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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. |
Vectors `vecx,vecy,vecz` each of magnitude `sqrt(2)` make angles of `60^0` with each other. If `vecxxx(vecyxx(veczxxvecx)=vecb nd vecxxxvecy=vecc, find vecx, vecy, vecz` in terms of `veca,vecb and vecc`.A. `1/2[(veca-vecb)xx vecc+ (veca + vecb)]`B. `1/2[(veca+vecb)xx vecc+ (veca - vecb)]`C. `1/2[-(veca+vecb)xx vecc+ (veca + vecb)]`D. `1/2[(veca+vecb)xx vecc- (veca + vecb)]` |
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Answer» Correct Answer - d Given that `|vecx|= |vecy|=|vecz|=sqrt2` and they are inclined at an angle of `60^(@)` with each other. `vecx.vecy=vecy.vecz=vecz.vecx=sqrt2.sqrt2cos 60^(@)=1 vecx xx (vecyxxvecz)=veca` `or (vecx.vecz)vecy-(vecx.vecy)vecz=vecaor vecy-vecz=veca` (i) similarly `vecyxx(vecz xxvecx)=vecb Rightarrow vecz-vecx=vecb` `vecy=veca+vecz,vecx=vecz-vecb` Now , ` vecx, xx vecy=vecc` ` Rightarrow (vecz - vecb) xx (vecz + veca) = vecc` ` or vecz xx (veca xx vecb) = vecc + (vecb xxx veca)` ` or (veca + vecb) xx {vecz xx (veca + vecb)} ` `= (veca xx vecb) xx vecc+ (veca +vecb) xx (vecbxxveca)` `or (veca + vecb) ^(2)vecz - {(veca + vecb).vecz} (veca + vecb)` `= (veca + vecb) xx vecc + |veca|^(2)vecb-|vecb|^(2)veca` `+ (veca.vecb) (vecb.veca)` `Now , (i) Rightarrow |veca|^(2)= |vecy-vecz|^(2)=2 +2-2=2` similarly , (ii) `Rightarrow |vecb|^(2)=2` Also (i) and (ii) `Rightarrow veca+vecb=vecy-vecx` `Rightarrow |veca+vecb|^(2)=2` `Also (veca +vecb).vecz= (vecy -vecx).vecz = vecy.vecz-vecx.vecz` 1-1=0 `and veca.vecb= (vecy.vecz). (vecz-vecx)` ` =vecy.vecz-vecx.vecy-|vecz|^(2)+vecx.vecz= -1` Thus from (v) , we have `2vecz=(veca+vecb)xxvecc+2(vecb-veca)-(vecb-veca)` `or vecz= (1//2)[(veca + vecb) xx vecc + vecb-veca]` `vecy= veca+vecz= (1//2)[(veca+vecb)xxvecc+vecb+veca]` `and vecx=vecz-vecb=(1//2)[(veca+vecb)xxvecc-(veca+vecb)]` |
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| 2. |
If `veca=xhati + y hatj + zhatk, vecb= yhati + zhatj + xhatk and vecc= zhati + xhatj+ yhatk ,, " then " vecaxx (vecbxx vecc) `isA. parallel to ` ( y- z) hati + (z - x) hatj + (x -y) hatk`B. orthogonal to `hati + hatj + hatk`C. orthogonal to ` ( y+z) hati + ( z + x) hatj + ( x + y) hatk`D. orthogonal to `xhati + yhatj + zhatk` |
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Answer» Correct Answer - a,b,c,d `veca xx (vecbxxvecc0 = (veca .vecc) vecb- (veca .vecb)vecc` `= (yz+yx +zx) {(y-z)hati+(z-x)hatj +(x-y)hatk}` Clearly vector is parallel to `(y-z) hati + (z-x) hatj + (x-y) hatk` it is orthogonal to `hati+hatj +hatk as (y-z) (1) + (z-x) (1) + (x-y) (1) =0` it is orhtogonal to `( y+z) hati+ (z+x) hatj + (x +y) hatk` as (y-z) (y+z) + (z+x) + (x-y) (x+y) ` = y^(2) -z^(2) +z^(2)- x^(2) +x^(2) -y^(2) =0` Also it is orthogonal to `xhati + yhatj + zhatk` |
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| 3. |
Let `veca=alphahati+2hatj- 3hatk, vecb=hati+ 2alphahatj - 2hatk and vecc = 2hati - alphahatj + hatk`. Find the value of `6 alpha`. Such that `{(vecaxxvecb)xx(vecbxx vecc)}xx(veccxxveca)=0` |
