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Statement 1: Let `veca, vecb, vecc` be three coterminous edges of a parallelopiped of volume 2 cubic units and `vecr` is any vector in space then `|(vecr.veca)(vecbxxvecc)+(vecr.vecb)(veccxxveca)+(vecc.vecc)(vecaxxvecb|=2|vecr|` Statement 2: Any vector in space can be written as a linear combination of three non-coplanar vectors.A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - A Clearly, statement -2 is true We have `|[(veca, vecb, vecc)]|=2` `:.[(vecaxxvecb, vecbxxvecc, veccxxveca)]=[(veca,vecb, vecc)]^(2)=4!=0` `implies vecaxxvecb, vecbxxvecc, veccxxveca` are non coplanar. Using statement -2 we have `vecr=x(vecaxxvecb)+y(vecbxxvecc)+z(veccxxveca)`...............i Taking dot products with `veca,vecb` and `vecc` respectively, we get `vecr.veca=y[(veca, vecb, vecc)],vecr.vecb=z[(veca, vecb, vecc)]` and `vecr.vecc=x[(veca,vecb,vecc)]` Substituting the values of `x,y` in i we get `vecr[(veca, vecb,vecc)]=(vecr.veca)(vecbxxvecc)+(vecr.vecb)(veccxxveca)+(vecr.vecc)(vecaxxvecb)` `implies|(vecr.veca)(vecbxxvecc)+(vecr.vecb)(veccxxveca)+(vecr.veca)(vecaxxvecb)|` `=|vecr||(veca, vecb,vecc)|` `implies|(vecr.veca)(vecbxxvecc)+(vecr.vecb)(veccxxveca)+(vecr.vecc)(vecaxxvecb)|=2|vecr|` |
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