InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
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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Answer» Correct Answer - A Let `P,Q,R` and `S` be the given points with position vectors `alpha hati+hatj+hatk, hati-hatj-hatk, hati+2hatj-hatk` and `hati+hatj+betahatk` respectively. Then, `vec(QP)=(alpha-1)hati+2hatj+2hatk, vec(QR)=0hati+3hatj+0hatk` and `vec(QS)=0hati+2hatj+(beta+1)hatk` are coplanar. `:.[vec(QP),vec(QR),vec(QS)]=0` `implies|(alpha-1,2,2),(0,3,0),(0,2,beta+1)|=0` `implies (alpha-1)(beta+1)=0implies(1-alpha)(1+beta)=0` |
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| 2. |
The value of `veca.(vecb+vecc)xx(veca+vecb+vecc)`, isA. `2[veca vecb vecc]`B. `[veca vecb vecc]`C. 0D. None of these |
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Answer» Correct Answer - C We have `veca+vecb+vecc=veca+(vecb+vecc)` `implies veca, vecb+vecc` and `veca+vecb+vecc` are coplanar `impliesveca.(vecb+vecc)xx(veca+vecb+vecc)=0` i.e. `[(veca, vecb+c, veca+vecb+vecc)]=0` |
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| 3. |
Let `veca=2hati+3hatj-hatk` and `vecb=hati-2hatj+3hatk`. Then , the value of `lamda` for which the vector `vecc=lamdahati+hatj+(2lamda-1)hatk` is parallel to the plane containing `veca` and `vecb`. IsA. `1`B. `0`C. `-1`D. `2` |
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Answer» Correct Answer - B We are given that `vecc` is parallel to the plane containing `veca` and `vecb` `impliesvecc_|_(vecaxxvecb)impliesvecc.(vecaxxvecb)=0implies[(veca, vecb, vecc)]=0` `implies |(2,3,-1),(1,-2,3),(lamda, 1, 2lamda-1)|=0implieslamda=0` |
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| 4. |
If `veca=hati-2hatj+3hatk, vecb=2hati+3hatj-hatk` and `vecc=rhati+hatj+(2r-1)hatk` are three vectors such that `vecc` is parallel to the plane of `veca` and `vecb` then`r` is equal to,A. `1`B. `0`C. `2`D. `-1` |
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Answer» Correct Answer - B It is given `vecc` is parallel to the plane of `veca` and `vecb` `:.vecc_|_vecaxxvecb` `impliesvecc(vecaxxvecb)=0implies[(veca, vecb, vecc)]=0` `implies|(1,-2,3),(2,3,-1),(4,1,2r-1)|=0` `implies(6r-2)+2(5r-2)+(6-9r)=0implies7r=0impliesr=0`. |
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| 5. |
Let `vecr, veca, vecb` and `vecc` be four non-zero vectors such that `vecr.veca=0, |vecrxxvecb|=|vecr||vecb|,|vecrxxvecc|=|vecr||vecc|` then `[(veca, vecb, vecc)]=`A. `-1`B. `0`C. `1`D. `2` |
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Answer» Correct Answer - B We have `{:(vecr.veca=0impliesvecr_|_veca),(|vecrxxvecb|=|vecr||vecb|impliesvecr_|_vecb),("and"|vecrxxvecr|=|vecr||vecc|impliesvecr_|_vecc):}}impliesveca, vecb vecc` are coplanar. Hence `[(veca, vecb, vecc)]=0` |
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| 6. |
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` |
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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 |
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| 7. |
A solution of the vector equation `vecrxxvecb=vecaxxvecb`, where `veca, vecb` are two given vectors is where `lamda` is a parameter.A. `vecr=lamdavecb`B. `vecr=veca+lamdavecb`C. `vecr=vecb+lamdaveca`D. `vecr=lamdaveca` |