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Answer» Correct Answer - 4 `veca=alphahati+2hati-3hatk,vecb=hati+2alphahati-2hatk,` `vecc= 2hati - alphahatj + hatk {(veca xx vecb) xx (vecb xx vecc)}xx (vecc xx veca)=vec0` `or {[veca vecb vecc] vecb-[veca vecbvecb]vecc}xx (veccxxveca)=0` `or [veca vecb vecc] vecb xx (vecc xx veca)=vec0` `or [veca vecb vecc] ((veca.vecb)vecc- (vecb.vecc)veca)=vec0` `[ veca vecb vecc] =0 ` ( `veca and vecc` are not collinear) `Rightarrow |{:(alpha,2,-3),(1,2alpha,-2),(2,-alpha,1):}|` `or alpha(2alpha - 2alpha) -2(1+4) -3 (-alpha - 4alpha)=0` `or 10 -15 alpha=0` `alpha 2//3` |
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| 4. |
If `veca,vecb,vecc and vecd` are unit vectors such that `(vecaxxvecb).(veccxxvecd)=1 and veca.vecc=1/2` then (A) `veca,vecb,vecc` are non coplanar (B) `vecb,vecc, vecd` are non coplanar (C) `vecb, vecd` are non paralel (D) `veca, vecd` are paralel and `vecb, vecc` are parallelA. `veca, vecb and vecc` are non- coplanarB. `vecb, vecc and vecd` are non-coplanarC. `vecb and vecd` are non- parallelD. `veca and vecd` are parallel and `vecb and vecc` are parallel |
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Answer» Correct Answer - c ` (veca xx vecb) . (vecc xx vecd) =1 ` is possible only when ` |veca xx vecb|= |vecc xx vecd|=1 and (veca xx vecb) || (vecc xx vecd)` since `veca . Vecc = 1/2 and if vecb||vecd , " then " |vecc xx vecd|ne 1` |
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| 5. |
Let `veca=hati + hatj +hatk,vecb=hati- hatj + hatk and vecc= hati-hatj - hatk` be three vectors. A vectors `vecv` in the plane of `veca and vecb` , whose projection on `vecc is 1/sqrt3` is given byA. `hati-3hatj + 3hatk`B. `-3hati-3hatj +hatk `C. `3hati -hatj + 3hatk`D. `hati+ 3hatj -3hatk` |
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Answer» Correct Answer - c `vecv= lamdaveca + muvecb` `= lambda( hati +hatj +hatk) + mu(hati - hatj +hatk)` Projection of `vecv " on " vecc` `(vecv.vecc)/(|vecc|)= 1/sqrt3` `or ([(lamda+mu)hati+(lamda-mu)hatj+(lamda+mu)hatk].(hati-hatj-hatk))/sqrt3= 1/sqrt3` `or lamda + mu - lamda+ mu- lamda- mu =1` ` or mu - lamda =1` `or lamda = mu -1` `Rightarrow vecv= (mu-1) (hati+hatj+hatk)+mu(hati-hatj+hatk)` ` = (2mu -1) hati -hati -hatj + (2mu -1) hatk` At `mu-2 , vecv = 3hati -hatj +3hatk` |
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| 6. |
Statement 1: Vector `vecc = -5hati + 7 hatj + 2hatk ` is along the bisector of angle between `veca = hati + 2hatj + 2hatk and vecb = 8 hati + hatj - 4hatk`. Statement 2 : `vecc` is equally inclined to `veca and vecb`.A. Both the statements are true and statement 2 is the correct explanation for statement 1.B. Both statements are true but statement 2 is not the correct explanation for statement 1.C. Statement 1 is true and Statement 2 is falseD. Statement 1 is false and Statement 2 is true. |
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Answer» Correct Answer - b A vector along the bisector is ` veca/(|veca|)+vecb/|vecb|= (-5hati+7hatj+2hatk)/9` Hence `vecc = -5hati + 7hatj +2hatk` is along the bisectior. It is obvious that `vecc` makes an equal with `veca and vecb` However, statement 2 does not explain statment 1, as a vector equally inclined to given two vectors is not necessarily coplanar. |
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| 7. |
