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For any three vectors `veca, vecb, vecc` the value of `[(veca-vecb, vecb-vecc, vecc-veca)]`, isA. `0`B. `[(veca, vecb, vecc)]`C. `-[(veca, vecb, vecc)]`D. `-2[(veca, vecb, vecc)]` |
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Answer» Correct Answer - A Let `vec(alpha)=veca-vecb, vec(beta)=vecb-vecc` and `vec(gamma)=vecc-veca` or `vec(alpha)=veca-vecb+0vecc, vec(beta)=0veca+vecb-vecc` and `vec(gamma)=-veca+0vecb+vecc` Then `[(vec(alpha), vec(beta), vec(gamma))]=|(1,-1,0),(0,1,-1),(-1,0,1)|[(veca, vecb, vecc)]` `implies[(vec(alpha),vec(beta),vec(gamma))]=0[(veca, vecb, vecc)]=0` Alter 1 We have `vec(alpha)+vec(beta)+vec(gamma)=veca-vecb+vecb-vecc+vecc-veca=vec0` `impliesvec(alpha),vec(beta),vec(gamma)` are coplanar. `implies[(vec(alpha), vec(beta), vec(gamma))=0` or `[(veca-vecb,vecb-vecc, vecc-veca)]=0` |
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