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Answer» Correct Answer - B We have `vecrxxvecb=vecaxxvecb` `impliesvecrxxvecb-vecaxxvecb=vec0` `impliesvecrdxvecb-vecaxxvecb=vec0` `implies (vecr-veca)xxvecb=vec0` `impliesvecr-veca` is aparallel to `vecb`. `impliesvecr-veca=lamdavecb` for some scalar `lamda` `impliesvecr=veca+lamdavecb` is the required solution. |
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| 8. |
If `veca, vecb, vecc` are three non-zero non-null vectors are `vecr` is any vector in space then `[(vecb, vecc, vecr)]veca+[(vecc, veca, vecr)]vecb+[(veca, vecb, vecr)]vecc` is equal toA. `2[(veca, vecb, vecc)]vecr`B. `3[(veca, vecb, vecc)]vecr`C. `[(veca, vecb, vecc)]`D. None of these |
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Answer» Correct Answer - C We have `vecr=xveca+yvecb+zvecc`………………i Taking product successively with `vecbxxvecc, veccxxveca` and `vecaxxvecb` we obtain `x=([(vecb, vecc, vecr)])/([(veca, vecb, vecc)]),y=([(vecc, veca, vecr)])/([(veca, vecb, vecc)]),z=([(veca, vecb, vecr)])/([(veca, vecb, vecc)])` Substituting the values of `x,y,z` in (i) we get `[(vecb, vecc, vecr)]veca+[(vecc, veca, vecr)]vecb+[(veca, vecb, vecr)]=vecc=[(veca, vecb, vecc)]vecr` |
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| 9. |
Which of the following expressions are meaningful?A. `vecu.(vecvxxvecw)`B. `(vecu.vecv).vecw`C. `(vecu.vecv)vecw`D. `vecuxx(vecv.vecw)` |
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Answer» Correct Answer - A::C Since `vecvxxvecw` is a vector. Therefore `vecu.(vecvxxvecw)` is a scalar quantity. As `(vecu.vecv)` is a scalar. Therefore `(vecu.vecv).vecw` is not meaningful. But`(vecu.vecv)vecw` is a scalar multiple of `vecw`. So it is meaningful. Since `vecv.vecw` is a scalar. So `vecuxx(vecvxxvecw)` is not meaningful. |
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| 10. |
The number of distinct values of `lamda`, for which the vectors `-lamda^(2)hati+hatj+hatk, hati-lamda^(2)hatj+hatk` and `hati+hatj-lamda^(2)hatk` are coplanar, is |
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Answer» Correct Answer - C Given vectors will be coplanar if `|(-lamda^(2),1,1),(1,-lamda^(2),1),(1,1,-lamda^(2))|=0` `implieslamda^(6)-3lamda^(2)-2=0implies(1+lamda^(2))^(2)(lamda^(2)-2)=0implieslamda=+-sqrt(2)` |
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| 11. |
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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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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| 12. |
If `veca, vecb` and `vecc` are unit coplanar vectors, then`[(2veca-3vecb,7vecb-9vecc,10vecc-23veca)]` |
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Answer» Correct Answer - A Since `veca,vecb,vecc`are coplanar vectors. Therefore, `2veca-3vecb,7vecb-9vecc` and `12vecc-23veca` are also coplanar vectors. Hence `[(2veca-3vecb, 7vecb-9vecc,12vecc-23veca)]=0` |
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| 13. |