Which of the following expressionsare meaningful?` vec udot( vec vxx vec w)`b. `( vec udot vec v)dot vec w`c. `( vec udot vec v)dot vec w`d. ` vec uxx( vec vdot vec w)`A. `vecu.(vecvxx vecw)`B. `(vecu.vecv).vecw`C. `(vecu.vecv)vecw`D. `vecu xx (vecv . Vecw)` |
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Answer» Correct Answer - a,c Dot product of two vectors gives a scalar quantity. |
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| 8. |
Let `veca=a_(1)hati+a_(2)hatj+a_(3)hatk, vecb=b_(1)hati+b_(2)hatj+b_(3)hatk` and `vecc=c_(1)hati+c_(2)hatj+c_(3)hatk` be three non zero vectors such that `vecc` is a unit vector perpendicular to both `veca` and `vecb` . If the angle between `veca` and `vecb` is `(pi)/6`, then `|(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3))|^(2)` is equal to |
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Answer» Correct Answer - c We are given that `veca = a_(1)hati+a_(2)hatj +a_(3)hatk` `vecb = b_(1)hati +b_(2)hatj +b_(3)hatk` `vecc =c_(1)hati +c_(2)hatj +c_(3)hatk` `"then"|{:(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3)):}|^(2)=[veca vecbvecc]^(2)` ` (veca xx vecb.vecc)^(2)` `(|veca xx vecb|.1cos)^(@2)` (since `vecc` is `bot "to" veca and vecb, vecc "is " bot "to" vecaxx vecb)` `(|veca xx vecb|)^(2)` `(|veca||vecb|.sin""pi/6)^(2)` `(1/2sqrt(a_(1)^(2)+a_(2)^(2)+a_(3)^(2))sqrt(b_(1)^(2)+b_(2)^(2)+b_(3)^(2)))^(2)` `1/4(a_(1)^(2)+a_(2)^(2)+a_(2)^(2))(b_(1)^(2)+b_(2)^(2)+b_(3)^(2))` |
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| 9. |
Let `veca=hati + 2hatj +hatk, vecb=hati - hatj +hatk and vecc= hati+hatj-hatk` A vector in the plane of `veca and vecb` whose projections on `vecc`` `is` `1//sqrt3` isA. `4 hati - hatj + 4hatk`B. `3hati+hatj - 3hatk`C. `2hati+hat- 2hatk`D. `4hati + hatj -4hatk` |
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Answer» Correct Answer - a A vector in the plane of `veca and vecb` is `vecu=muveca +lambdavecb= (mu +lamda)hati+ (2mu - lamda) hatj + ( mu +lambda)hatk` projection of `vecu` on ` vecc = 1/sqrt3` ` Rightarrow (vecu.vecc)/(|vecc|)= 1/sqrt3` `or vecu.vecc=1` `or |mu + lambda+ 2mu -lamda-mu-lambda|=1` `or |2mu -lamda|=1` `or lamda = 2mu +- 1` `Rightarrow vecu = 2hati +hatj + 2hatk or 4 hati -hatj + 4hatk` j |
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| 10. |
For three vectors `vecu,vecv,vecw` which of the following expressions is not eqal to any of the remaining three?A. `vecu.(vecvxx vecw)`B. `(vecvxx vecw).vecu`C. `vecv.(vecuxx vecw)`D. `(vecuxx vecv).vecw` |
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Answer» Correct Answer - c `[ vecu vecv vecw] = [vecv vecw vecu] = [vecw vecu vecv]` but `[vecv vecu vecw]=- [vecu vecv vecw]` |
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| 11. |
The number of vectors ofunit length perpendicular to vectors ` vec a=(1,1,0)a n d vec b=(0,1,1)`isa. one b. two c.three``d. infiniteA. oneB. twoC. threeD. infinite |
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Answer» Correct Answer - b we know that if `hatn` is perpendicular to `veca` as we as `vecb` .then `hatn=(vecaxxvecb)/(|vecaxxvecb|)or (vecbxxveca)/(|vecbxxveca|)` As ` veca xx vecb and vecb xx veca` represent two vectors in opposite directions , we have two possible value of `hatn`. |
