3)๐‘–ย ฬ‚ + (x โˆ’ 2)๐‘—ย ฬ‚ + (5 โ€“ 1) ๐‘˜ย ฬ‚ = 1๐’Šย ฬ‚ + (x โˆ’ 2)๐’‹ย ฬ‚ + 4๐’Œย ฬ‚ A (3, 2, 1), C(4, 2, โ€“2) (๐‘จ๐‘ช)ย โƒ— = (4 โ€“ 3)๐‘–ย ฬ‚ + (2 โ€“ 2)๐‘—ย ฬ‚ + (โ€“2 โ€“ 1) ๐‘˜ย ฬ‚ = 1๐’Šย ฬ‚ + 0๐’‹ย ฬ‚ โˆ’ 3๐’Œย ฬ‚ A (3, 2, 1), D (6, 5, โ€“1) (๐‘จ๐‘ซ)ย โƒ— = (6 โ€“ 3)๐‘–ย ฬ‚ + (5 โ€“ 2)๐‘—ย ฬ‚ + (โ€“1 โ€“ 1) ๐‘˜ย ฬ‚ = 3๐’Šย ฬ‚ + 3๐’‹ย ฬ‚ โ€“ 2๐’Œย ฬ‚ A (3, 2, 1), D (6, 5, โ€“1) (๐‘จ๐‘ซ)ย โƒ— = (6 โ€“ 3)๐‘–ย ฬ‚ + (5 โ€“ 2)๐‘—ย ฬ‚ + (โ€“1 โ€“ 1) ๐‘˜ย ฬ‚ = 3๐’Šย ฬ‚ + 3๐’‹ย ฬ‚ โ€“ 2๐’Œย ฬ‚ A (3, 2, 1), D (6, 5, โ€“1) (๐‘จ๐‘ซ)ย โƒ— = (6 โ€“ 3)๐‘–ย ฬ‚ + (5 โ€“ 2)๐‘—ย ฬ‚ + (โ€“1 โ€“ 1) ๐‘˜ย ฬ‚ = 3๐’Šย ฬ‚ + 3๐’‹ย ฬ‚ โ€“ 2๐’Œย ฬ‚ A (3, 2, 1), D (6, 5, โ€“1) (๐‘จ๐‘ซ)ย โƒ— = (6 โ€“ 3)๐‘–ย ฬ‚ + (5 โ€“ 2)๐‘—ย ฬ‚ + (โ€“1 โ€“ 1) ๐‘˜ย ฬ‚ = 3๐’Šย ฬ‚ + 3๐’‹ย ฬ‚ โ€“ 2๐’Œย ฬ‚ [(๐ด๐ต)ย โƒ—, (๐ด๐ถ)ย โƒ—, (๐ด๐ท)ย โƒ— ] = |โ– 8(1&(๐‘ฅโˆ’2)&4@1&0&โˆ’3@3&3&โˆ’2)| 0 = 1[(0ร—โˆ’2)โˆ’(3ร—โˆ’3) ] โˆ’ (x โˆ’ 2) [(1ร—โˆ’2)โˆ’(3ร—โˆ’3)] + 4[(1ร—3)โˆ’(3ร—0) ] 0 = 1[0โˆ’(โˆ’9)]โˆ’"(x โˆ’ 2)" [โˆ’2โˆ’(โˆ’9)]+4[3โˆ’0] 0 = 1(9) โ€“ "(x โˆ’ 2)"(โ€“2 + 9) + 4(3) 0 = 9 โ€“ (x โ€“ 2) (7) + 12 0 = 9 โ€“ 7x + 14 + 12 0 = 35 โ€“ 7x 7x = 35 x = 35/7 x = 5 Ex 10.5, 6 (Supplementary NCERT) Find x such that A (3, 2, 1), B = (4, x, 5), C(4, 2, โ€“2) and D (6, 5, โ€“1) are coplanar 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 (3, 2, 1), B (4, x, 5) (๐‘จ๐‘ฉ)ย โƒ— = (4 โ€“ 3)๐‘–ย ฬ‚ + (x โˆ’ 2)๐‘—ย ฬ‚ + (5 โ€“ 1) ๐‘˜ย ฬ‚ = 1๐’Šย ฬ‚ + (x โˆ’ 2)๐’‹ย ฬ‚ + 4๐’Œย ฬ‚ A (3, 2, 1), C(4, 2, โ€“2) (๐‘จ๐‘ช)ย โƒ— = (4 โ€“ 3)๐‘–ย ฬ‚ + (2 โ€“ 2)๐‘—ย ฬ‚ + (โ€“2 โ€“ 1) ๐‘˜ย ฬ‚ = 1๐’Šย ฬ‚ + 0๐’‹ย ฬ‚ โˆ’ 3๐’Œย ฬ‚ A (3, 2, 1), C(4, 2, โ€“2) (๐‘จ๐‘ช)ย โƒ— = (4 โ€“ 3)๐‘–ย ฬ‚ + (2 โ€“ 2)๐‘—ย ฬ‚ + (โ€“2 โ€“ 1) ๐‘˜ย ฬ‚ = 1๐’Šย ฬ‚ + 0๐’‹ย ฬ‚ โˆ’ 3๐’Œย ฬ‚ A (3, 2, 1), D (6, 5, โ€“1) (๐‘จ๐‘ซ)ย โƒ— = (6 โ€“ 3)๐‘–ย ฬ‚ + (5 โ€“ 2)๐‘—ย ฬ‚ + (โ€“1 โ€“ 1) ๐‘˜ย ฬ‚ = 3๐’Šย ฬ‚ + 3๐’‹ย ฬ‚ โ€“ 2๐’Œย ฬ‚ [(๐ด๐ต)ย โƒ—, (๐ด๐ถ)ย โƒ—, (๐ด๐ท)ย โƒ— ] = |โ– 8(1&(๐‘ฅโˆ’2)&4@1&0&โˆ’3@3&3&โˆ’2)| 0 = 1[(0ร—โˆ’2)โˆ’(3ร—โˆ’3) ] โˆ’ (x โˆ’ 2) [(1ร—โˆ’2)โˆ’(3ร—โˆ’3)] + 4[(1ร—3)โˆ’(3ร—0) ] 0 = 1[0โˆ’(โˆ’9)]โˆ’"(x โˆ’ 2)" [โˆ’2โˆ’(โˆ’9)]+4[3โˆ’0] 0 = 1(9) โ€“ "(x โˆ’ 2)"(โ€“2 + 9) + 4(3) 0 = 9 โ€“ (x โ€“ 2) (7) + 12 0 = 9 โ€“ 7x + 14 + 12 0 = 35 โ€“ 7x 7x = 35 x = 35/7 x = 5

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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.