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given that veca. vecb = veca.vecc, veca ...

given that `veca. vecb = veca.vecc, veca xx vecb= veca xx vecc and veca ` is not a zero vector. Show that `vecb=vecc`.

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we have er`veca. Vecb = veca .vecc.` therefore,
`veca.vecb-veca .vecc = 0 or veca. (vecb -vecc) = 0 `
Therefore, there are three possibilities : (i) ,
(ii) `vecb - vecc = vec0 and (iii) veca` is perpendicu
`vecb - vecc`
Again, `veca xx vecb = veca xx vecc`, therefore, `veca xx vecb - veca xx vecc = vec0`
`or veca xx ( vecb - vecc) = vec0`
Therefore, again there are three posibilities,
`(i) veca= vec0, (ii) vecb - vecc = vec0 and (iii) veca` is parallel to `vecb - vecc`.
now ` veca` is given to be a non-zero vector. therefore, we have the following possibilities left :
`1. vecb -vecc= vec0`
2. `veca` is -perendicular to `vecb - vecc and veca` is parallel to `vecb - vecc`, which is absurd.
Therefore, the only possibility , left is `vecb -vecc = vec0 or vecb = vecc`
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