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| 12. |
If `veca=hati+hatj+hatk,hatb=hati-hatj+hatk,vecc=hati+2hatj-hatk`, then find the value of `|{:(veca.veca,veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc),(vecc.veca,vecc.vecb,vecc.vecc):}|` |
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Answer» `|{:(veca.veca,veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc),(vecc.veca,vecc.vecb,vecc.vecc):}|=[veca vecbvecc][vecavecbvecc]=[vecavecbvecc]^(2)` `|{:(veca.veca,veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc),(vecc.veca,vecc.vecb,vecc.vecc):}|= 4^(2)=16` |
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| 13. |
Let `veca=2hati=hatj+hatk, vecb=hati+2hatj-hatk and vecc=hati+hatj-2hatk` be three vectors . A vector in the pland of `vecb and vecc` whose projection on `veca` is of magnitude `sqrt((2/3))`is (A) `2hati+3hatj+3hatk` (B) `2hati+3hatj-3hatk` (C) `-2hati-hatj+5hatk` (D) `2hati+hatj+5hatk`A. `2hati+3hatj-3hatk`B. `2hati+3hatj +3hatk`C. `-2hati - hatj + 5hatk`D. `2hati + hatj + 5hatk` |
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Answer» Correct Answer - a,c we have `veca =2hati-hatj+hatk,vecb=hati+2hatj-hatk,vecc=hati+hatj=2hatk` Any vector in the plane of `vecb and vecc` is `vecu=muvecb+lamdavecc` ` mu(hati+2hatj-hatk)+lamda(hati +hatj-2hatk)` `=(mu+lamda)hati+(2mu+lamda)hatj-(mu + 2lamda)hatk` Given that the magnitude of projection of `vecu "on " veca` is `sqrt(2//3)` thus `sqrt(2/3)=|(vecu.veca)/(|veca|)|` `|(2(mu+lamda)-(2mu+lamda) -(mu+2lamda))/sqrt6|` `or |-lamda- mu|2` ` Rightarrow lamda + mu=2 or lamda+mu =-2` |
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| 14. |
If `vecb and vecc` are two non-collinear such that `veca ||(vecbxxvecc)`. Then prove that `(vecaxxvecb).(vecaxxvecc)` is equal to `|veca|^(2)(vecb.vecc)` ` |
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Answer» `veca||(vecbxxvecc)` `veca=lambda(vecbxxvecc)and veca=vecabotvecband vecabotvecc` `Now, (vecaxxvecb).(vecaxxvecc)=|{:(veca.a,veca.vecc),(vecb.veca,vecb.vecc):}|` `=|{:(veca.veca,0),(0,vecb.vecc):}|=|veca|^(2)(vecb.vecc)` |
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| 15. |
`vecA=(2veci+veck),vecB=(veci+vecj+veck) and vecC=4veci-vec3j+7veck` determine a vector `verR` satisfying `vecRxxvecB=vecCxxvecB and vecR.vecA=0` |
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Answer» Correct Answer - `-hati-8hatj+2hatk` we are given that `vecA=2hati + hatk,vecB=hati+hatj+hatk and vecC= 4hati -3hatj +7hatk and ` to determine a vector `vecR` such that `vecR xx vecB = vecC xx vecB and vecR.vecA =0` Let `vecR =x hati + yhatj + zhatk` then `vceR xx vecB = vecC xx vecB` `Rightarrow |{:(hati,hatj,hatk),(x,y,z),(1,1,1):}|=|{:(hati,hatj,hatk),(4,-3,7),(1,1,1):}|` `or (y-z) hati - (x-z) hatj + (x-y) hatk` ` =-10 hati + (x -z) hatj + 7 hatk` y-z= -10 x-z =-3 x-y= 7 Also ` vecR.vecA=0` ` Rightarrow 2x +z=0` Subsituting y =x-7 and z =-2x from (ii) and (iv), respectively in (i) , we get x-7 +2x =-10 3x=-3 x=-1,y =-8 and z=2 |
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| 16. |
If `veca and vecb` are any two unit vectors, then find the greatest postive integer in the range of `(3|veca + vecb|)/2+2|veca-vecb|` |
