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If bar(a),bar(b)andbar(c) are any three ...

If `bar(a),bar(b)andbar(c)` are any three vectors, prove that (1) `[bar(a)+bar(b) bar(b)+bar(c) bar(c)+bar(a)]=2[bar(a)bar(b)bar(c)]` (2) `[bar(a) bar(b)+bar(c) bar(a)+bar(b)+bar(c)]=0`

Text Solution

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We use the following results :
(i) If any vector in a scalar triple product `[bar(a)bar(b)bar(c)]` is repeated, then `[bar(a)bar(b)bar(c)]=0`,
(ii) `bar(a)*(bar(b)xxbar(c))=bar(b)*(bar(c)xxbar(a))=bar(c)*(bar(a)xxbar(b))`
(1) `[bar(a)+bar(b)" "bar(b)+bar(c)" "bar(c)+bar(a)]`
`=(bar(a)+bar(b))*[(bar(b)+bar(c))xx(bar(c)+bar(a))]`
`=(bar(a)+bar(b))*[bar(b)xxbar(c)+bar(b)xxbar(a)+bar(c)xxbar(c)+bar(c)xxbar(a)]`
`=bar(a)*(bar(b)xxbar(c))+bar(a)*(bar(b)xxbar(a))+bar(a)*(bar(c)xxbar(c))+bar(a)*(bar(c)xxbar(a))+bar(b)*(bar(b)xxbar(c))+bar(b)*(bar(b)xxbar(a))+bar(b)*(bar(c)xxbar(c))+bar(b)*(bar(c)xxbar(a))`
`=bar(a)*(bar(b)xxbar(c))+0+0+0+0+0+0+bar(a)*(bar(b)xxbar(c))`
`=[bar(a)bar(b)bar(c)]+[bar(a)bar(b)bar(c)]=2[bar(a)bar(b)bar(c)]`.
(2) `[bar(a)" "bar(b)+bar(c)" "bar(a)+bar(b)+bar(c)]`
`=bar(a)*[(bar(b)+bar(c))xx(bar(a)+bar(b)+bar(c))]`
`=bar(a)*(bar(b)xxbar(a)+bar(b)xxbar(b)+bar(b)xxbar(c)+bar(c)xxbar(a)+bar(c)xxbar(b)+bar(c)xxbar(c))`
`=bar(a)*(bar(b)xxbar(a))+bar(a)*(bar(b)xxbar(b))+bar(a)*(bar(b)xxbar(c))+bar(a)*(bar(c)xxbar(a))+bar(a)*(bar(c)xxbar(b))+bar(a)*(bar(c)xxbar(c))`
`=0+0+bar(a)*(bar(b)xxbar(c))+0-bar(a)*(bar(b)xxbar(c))+0=0`.
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