Statement 1: Any vector in space can be uniquely written as the linear combination of three non-coplanar vectors. Stetement 2: If `veca, vecb, vecc` are three non-coplanar vectors and `vecr` is any vector in space then `[(veca,vecb, vecc)]vecc+[(vecb, vecc, vecr)]veca+[(vecc, veca, vecr)]vecb=[(veca, vecb, vecc)]vecr`A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B Clearly statemnet -1 is true We have `vecr=xveca+yvecb+zvecc` ……………..i Taking product successively with `vecbxxvecc, veccxxveca` and `vecaxxvecb`, we obtain `x=([(vecb, vecc,vecr)])/([(veca, vecb, vecc)]),y=([(vecc, veca, vecr)])/([(veca, vecb, vecc)]),z=([(veca, vecb, vecr)])/([(veca, vecb, vecc)])` Substituting the values of `x,y,z` in (i) we get `[(vecb, vecc, vecr)]veca+[(vecc, veca, vecr)]vecb+[(veca, vecb, vecr)]vecr=[(veca, vecb, vecc)]vecr` So Statement 2 is true. But, statement2 is not a correct explanation for statement -1 |
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| 14. |
For any three vectors `veca, vecb, vecc` the value of `[(veca+vecb,vecb+vecc,vecc+veca)]` isA. `0`B. `2[(veca, vecb, vecc)]`C. `[(veca, vecb, vecc)]`D. `-[(veca, vecb, vecc)]` |
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Answer» Correct Answer - B Let `vec(alpha)=veca+vecb, vecbeta=vecb+vecc`, and `vec(gamma)=vecc+veca`. Then `vec(alpha)=veca+vecb+0vecc,vec(beta)=0veca+vecb+vecc` and `vec(gamma)=veca+0vecb+vecc` `[(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))]=2[(veca, vecb, vecc)]` |
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| 15. |
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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| 16. |
Let `veca,vecb` and `vecc` be three vectors. Then scalar triple product `[veca vecb vecc]` is equal toA. `[(vecb, veca, vecc)]`B. `[veca vecc vecb]`C. `[vecc vecb veca]`D. `[vecb vecc veca]` |
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Answer» Correct Answer - D We have `[veca vecb vecc]=[vecb vecc veca]-[vecc veca vecb]` So option d is correct. |
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| 17. |
If `vecu, vecv, vecw` are three non-coplanar vectors, the `(vecu+vecv-vecw).(vecu-vecv)xx(vecv-vecw)` equalsA. `vecu.(vecvxxvecw)`B. `vecu.(vecwxxvecv)`C. `3vecu.(veccxxvecw)`D. 0 |
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Answer» Correct Answer - A We have `(vecu+vecv-vecw).(vecu-vecv)xx(vecv-vecw)` `=[(vecu+vecv-vecw,vecu-vecv,vecv-vecw)]` `=[(vecu+vecv-vecw,vecu-vecv+0vecw,0vecu+vecv-vecw)]` `=|(1,1,-1),(1,-1,0),(0,1,-1)|[(vecu, vecv, vecw)]` `=[(vecu, vecv, vecw)]=vecu.(vecvxxvecw)` |
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| 18. |
If `[veca vecb vecc]=1` then value of `(veca.vecbxxvecc)/(veccxxveca.vecb)+(vecb.veccxxveca)/(vecaxxvecb.vecc)+(vecc.vecaxxvecb)/(vecbxxvecc.veca)` isA. `3`B. `1`C. `-1`D. None of these |
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Answer» Correct Answer - A `(veca.(vecbxxvecc))/((veccxxveca).vecb)+(vecb(veccxxveca))/((vecaxxvecb).vecc)+(vecc.(vecaxxvecb))/((vecbxxvecc).veca)` `=([veca vecb vecc])/([vecc veca vecb])+([vecc veca vecb])/[(vecb vecc veca])=1/1+1/1+1/1=3` |
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| 19. |
If `vecu, vecv, vecw` are three vectors such that `[vecu vecv vecw]=1`, then `3[vecu vecv vecw]-[vecv vecw vecu]-2[vecw vecv vecu]=` |
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Answer» Correct Answer - D `3[vecu vecv vecw]-[vecv vecw vecu]-2[vecw vecv vecu]` `=3[vecu vecv vecw]-[vecu vecv vecw]+2[vecw vecu vecv]` `=3[vecu vecv vecw]-[vecu vecv vecw]+2[vecu vecv vecw]` `=4[vecu vecv vecw]=4xx1=4` |
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| 20. |