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Answer» Correct Answer - 5 Let angle between `veca and vecb br theta` we have ` |veca| = |vecb|=1` Now `|veca + vecb|= 2cos theta/2 and |veca - vecb|=2 sin theta/2` consider `F(theta) = 3 cos theta/2 + 4 sin theta/2 theta in [ 0,pi]` |
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| 17. |
Let `vecu` be a vector on rectangular coodinate system with sloping angle `60^(@)` suppose that `|vecu-hati|` is geomtric mean of `|vecu| and |vecu-2hati|`, where `hati` is the unit vector along the x-axis . Then find the value of `(sqrt2+ 1) |vecu|` |
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Answer» Correct Answer - 1 since angle between `vecu and hati is 60^(@)` we have `vecu. I = |vecu||hati|cos 60^(@) = (|vecu|)/2` Given that ` |vecu - hati| ,|vecu| , |vecu -2hati|` are in G.P. so `|vecu - hati|^(2)= |vecu| |vecu -2 hati|` squaring both sides, `[|vecu|^(2)+|hati|^(2)-2vecu.hati]^(2)=|vecu|^(2)[|vecu|^(2)+4|hati|^(2)-4vecu.hati]` `[|vecu|^(2)+1-(2|vecu|)/2]^(2)=|vecu|^(2)[|vecu|^(2)+4-4(|vecu|)/2]` `or |vecu|^(2)+ 2|vecu|-1=0Rightarrow|vecu|=-(2+-2sqrt2)/2` `or |vecu|= sqrt2-1` |
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| 18. |
If `(veca xxv vecb) xx (vecc xx vecd) . (veca xx vecd) =0` then which of the following may be true ?A. `veca, vecb and vecd` are nenessarily coplanarB. `veca` lies iin the plane of `vecc and vecd`C. `vecvb` lies in the plane of `veca and vecd`D. `vecc` lies in the plane of `veca and vecd` |
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Answer» Correct Answer - b,c,d `(vecaxxvecb )xx(vecc xx vecd). (vecaxxvecd) =0` `or ([veca vecc vecd] vecb- [vecb vecc vecd]veca) . (veca xx vecd)=0` `or [veca veccvecd] [vecb veca vecd] =0` Hence, either `vecc or vecb` must lie in the plane of ` veca and vecd`. |
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| 19. |
Vector ` vec O A= hat i+2 hat j+2 hat k`turns through a right anglepassing through the positive x-axis on the way. Show that the vector in its new position is`(4 hat i- hat j- hat k)/(sqrt(2))dot` |
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Answer» Let the new vector be `vec(OB)= xhati+yhatj=zhatk` According to the given condition, we have `|vec(OB)|=|vec(OA)|=3 Rightarrowx^(2)+y^(2)+z^(2)=9` `vec(OA)vec(OB)Rightarrowx+2y+2z=0` Since with turing `vec(OA)`, it passes through the postivie x-axis on the way, vectors `vec(OA),vec(OB)and lambdahati` coplanar . thus, `|{:(x,y,z),(1,2,2),(lambda,0,0):}|=0` or y-z=0 solving (i) (ii) and (iii) for x,y and z. we have x-4y=-4z `Rightarrow 16y^(2)+y^(2)+y^(2)=9` `Rightarrowy=+-1/sqrt2` `Rightarrow vec(OB) = +-(4/sqrt2hati-1/sqrt2hatj-1/sqrt2hatk)` since angle between `vec(OB) and hati` is acute, `vec(OB)=4/sqrt2hati-1/sqrt2hatj-1/sqrt2hatk` |
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| 20. |
Find `vecaxxvecb and |vecaxxvecb|` if `vec=hati-7hatj+7hatk vecb=3hati-2hat+2hatk`` |
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Answer» `veca=hati-7hatjandvecb=3hati+2hatk` `vecaxxvecb=|{:(hati,hatj,hatk),(1,-7,7),(3,-2,2):}|` `= hati(-14+14)-hatj(2-21)+hatk(-2+21)=19hatj+19hatk` `|vecaxxvecb|=sqrt((19)^(2)+(19)^(2))=sqrt(2xx(19)^(2))=19sqrt2` |
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| 21. |
