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If vec axx( vec bxx vec c) is perpendic...

If ` vec axx( vec bxx vec c)` is perpendicular to `( vec axx vec b)xx vec c ,` we may have a.`( vec a . vec c)| vec b|^2=( vec a . vec b)( vec b . vec c)` b. ` vec adot vec b=0` c. ` vec adot vec c=0` d. ` vec bdot vec c=0`

A

`(veca.vecb)|vecb|^(2)= (veca.vecb)(vecb.vecc)`

B

`veca.vecb=0`

C

`veca.vecc=0`

D

`vecb.vecc=0`

Text Solution

Verified by Experts

The correct Answer is:
a,c
`vecaxx(vecbxxvecc)= (veca.vecc)vecb-(veca.vecb)vecb]and (vecaxx vecb) xxvecc=-(vecc. vecb) veca + (veca.vecc)vecb`
we have been givn
`(veca xx (vecbxx vecc)). ((vecaxxvecb)xxvecc)=0`
` or (veca . vecc)^(2)|vecb|^(2)- (veca.vecc)(vecb.vecc) (veca.vecb)`
` - (veca.vecb) (veca.vecc)(vecb.vecc)+ (veca.vecb)(vecb.vecc)(vecc.veca)=0`
`or (veca.vecc)^(2)|vecb|^(2)= (veca.vecc)(veca.vecb)(vecb.vecc)`
`or (veca.vecc)((veca.vecc)(vecb.vecb)- (veca.vecb) (vecb.vecc))=0`
`veca.veca=0 or (veca.vecc) |vecb|^(2) = (veca.vecb)(vecb.vecc)`
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