If `veca, vecb, vecc` are non coplanar non null vectors such that `[(veca, vecb, vecc)]=2` then `{[(vecaxxvecb, vecbxxvecc, veccxxveca)]}^(2)=`A. 4B. 16C. 8D. none of these |
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Answer» Correct Answer - B We have `{[(vecaxxvecb, vecbxxvecc, veccxxveca)]}={[(veca, vecb, vecc)]}^(4)=2^(4)=16` |
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| 21. |
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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| 22. |
If `vec(alpha)=x(vecaxxvecb)+y(vecbxxvecc)+z(veccxxveca)` and `[veca vecb vecc]=1/8`, then `x+y+z=`A. `8vec(alpha).(veca+vecb+vecc)`B. `vec(alpha).(veca+vecb+vecc)`C. `8(veca+vecb+vecc)`D. None of these |
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Answer» Correct Answer - A We have `vec(alpha)=x(vecaxxvecb)+y(vecbxxvecc)+z(veccxxveca)` Taking dot products with `veca, vecb` and `vecc` repectively, we get `vec(alpha).veca=y[veca vecb vecc]impliesy=8(vec(alpha).veca)` `vec(alpha).vecb=z[veca vecb vecc]impliesz=8(vec(alpha).vecb)` and `vec(alpha).vecc=x[veca vecb vecc]=x=8(vec(alpha).vecc)` `:.x+y+z=8vec(alpha).(veca+vecb+vecc)` |
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| 23. |
Statement 1: For any three vectors `veca,vecb,vecc` `[(vecaxxvecb,vecbxxvecc,veccxxveca)]=0` Statement 2: If `vecp,vecq,vecr` are linear dependent vectors then they are coplanar.A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - D If `vecp,vecq,vecr` are linearly independent vectors, then there exist scalars `x,y,z` not all zero such that `xvecp+yvecq+zvecr=vec0` `impliesvecp=(-y/x)vecq+((-z)/x)vecr` `impliesvecp,vecq,vecr` are coplanar. So, statement 2 is true. We know that `[(vecpxxvecb,vecbxxvecc,veccxxveca)]=[(veca,vecb,vecc)]!=0` unless `veca,vecb,vecc` are coplanar. So, statement -2 is not true. |
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| 24. |
The three vectors `hati+hatj, hatj+hatk, hatk+hati` taken two at a time form three planes. The three unit vectors drawn perpendicular to these three planes form a parallelopiped of volume.A. `1/3`B. `4`C. `(3sqrt(3))/4`D. `4/(3sqrt(3))` |
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Answer» Correct Answer - B Let `veca=hati+hatj, vecb=hatj+hatk, vecc=hatk+hati` Let `vecn_(1),vecn_(2),vecn_(3)` be the normals to the given planes. Then, `vecn_(1)=vecaxxvecb, vecn_(2)=vecbxxvecc` and `vecn_(3)=veccxxveca` `:.` Volume of the given parallelopiped is given by `[(vecn_(1), vecn_(2), vecn_(3))]=[(veca, vecb, vecc)]^(2)=|(1, 1, 0),(0, 1,1),(1,0, 1)|^(2)=4` |
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| 25. |
Let `veca, vecb, vecc` be three non-zero non coplanar vectors and `vecp, vecq` and `vecr` be three vectors given by `vecp=veca+vecb-2vecc, vecq=3veca-2vecb+vecc` and `vecr=veca-4vcb+2vecc` If the volume of the parallelopiped determined by `veca, vecb` and `vecc` is `V_(1)` and that of the parallelopiped determined by `veca, vecq` and `vecr` is `V_(2)`, then `V_(2):V_(1)=`A. `3:1`B. `7:1`C. `11:1`D. `15:1` |
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Answer» Correct Answer - D We have `V_(1)=[(veca, vecb, vecc)]` and `V_(2)=[(vecp, vecq, vecr)]` Now, `V_(2)=[(vecp, vecp, vecr)]` `impliesV_(2)=|(1,1,-2),(3,-2,1),(1,-4,2)|[(veca, vecb, vecc)]` `impliesV_(2)=15V_(1)impliesV_(2):V_(1)=15:1` |
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| 26. |