Prove that `(veca-vecb)xx(veca+vecb)=2(vecaxxvecb)` also interpret this result. |
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Answer» `(veca-vecb)xx(veca+vecb)=(veca-vecb) xxveca+(veca-vecb)xxvecb` [ By distributivity of vector product over addition] `= vecaxxveca-vecbexxveca+vecaxxvecb-vecbxxvecb` [ Again , by distributivity of vector product over addition ] `=vec0 +vecaxxvecb+vecaxxvecb-vec0` `= 2vecaxxvecb` |
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| 22. |
Find a unit vectorperpendicular to the plane determined by the points `(1,-1,2),(2,0,-1)a n d(0,2,1)dot` |
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Answer» Given points are A(1,-1,2),B(2,0,-1) and C(0,2,1) `Rightarrow vec(AB)=veca=hati+hatj-3hatk,vec(BC)=vecb=-2hati+2hatj+2hatk` `vecaxxvecb=|{:(hati,hatj,hatk),(1,1,-3),(-2,2,2):}|=8hati+4hatj+4hatk` Hence, Unit vector `vecaxxvecb=|{:(hati,hatj,hatk),(1,1,-3),(-2,2,2):}|=8hati+4hatj+4hatk` |
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| 23. |
Let the vectors `veca and vecb` be such that `|veca|=3and|vecb|=sqrt2/3"then|" vecaxxvecb`is a unit vector. If the angle between `veca and vecb` is ? |
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Answer» It is given that `|veca|=3and |vecb|=sqrt2/3` we know that `vecaxxvecb= |veca|vecb|sin theta hatn "where" hatn` is a unit vector peroendicular to both `veca and vecb and theta` is the angle between `veca and vecb`. `now veca xx vecb` is a unit vector if `|vecaxxvecb|=1` ` or , ||veca||vecb|sin theta|=1`. `or , ||veca||vecb| sin theta|=1` `or 3xxsqrt2/3xxsintheta=1` `or theta=pi/4` Hence , `veca xx vecb` is a unit vector if the angle between `veca and vecb is pi/4`. |
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| 24. |
If `veca and vecb` are two vectors , then prove that `(vecaxxvecb)^(2)=|{:(veca.veca" ",veca.vecb),(vecb.veca" ",vecb.vecb):}|` |
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Answer» `(vecaxxvecb)^(2)=(ab sintheta.hatn)^(2)` `=a^(2)b^(2)sin^(2)theta` =`(veca.veca) (vecb.vecb)-(veca.vecb)^(2)` `=|{:(veca.veca" ",veca.vecb),(vecb.veca" ",vecb.vecb):}|` |
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| 25. |
A unit vector a makes an angle `pi/ 4` with z-axis. If `a + i + j` is a unit vector, then a can be equal to |
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Answer» `Let veca= xhati+yhatj+zhatk` Given `|veca|=1` therefore, `x^(2)+y^(2)+z^(2)=1` Angle between `veca` and z-axis is `pi//4` , therefore, `cos (pi/4)= (veca.hatk)/(|veca||hatk|)` `z=1/sqrt2` `Now veca+hati+ahtj= (x+1)hati+(y+1)hatj+zhatk` Given that `veca+hati+hatj` is a unit vector. therefore, `|veca+hati+hatj|=sqrt([(x+1)^(2)+(y+1)^(2)z^(2)])=1` `x^(2)+y^(2)=z^(2)+2x+2y+1=0` 1+2x+2y+1=0 y=-(x+1) from (i), we have `x^(2)+(x+1)^(2)+(1//2)=1` `Rightarrow 4x^(2)+4x+1=0or(2x + 1)^(2)=0` `x=1/2 Rightarrowy=1/2` Hence,` veca=-1/2hati-1/2hatj+1/sqrt2hatk` |
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| 26. |
If `a+2b+3c=4,`then find the least valueof `a^2+b^2+c^2dot` |
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Answer» consider vectors `vecP=ahati+bhatj+chatkandvecq=hati+2hatj+3hatk` `costheta= (a+2b+3c)/(sqrt(a^(2)b^(2)=c^(2))sqrt(1^(2)+2^(2)+3^(2)))` `cos^(2)theta= ((a+2b+3c)^(2))/(14(a^(2)+b^(2)+c^(2)))le1` `Rightarrow a^(2)+b^(2)+c^(2)ge8/7` Hence, least value of `a^(2) + b^(2) + c^(2) is 8/7` |
