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  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise
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Example 30 (Supplementary NCERT) Prove that [โ– 8(๐‘Žย โƒ—" + " ๐‘ย โƒ—&๐‘ย โƒ—+๐‘ย โƒ—&๐‘ย โƒ—+๐‘Žย โƒ— )] = 2 [๐‘Žย โƒ— , ๐‘ย โƒ—, ๐‘ย โƒ—] To prove [โ– 8(๐‘Žย โƒ—" + " ๐‘ย โƒ—&๐‘ย โƒ—+๐‘ย โƒ—&๐‘ย โƒ—+๐‘Žย โƒ— )] = 2 [๐‘Žย โƒ— , ๐‘ย โƒ—, ๐‘ย โƒ—] Solving LHS [โ– 8(๐‘Žย โƒ—" + " ๐‘ย โƒ—&๐‘ย โƒ—+๐‘ย โƒ—&๐‘ย โƒ—+๐‘Žย โƒ— )] = (๐‘Žย โƒ—" + " ๐‘ย โƒ—). ["(" ๐‘ย โƒ—+๐‘ย โƒ—") " ร— " (" ๐‘ย โƒ—+๐‘Žย โƒ—)] = (๐‘Žย โƒ—" + " ๐‘ย โƒ—). ["(" ๐‘ย โƒ—ร—๐‘ย โƒ—") + (" ๐‘ย โƒ—ร—๐‘Žย โƒ—)+"(" ๐’„ย โƒ—ร—๐’„ย โƒ—") + (" ๐‘ย โƒ—ร—๐‘Žย โƒ—)] = "(" ๐‘Žย โƒ—+๐‘ย โƒ—")." [(๐‘ย โƒ—ร—๐‘ย โƒ—") + (" ๐‘ย โƒ—ร—๐‘Žย โƒ— ) "+ 0 + (" ๐‘ย โƒ—" ร— " ๐‘Žย โƒ—") " ] = ๐‘Žย โƒ—. (๐‘ย โƒ— ร— ๐‘ย โƒ—) + ๐‘ย โƒ—.(๐‘ย โƒ— ร— ๐‘ย โƒ—) + ๐‘Žย โƒ—. (๐‘ย โƒ— ร— ๐‘Žย โƒ—) + ๐‘ย โƒ—.(๐‘ย โƒ— ร— ๐‘Žย โƒ—) + ๐‘Žย โƒ—.(๐‘ย โƒ— ร— ๐‘Žย โƒ—) + ๐‘ย โƒ—.(๐‘ย โƒ— ร— ๐‘Žย โƒ—) = [๐‘Žย โƒ—", " ๐‘ย โƒ—", " ๐‘ย โƒ— ] + [๐’ƒย โƒ—", " ๐’ƒย โƒ—", " ๐’„ย โƒ— ] + [๐’‚ย โƒ—", " ๐’ƒย โƒ—", " ๐’‚ย โƒ— ] + [๐’ƒย โƒ—", " ๐’ƒย โƒ—", " ๐’‚ย โƒ— ] + [๐’‚ย โƒ—", " ๐’„ย โƒ—", " ๐’‚ย โƒ— ] + [๐‘ย โƒ—", " ๐‘ย โƒ—", " ๐‘Žย โƒ— ] [๐‘ย โƒ—", " ๐‘ย โƒ—", " ๐‘ย โƒ— ] = [๐‘ย โƒ—", " ๐‘ย โƒ—", " ๐‘ย โƒ— ] = ๐‘ย โƒ—. (๐‘ย โƒ— ร— ๐‘ย โƒ—) As (๐‘ย โƒ— ร— ๐‘ย โƒ—) = 0ย โƒ— = ๐‘ย โƒ— . 0ย โƒ— = 0ย โƒ— Using Prop: [๐‘ย โƒ—", " ๐‘ย โƒ—", " ๐‘ย โƒ— ] = 0 [๐‘Žย โƒ—", " ๐‘ย โƒ—", " ๐‘Žย โƒ— ] = 0 [๐‘ย โƒ—", " ๐‘ย โƒ—", " ๐‘Žย โƒ— ] = 0 [๐‘Žย โƒ—", " ๐‘ย โƒ—", " ๐‘Žย โƒ— ] = 0 = [๐‘Žย โƒ—", " ๐‘ย โƒ—", " ๐‘ย โƒ— ] + 0 + 0 + 0 + 0 + [๐‘ย โƒ—", " ๐‘ย โƒ—", " ๐‘Žย โƒ— ] = [๐‘Žย โƒ—", " ๐‘ย โƒ—", " ๐‘ย โƒ— ] + [๐’ƒย โƒ—", " ๐’„ย โƒ—", " ๐’‚ย โƒ— ] = [๐‘Žย โƒ—", " ๐‘ย โƒ—", " ๐‘ย โƒ— ] + [๐’‚ย โƒ—", " ๐’ƒย โƒ—", " ๐’„ย โƒ— ] = 2[๐‘Žย โƒ—", " ๐‘ย โƒ—", " ๐‘ย โƒ— ] = R.H.S Hence proved

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