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If `vecr=x(vecaxxvecb)+y(vecbxxvecc)+z(vecc+veca)` Such that `x+y+z!=0` and `vecr.(veca+vecb+vecc)=x+y+z`, then `[veca vecb vecc]=` |
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Answer» Correct Answer - B We have `vecr=x(vecaxxvecb)+y(vecbxxvecc)+z(vecc xxveca)` `:.vecr.(veca+vecb+vecc)=x+y+z` `implies{x(vecaxxvecb)+y(vecbxxvecc)+z(veccxxveca)}.(veca+vecb+vecc)=x+y+z` `impliesy[vecb vecc veca]+z[vecc veca vecb]+x[veca vecb vecc]=x+y+z` `implies (x+y+z)[veca vecb vecc]=x+y+z` `implies [veca vecb vecc]=1` |
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