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| 27. |
Prove that: `[vecaxxvecb vecb xx vecc veccxxveca]=[vecavecbvecc]^2` |
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Answer» `[vecaxxvecb" "vecbxxvecc " " veccxxveca]=(vecaxxvecb).((vecbxxvecc)xx(veccxxveca))` `= (veca xxvecb).[[vecbveccveca]vecc-[vecbvecc vecc]veca]` `= [veca vecbvecc]^(2)` |
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| 28. |
Prove that `hati xx(vecaxxveci)+hatjxx(vecaxxvecj)+hatkxx(vecaxxveck)=2veca` |
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Answer» `hatixx(veca xxhati)=(hati.hati)veca-(veca.veci)=veca-(veca.veci)hati` `hatjxx(vecaxxhatj)=veca-(veca.hatj)and hatkxx(vecaxxhatk)veca-(veca.hatk)hatk`. `hatixx(vecaxxhati)+hatjxx(vecaxxhatj)+hatkxx(vecaxxhatk)=3veca-((veca.hati)hati+(veca.hatj)+(vecaxx hatk)hatk)=2veca` |
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| 29. |
If ` |veca|=2` then find the value of `|vecaxxveci|^(2)+|vecaxxvecj|^(2)+|vecaxxveck|^2` |
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Answer» `|vecaxxveci|^(2)=|{:(hati,hatj,hatk),(a_(1),a_(2),a_(3)),(1,0,0):}|^(2)` `=|a_(3)hatj-a_(2)hatk|^(2)=a_(3)^(2)+a_(2)^(2)` similarly, `|vecaxxvecj|^(2)=a_(1)^(2)+a_(3)^(2) and|vecaxxveck|^(2)=a_(1)^(2)+a_(2)^(2)` Hence, the required result can be given as `2(a_(1)^(2)+a_(2)^(2)+a_(3)^(2))=2|veca|^(2)=8` |
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| 30. |
If `veca.veci=veca.(hati+hatj)=veca.(hati+hatj+hatk)` . Then find the unit vector `veca`. |
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Answer» Let `veca=xhati+yhatj+zhatk` then `veca.hati=(xhati+yhatj+zhatk).hati=x and veca. (hati+hatj)=x+y` `and veca.(hati+hatj=hatk)=x+y+z("given that" x=x+y=x+y+z)` now, `x=x+y Rightarrow y =0 and x+y =x+y+z Rightarrow z=0` Hence, x =1 ( sicne `veca` is a unit vector) `veca=veci` |
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| 31. |
If `hati xx[(veca-hatj)xxhati]+hatjxx[(veca-hatk)xxhatj]+veckxx[(veca-veci)xxhatk]=0` , then find vector `veca`. |
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Answer» `hatixx[(veca-vecj)xxhati]=(hati.hati)(veca-vecj)-(hati.(veca-hatj))hati` `= veca-hatj-(hati.veca)hati` similarly, `hatjxx[(veca-hatk)xxvecj]=veca-veck+(vecj.veca)hatj` `hatkxx[(veca-hati)xxhatk]=veca-hati-(hatk.veca)hatj` `hatixx[(veca-hatj)xxhati]xx[(veca-veck)xxhatj]+hatkxx[(veca-hati)xxhatk]` `=veca-hatj-(hati.veca)hati+veca-hatk+(hatj.veca)hatj+veca-hati-(hatk.veca)hatk=0` `3veca-(hati+hatj+hatk)-veca=0` `veca=1/2(hati+hatj+hatk)` |
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| 32. |
Let `veca= 2 hati + 3hatj - 6hatk, vecb = 2hati - 3hatj + 6hatk and vecc = -2 hati + 3hatj + 6hatk`. Let `veca_(1)` be the projection of `veca on vecb and veca_(2)` be the projection of `veca_(1) on vecc` . Then`veca_(1).vecb` is equal toA. `-41`B. `-41//7`C. 41D. 287 |
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Answer» Correct Answer - a `veca_(1).vecb= (-41)/49 (2hati - 3hatj + 6hatk) . (2hati - 3hatj + 6hatk) =-41` |
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| 33. |
