If ` veca, vecb, vecc and vecd ` ar distinct vectors such that `veca xx vecc = vecb xx vecd and veca xx vecb = vecc xx vecd`. Prove that `(vecaxxvecd).(vecb.vecc)ne 0, i.e., veca.vecb + vecd.vecc nevecd.vecb + veca.vecc.`

Explanation

Given that `veca xx vecc = vecb xx vecd`
and `veca xx vecb = vecc xx vecd`
subtracting (ii) from (i), we get
`vecaxx (vecc-vecb)=(vecb-vecc)xxvecd`
`vecaxx(vecc-vecb)=vecd xx(vecc-vecb)`
`vecaxx(vecc-vecb)-vecd xx (vecc-vecd)=0`
`or (veca -vecd) xx (vecc-vecd)=0`
` or (veca-vecd)xx (vecc-vecd)=0`
`or (veca -vecd)||vecc-vecd) (because veca -vecd ne0, vecc-vecb ne 0)`
Hence the angle between ` veca -vecd d and vecc-vecb` is either 0 is `180^(@)`
`Rightarrow (veca-vecd). (vecc-vecb)=|veca-vecd||vecc-vecb|cos0 ne 0`
as `veca, vecb , vecc and vecd` all are different.