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vertors vecx , vecy and vecz each of ma...

vertors `vecx , vecy and vecz ` each of magnitude `sqrt2` , make an angle of `60^(@)` with each other . `vecx xx ( vecy xx vecz) =veca , vecy xx (vecz xx vecx) = vecb and vecx xx vecy =vecc `
Vector `vecx` is

A

`1/2[(veca-vecc) xx vecc-vecb+veca]`

B

`1/2[(veca-vecb) xx vecc+vecb-veca]`

C

`1/2[veccxx(veca-vecb) + vecb +veca]`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
b
Given that `|vecx|= |vecy|=|vecz|=sqrt2` and they are inclined at an angle of `60^(@)` with each other.
`vecx.vecy=vecy.vecz=vecz.vecx=sqrt2.sqrt2cos 60^(@)=1 vecx xx (vecyxxvecz)=veca`
`or (vecx.vecz)vecy-(vecx.vecy)vecz=vecaor vecy-vecz=veca` (i)
similarly `vecyxx(vecz xxvecx)=vecb Rightarrow vecz-vecx=vecb`
`vecy=veca+vecz,vecx=vecz-vecb`
Now , ` vecx, xx vecy=vecc`
` Rightarrow (vecz - vecb) xx (vecz + veca) = vecc`
` or vecz xx (veca xx vecb) = vecc + (vecb xxx veca)`
` or (veca + vecb) xx {vecz xx (veca + vecb)} `
`= (veca xx vecb) xx vecc+ (veca +vecb) xx (vecbxxveca)`
`or (veca + vecb) ^(2)vecz - {(veca + vecb).vecz} (veca + vecb)`
`= (veca + vecb) xx vecc + |veca|^(2)vecb-|vecb|^(2)veca`
`+ (veca.vecb) (vecb.veca)`
`Now , (i) Rightarrow |veca|^(2)= |vecy-vecz|^(2)=2 +2-2=2`
similarly , (ii) `Rightarrow |vecb|^(2)=2`
Also (i) and (ii) `Rightarrow veca+vecb=vecy-vecx`
`Rightarrow |veca+vecb|^(2)=2`
`Also (veca +vecb).vecz= (vecy -vecx).vecz = vecy.vecz-vecx.vecz`
1-1=0
`and veca.vecb= (vecy.vecz). (vecz-vecx)`
` =vecy.vecz-vecx.vecy-|vecz|^(2)+vecx.vecz= -1`
Thus from (v) , we have
`2vecz=(veca+vecb)xxvecc+2(vecb-veca)-(vecb-veca)`
`or vecz= (1//2)[(veca + vecb) xx vecc + vecb-veca]`
`vecy= veca+vecz= (1//2)[(veca+vecb)xxvecc+vecb+veca]`
`and vecx=vecz-vecb=(1//2)[(veca+vecb)xxvecc-(veca+vecb)]`
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