If `veca, vecb, vecc` are three non-coplanar vectors, then a vector `vecr` satisfying `vecr.veca=vecr.vecb=vecr.vecc=1`, isA. `vecaxxvecb+vecbxxvecc+veccxxveca`B. `1/([(veca, vecb, vecc)]){vecaxxvecb+vecbxxvec+veccxxveca}`C. `[(veca, vecb, vecc)]{vecaxxvecb+vecbxxvecc+vecxxveca}`D. none of these

If `veca, vecb, vecc` are three non-coplanar vectors, then a vector `v

1.	If `veca, vecb, vecc` are three non-coplanar vectors, then a vector `vecr` satisfying `vecr.veca=vecr.vecb=vecr.vecc=1`, isA. `vecaxxvecb+vecbxxvecc+veccxxveca`B. `1/([(veca, vecb, vecc)]){vecaxxvecb+vecbxxvec+veccxxveca}`C. `[(veca, vecb, vecc)]{vecaxxvecb+vecbxxvecc+vecxxveca}`D. none of these
Answer» Correct Answer - B Since `veca, vecb, vecc` are three non-coplanar vectors. Therefore `vecaxxvecb, vecbxxvecc` and `veccxxveca` are also non coplanar vectors. Let `vecr=x(vecaxxvecb)+y(vecbxxvecc)+z(veccxxveca)`. Then, `vecr.veca=1implies1=y{((vecbxxvecc).veca)}impliesy=1/([(veca, vecb, vecc)])` Similarly `vecr.vecb=1` and `vecr.vecc=1` will give `z=1/([(veca, vecb, vecc)])` and `x=1/([(veca, vecb, vecc)])` Hence `vecr=1/([(veca, vecb, vecc)]){vecaxxvecb+vecbxxvecc+veccxxveca}` is the required solution.

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The points with position vectors `alpha hati+hatj+hatk, hati-hatj-hatk, hati+2hatj-hatk, hati+hatj+betahatk` are coplanar ifA. `(1-alpha)(1+beta)=0`B. `(1-alpha)(1-beta)=0`C. `(1+alpha)(1+beta)=0`D. `(1+alpha)(1-beta)=0`
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If `veca, vecb, vecc` are three non-coplanar vectors, then a vector `vecr` satisfying `vecr.veca=vecr.vecb=vecr.vecc=1`, isA. `vecaxxvecb+vecbxxvecc+veccxxveca`B. `1/([(veca, vecb, vecc)]){vecaxxvecb+vecbxxvec+veccxxveca}`C. `[(veca, vecb, vecc)]{vecaxxvecb+vecbxxvecc+vecxxveca}`D. none of these

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