If the vectors `veca` and `vecb` are perpendicular to each other then a vector `vecv` in terms of `veca` and `vecb` satisfying the equations `vecv.veca=0, vecv.vecb=1` and `[(vecv, veca, vecb)]=1` isA. `(vecb)/(|vecb|^(2))+(vecaxxvecb)/(|vecaxxvecb|^(2))`B. `(vecb)/(|vecb|)+(vecaxxvecb)/(|vecaxxvecb|^(2))`C. `(vecb)/(|vecb|^(2))+(vecaxxvecb)/(|vecaxxvecb|)`D. none of these |
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Answer» Correct Answer - A If `veca, vecb, vecc` are three non coplanar vectors, then any vector `vecr` can be expressed as `vecr={(vecr.veca)/(|veca|^(2))}veca+{(vecr.vecb)/(|vecb|^(2))}+vecb+{(vecr.vecc)/(|vecc|^(2))}vecc` Clearly `veca, vecb` and `vecaxxvecb` are three non-coplanar vectors. `:.vecv={(vecv.veca)/(|veca|^(2))}veca+{(vecv.vecb)/(|vecb|^(2))}vecb+{(vecv.(vecaxxvecb))/(|vecaxxvecb|^(2))}(vecaxxvecb)` `implies vecv=0veca+1/(|vecb|^(2)).vecb+1/(|vecaxxvecb|^(2)).(vecaxxvecb)` |
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| 27. |
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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| 28. |
Let `vecx, vecy` and `vecz` be three vectors each of magnitude `sqrt(2)` and the angle between each pair of them is `pi/3`. If `veca` is a non-zero vector perpendicular to `vecx` and `vecyxxvecz` and `vecb` is a non zero vector perpendicular to `vecy` and `veczxxvecx` thenA. `vecb=(vecb.vecz)(vecz-vecx)`B. `veca=(veca.vecy)(vecy-vecz)`C. `veca.vecb=-(veca.vecy)(vecb.vecz)`D. `veca=(veca.vecy)(vecz-vecy)` |
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Answer» Correct Answer - A::B::C We have `|vecx|=|vecy|=|vecz|=sqrt(2)` `vecx.vecy=|vecx||vecy|"cos"(pi)/3=1,vecy.vecz=1` and `vecz.vecx=1` It is given that `veca` is perpendicular to `vecx` and `vecyxxvecz` and `vecb` is perpendicular to `vecy` and `veczxxvecx`. Therefore `veca||vecx xx (vecyxxvecz)` and `vecb||vecyxx(veczxxvecx)` `veca=lamda_(1){vecxxx(vecyxxvecz)}` and `vecb=lamda_(2){vecyxx(veczxx vecx)}` for scalar `lamda_(1)` and `lamda_(2)` `implies veca=lamda_(1){(vecx.vecz)vecy-(vecx.vecy)vecz}` and `vecb=lamda_(2){(vecy.vecx)vecz-(vecy.zvecz)vecx}` `impliesveca=lamda_(1)(vecy-vecz)` and `vecb=lamda_(2)(vecz-vecx)` `impliesveca.vecy=lamda_(1)(vecy.vecy-vecy.vecz)` and `vecb.vecz=lamda_(2)(vecz.vecz-vecx.vecz)` `impliesveca.vecy=lamda_(1)(2-1)` and `vecb.vecz=lamda_(2)(2-1)`............i `implies lamda_(1)=veca.vecy` and `lamda_(2)=vecb.vecz` Sunstituting the values of `lamda_(1)` and `lamda_(2)` in i we get `impliesveca=(veca.vecy)(vecy-vecz)` and `vecb=(vecb.vecz)(vecz-vecx)` Thus option b and c are correct. Now `veca.vecb=(veca.vecy)(vecy-vecz).(vecb.vecz)(vecz-vecx)` `impliesveca.vecb=(veca.vecy)(vecb.vecz){(vecy-vecz).(vecz-vecx)}` `impliesveca.vecb=(veca.vecy(vecb.vecz)){(vecy.vecz-vecy.vecx-vecz.vecz+vecz.vecx)}` `implies veca.vecb=(veca.vecy(vecb.vecz)(1-1-2+1)` `implies veca.vecb=-(veca.vecy)(vecb.vecz)` so option c is correct. |
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| 29. |
Let `G_(1),G_(2),G_(3)` be the centroids of the triangular faces `OBC, OCA, OAB` of a tetrahedron `OABC`. If `V_(1)` denote the volume of the tetrahedron `OABC` and `V_(2)` that of the parallelopiped with `OG_(1),OG_(2),OG_(3)` as three concurrent edges, thenA. `4V_(1)=9V_(2)`B. `9V_(1)=4V_(2)`C. `3V_(1)=2V_(2)`D. `3V_(2)=2V_(1)` |
