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veca , vecb and vecc are three non-copla...

`veca , vecb and vecc` are three non-coplanar vectors and `vecr`. Is any arbitrary vector. Prove that `[vecbvecc vecr]veca+[vecc veca vecr]vecb+[vecavecbvecr]vecc=[veca vecb vecc]vecr`.

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Since a vector can be expressend as a linear combination of three non-coplanar vectors, let
`vecr=xveca+yvecb+zvecc`
where, x,y and z are scalars.
Mutiplying both sides, of (i) scalarly by `veca'` we get
`vecr.veca'=xveca.veca'+yvecb.veca'+zvecc.veca'=x.1=x " " (veca.veca' =1, vecb,veca'=0 =vecc.veca')`
Similarly , multiplying both sides of (i) scalarly by `vecb' and vecc'` successively we get
`y = vecr.vecb' and z = vecr vecc'`
Putting in (i) , we get `vecr=(vecr.veca')veca+(vecr.vecb')vecb+(vecr.vecc')vecc`
ii. Since `veca', vecb' and vecc'` are three non- coplanar vectors,we can take `vecr=xveca'+yvecb' +zvecc'`
Multiplying both sides of (ii) scalarly by `veca` ,we get
`vecr.veca=x(veca'.veca)+y(vecb'.veca)+z(vecc'.veca)=x " " (veca'veca=1 ,vecb'veca=0=vecc'veca)`
Similarly multiplying both sides of (i) scalarly by `vecb and vecc` successively, we get
`y = vecr. vecb and z=vecr .vecc`
Putting in (ii), we get `vecr=(vecr.veca)veca'+(vecr.vecb)vecb'+(vecr.vecc)vecc'`
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