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  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise
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Example 27 (Supplementary NCERT) Show that the vectors ๐‘Žย โƒ— = ๐‘–ย ฬ‚ โˆ’ 2๐‘—ย ฬ‚ + 3๐‘˜ย ฬ‚, ๐‘ย โƒ— = โˆ’2๐‘–ย ฬ‚ + 3๐‘—ย ฬ‚ โˆ’ 4๐‘˜ย ฬ‚ and ๐‘ย โƒ— = ๐‘–ย ฬ‚ โˆ’ 3๐‘—ย ฬ‚ + 5๐‘˜ย ฬ‚ are coplanarThree vectors ๐‘Žย โƒ—, ๐‘ย โƒ—, ๐‘ย โƒ— are coplanar if [๐’‚ย โƒ—" " ๐’ƒย โƒ—" " ๐’„ย โƒ— ] = 0 Given, ๐‘Žย โƒ— = ๐‘–ย ฬ‚ โˆ’ 2๐‘—ย ฬ‚ + 3๐‘˜ย ฬ‚ ๐‘ย โƒ— = โˆ’2๐‘–ย ฬ‚ + 3๐‘—ย ฬ‚ โˆ’ 4๐‘˜ย ฬ‚ ๐‘ย โƒ— = ๐‘–ย ฬ‚ โˆ’ 3๐‘—ย ฬ‚ + ฮป๐‘˜ย ฬ‚ [๐‘Žย โƒ—" " ๐‘ย โƒ—" " ๐‘ย โƒ— ] = |โ– 8(1&โˆ’2&3@โˆ’2&3&โˆ’4@1&โˆ’3&5)| = 1[(3ร—5)โˆ’(โˆ’3ร—โˆ’4) ] โˆ’ (โˆ’2) [(โˆ’2ร—5)โˆ’(1ร—โˆ’4) ] + 3[(โˆ’2ร—โˆ’3)โˆ’(1ร—3) ] = 1 [15โˆ’(3 ร—4)]+2[โˆ’10โˆ’(โˆ’4)]+3[(2 ร—3)โˆ’3] = 1 [15โˆ’12]+2[โˆ’10+4]+3[6โˆ’3] = 1 [3]+2[โˆ’6]+3[3] = 3 โ€“ 12 + 9 = 0 Since [๐‘Žย โƒ—" " ๐‘ย โƒ—" " ๐‘ย โƒ— ] = 0 Vectors ๐‘Žย โƒ—, ๐‘ย โƒ—, ๐‘ย โƒ— are coplanar

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