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Answer» Correct Answer - A Taking `O` as the origin let the position vectors of `A,B` and `C` be `veca,vecb` and `vecc` respectively. Thenthe position vectors of `G_(1),G_(2)` and `G_(3)` are `(vecb+vecc)/3,(vecc+veca)/3` and `(veca+vecb)/3` respectively. `:.V_(1)=1/6[(veca,vecb, vecc)]` and `V_(2)=[(vec(OG_(1)),vec(OG_(2)),vec(OG_(3)))]` `V_(2)=[(vec(OG_(1)),vec(OG_(2)),vec(OG_(3)))]` `implies V_(2)=[((vecb+vecc)/3,(vecc+veca)/3,(veca+vecb)/3)]` `impliesV_(2)=1/27[(vecb+vecc,vecc+veca,veca+vecb)]` `impliesV_(2)=2/27[(veca,vecb,vecc)]impliesV_(2)=2/27xx6V_(1)=9V_(2)=4V_(1)` |
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| 30. |
If the acute angle that the vector `alphahati+betahatj+gammahatk` makes with the plane of the two vectors `2hati+3hatj-hatk` and `hati-hatj+2hatk` is `tan^(-1)1/(sqrt(2))` thenA. `alpha(beta+gamma)=beta gamma`B. `beta(gamma+alpha)=gamma alpha`C. `gamma(alpha+beta)=alpha beta`D. `alpha beta =beta gamma +gamma alpha =0` |
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Answer» Correct Answer - A Let `theta` be the angle between `vecr=alpha hati+beta hatj+gammahatk` and the plane containig the vectors `veca=2hati+hatj-hatk` and `vecb=hati-hatj+2hatk`. Then `(pi//2-theta)` is the angle between `vecr` and `vecaxxvecb`. `:.cos((pi)/2-theta)=(vecr.(vecaxxvecb))/(|vecr||vecaxxvecb|)` `impliessin theta=([(vecr, veca, vecb)])/(|vecr||vecaxxvecb|)` `implies1/(sqrt(3))=([(vecr, veca, vecb)])/(|vecr|(5sqrt(3))) [ :. tan theta=1/(sqrt(2))impliessin theta=1/(sqrt(3), and vecaxxvecb=5hati-5hatj-5hatk]` `implies5|vecr|=[(vecr, veca, vecb)]` `implies5|vecr|=5(alpha-beta-gamma)` `implies|vecr|=alpha-beta-gamma` `impliessqrt(alpha^(2)+beta^(2)+gamma^(2))=alpha-beta-gamma` `impliesbeta gamma=alpha gamma +alpha betaimpliesbeta gamma=alpha (beta+gamma)` |
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| 31. |
The volume of the tetrahedron whose vertices are the points `hati, hati+hatj, hati+hatj+hatk` and `2hati+3hatj+lamdahatk` is `1//6` units, Then the values of `lamda`A. does not existB. is 7C. is -1D. is any real value |
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Answer» Correct Answer - D Let ABCD be the given tetradehron. Then `vec(AB)=hatj,vec(AC)=hatj+hatk` and `vec(AD)=hati+3hatj+lamdahatk` `:.` volume `=1/6` `implies1/6[(vec(AB), vec(AC),vec(AD))]=1/6` `implies[(vec(AB),vec(AC),vec(AD))]=1` `=(vec(AB)xxvec(AC).vec(AD))=1` `implieshati.(hati+K3hatj+lamda hatk)=1`, which is true for all values of `lamda` |
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| 32. |
The volume of the tetrahedron whose vertices are the points with positon vectors `hati-6hatj+10hatk, -hati-3hatj+7hatk, 5hati-hatj+hatk` and `7hati-4hatj+7hatk` is 11 cubic units if the value of `lamda` isA. `-1,7`B. `1,7`C. `-7`D. `-1,-7` |
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Answer» Correct Answer - B Let A,B, C and D be the points with the given position vectors. Then `vec(AB)=-2hati+3hatj-3hatk, vec(AC)=4hati+5hatj+(lamda-10)hatk` and `vec(AD)=6hati+2hatj-3hatk` `:.` Volume `=11` cubic units. `=1/6[(vec(AB), vec(AC), vec(AD))]=+-11` `implies1/6|(-2, 3, -3),(4,5,lamda-10),(6,2,-3)|=+-11` `implies-88+22lamda=+-66implieslamda=1` or `lamda=7` |
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