InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Statement 1: Let `vecr` be any vector in space. Then, `vecr=(vecr.hati)hati+(vecr.hatj)hatj+(vecr.hatk)hatk` Statement 2: If `veca, vecb, vecc` are three non-coplanar vectors and `vecr` is any vector in space then `vecr={([(vecr, vecb, vecc)])/([(veca, vecb, vecc)])}veca+{([(vecr, vecc, veca)])/([(veca, vecb, vecc)])}vecb+{([(vecr, veca, vecb)])/([(veca, vecb, vecc)])}vecc`A. `1`B. `2`C. `3`D. `4` |
|
Answer» Correct Answer - A Since `veca,vecb, vecc` are non coplanar vectors. Therefore, there exists scalars x,y,z such that `vecr=xveca+yvecb+zvecc`……………..i Taking dot products with `vecbxxvecc, veccxxveca` and `vecaxxvecb` successively, we get `vecr.(vecbxxvecc)=(xveca+yvecb+zvecc).(vecbxxvecc)` `vecr.(veccxxveca)=(xvecaxxyvecb+zvecc).(veccxxveca)` `vecr.(vecaxxvecb)=(xveca+yvecb+zvecc).(vecaxxvecb)` `implies[(vecr, vecb, vecc)]=x[(veca, vecb, vecc)]` `=[(vecr, vecc, veca)]=y[(vecb, vecc, veca)]` and `[(vecr, veca, vecb)]=z[(vecc, veca, vecb)]` `impliesx=([(vecr, vecb, vecc)])/([(veca, vecb, vecc)]),y=([(vecy, vecc, veca)])/([(veca, vecb, vecc)])` and `z=([(vecr, veca, vecb)])/([(veca, vecb, vecc)])` Substituting the alues of `x,y,z` in (i) we get `vecr={([(vecr, vecb, vecc)])/([(veca, vecb, vec)])}veca+{([(vecr, vecc, veca)])/([(veca, vecb, vecc)])}vecb+{([(vecr, veca, vecb)])/([(veca,vecb, vecc)])}vecc` So, statement 2 is true. On replacing `veca, vecb,` and `vecc` by `hati, hatj` and `hatk` respectively in statement -2 we obtain statement-1. So, statement 1 is true and statement -2 is a correct explanation for statement -1 |
|