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  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise
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Example 28 (Supplementary NCERT) Find ๐œ† if the vectors ๐‘Žย โƒ— = ๐‘–ย ฬ‚ + 3๐‘—ย ฬ‚ + ๐‘˜ย ฬ‚, ๐‘ย โƒ— = 2๐‘–ย ฬ‚ โ€“ ๐‘—ย ฬ‚ โˆ’ ๐‘˜ย ฬ‚ and ๐‘ย โƒ— = ๐œ†๐‘–ย ฬ‚ + 7๐‘—ย ฬ‚ + 3๐‘˜ย ฬ‚ are coplanar.Three 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‚ + 3๐‘—ย ฬ‚ + ๐‘˜ย ฬ‚ ๐‘ย โƒ— = 2๐‘–ย ฬ‚ โ€“ ๐‘—ย ฬ‚ โˆ’ ๐‘˜ย ฬ‚ ๐‘ย โƒ— = ฮป๐‘–ย ฬ‚ + 7๐‘—ย ฬ‚ + 3๐‘˜ย ฬ‚ [๐‘Žย โƒ—" " ๐‘ย โƒ—" " ๐‘ย โƒ— ] = |โ– 8(1&3&1@2&โˆ’1&โˆ’1@๐œ†&7&3)| 0 = 1[(โˆ’1ร—3)โˆ’(7ร—โˆ’1) ] โˆ’ 3[(2ร—3)โˆ’(ฮปร—โˆ’1) ] + 1[(2ร—7)โˆ’(ฮปร—โˆ’1) ] 0 = 1 [โˆ’3โˆ’(โˆ’7)]โˆ’3[6โˆ’(โˆ’ฮป)]+1[14โˆ’(โˆ’ฮป)] 0 = 1 [โˆ’3+7]โˆ’3[6+ฮป]+1[14+ฮป] 0 = 4 โ€“ 18 โ€“ 3ฮป + 14 + ฮป 0 = 18 โ€“ 18 โ€“ 3ฮป + ฮป 0 = โ€“ 2ฮป โ€“2ฮป = 0 ฮป = 0/(โˆ’2) ๐€ = 0 Therefore, ๐‘Žย โƒ—,๐‘,๐‘ย โƒ— are coplanar if ฮป = 0

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