If `vec(alpha)=x(vecaxxvecb)+y(vecbxxvecc)+z(veccxxveca)` and `[veca vecb vecc]=1/8`, then `x+y+z=`A. `8vec(alpha).(veca+vecb+vecc)`B. `vec(alpha).(veca+vecb+vecc)`C. `8(veca+vecb+vecc)`D. None of these

If `vec(alpha)=x(vecaxxvecb)+y(vecbxxvecc)+z(veccxxveca)` and `[veca v

1.	If `vec(alpha)=x(vecaxxvecb)+y(vecbxxvecc)+z(veccxxveca)` and `[veca vecb vecc]=1/8`, then `x+y+z=`A. `8vec(alpha).(veca+vecb+vecc)`B. `vec(alpha).(veca+vecb+vecc)`C. `8(veca+vecb+vecc)`D. None of these
Answer» Correct Answer - A We have `vec(alpha)=x(vecaxxvecb)+y(vecbxxvecc)+z(veccxxveca)` Taking dot products with `veca, vecb` and `vecc` repectively, we get `vec(alpha).veca=y[veca vecb vecc]impliesy=8(vec(alpha).veca)` `vec(alpha).vecb=z[veca vecb vecc]impliesz=8(vec(alpha).vecb)` and `vec(alpha).vecc=x[veca vecb vecc]=x=8(vec(alpha).vecc)` `:.x+y+z=8vec(alpha).(veca+vecb+vecc)`

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