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If axx(bxxc) is perpendicular to (axxb)x...

If `axx(bxxc)` is perpendicular to `(axxb)xxc`, we may have

A

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

B

`veca.vecb=0`

C

`veca.vecc=0`

D

`vecb.vecc=0`

Text Solution

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The correct Answer is:
A, C
`vecaxx(vecbxxvecc)=(veca.vecc)vecb-(veca.vecb)vecc` and
`(vecaxxvecb)xxvecc=-(vecc.vecb)veca+(veca.vecc)vecb`
We have been given
`(vecaxx(vecbxxvecc)).((vecaxxvecb)xxvecc)=0`
`because((veca.vecc)vecb-(veca.vecb)(vecc).((veca.vecc).(veca.vecc)vecb-(vecc.vecb)veca)=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)`
`veca.vecc=0` or `(veca.vec)|vecb|^(2)=(veca.vecb)(vecb.vecc)`
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