Let `veca= 2 hati + 3hatj - 6hatk, vecb = 2hati - 3hatj + 6hatk and vecc = -2 hati + 3hatj + 6hatk`. Let `veca_(1)` be the projection of `veca on vecb and veca_(2)` be the projection of `veca_(1) on vecc` . Then which of the following is true ?A. `veca and vcea_(2)` are collinearB. `veca_(1) and vecc` are collinearC. `veca m veca_(1) and vecb` are coplanarD. `veca, veca_(1) and a_(2)` are coplanar |
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Answer» Correct Answer - c `veca,veca_(1)andvecb` are coplanar because `veca_(1) and vecb` are collinear. |
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| 34. |
Let `veca= 2 hati + 3hatj - 6hatk, vecb = 2hati - 3hatj + 6hatk and vecc = -2 hati + 3hatj + 6hatk`. Let `veca_(1)` be the projection of `veca on vecb and veca_(2)` be the projection of `veca_(1) on vecc` . Then `veca_(2)` is equal toA. `943/49 (2 hati - 3hatj - 6hatk)`B. `943/(49^(2)) (2 hati - 3hatj - 6hatk)`C. `943/49 (-2 hati + 3hatj + 6hatk)`D. `943/(49^(2)) (-2 hati + 3hatj + 6hatk)` |
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Answer» Correct Answer - b `veca_(1)[(2hati+3hatj-6hatk)((2hati-3hatj + 6hatk))/7](2hati-3hatj + 6hatk)/7` `(-41)/49(hati-3hatj + 6hatk)` `veca_(2) (-41)/49((2hati-3hatj + 6hatk).((-2hati+3hatj+6hatk))/7)` `xx ((-2hati + 3hatj = 6hatk))/7` `(-41)/((49)^(2))(-4-9+36) (-2hati+3hatj + 6hatk)` `943/(49^(2))(2hati -3hatj - 6hatk)` |
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| 35. |
Given two orthogonal vectors `vecA and VecB` each of length unity. Let `vecP` be the vector satisfying the equation `vecP xxvecB=vecA-vecP`. then `(vecP xx vecB ) xx vecB` is equal toA. `vecP`B. `-vecP`C. `2vecB`D. `vecA` |
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Answer» Correct Answer - b `vecPxxvecB=vecA-vecP and |vecA|=|vecB|=1 and vecA.vecB=0`is given Now `vecP xx vecB = vecA-vecP` `(vecPxxvecB) xxvecB= (vecA-vecP)xxvecB` (taking cross product with `vecB` on both sides) `or (vecP.vecB)vecB- (vecB.vec B)vecP=vecAxxvec B -vecP xx vecB` `or (vecP.vecB) vecB-vecP=vecA xx vecB-vecA+vecP` `or 2vecP = vecA-vecA-vecAxxvec B-(vecP.vecB)vecB` `or vecP= (vecA-vecAxxvecB=-(vecP.vecB)vecB)/2` taking dot product with `vecB` on both sides of (i) get `vecP.vecB=vecA.vecB-vecP.vecB`ltbr gt `Rightarrow vecP.vecB=0` `Rightarrow vecP=( vecA + vecBxxvecA)/2` Now `(vecP xx vecB) xx vecB= (vecP.vecB) vecB-(vecB.vecB) vecP=-vecP` `vecP,vecA,vecPxxvecB(=vecA-vecP)` are depenment Also `vecP.vecB=0` ` and |vecP|^(2)=|(vecA-vecAxxvecB)/2|^(2)` `(|vecA|^(2)+|vecAxxvecB|^(2))/4` `(1+1)/4=1/2or |vecP|=1/sqrt2` |
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| 36. |
Let `veca vecb and vecc` be non- zero vectors aned `vecV_(1) =veca xx (vecb xx vecc) and vecV_(2) = (veca xx vecb) xx vecc`.vectors `vecV_(1) and vecV_(2)` are equal . ThenA. `veca and vecb` ar orthogonalB. `veca and vecc` are collinearC. `vecb and vecc` ar orthogonalD. `vecb= lambda (veca xx vecc) " when " lambda ` is a scalar |
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Answer» Correct Answer - b,d `vecV_(1) = vecV_(2)` `vecaxx (vecbxxvecc) = (vecaxxvecb)xxvecc` `or (veca.vecc)vecb - (veca.vecb)vecc= (veca.vecc)vecb- (vecb.vecc)veca` `or (veca .vecb)vecc = (vecb.vecc)veca` Thus , either `vecc and veca` ar collinear or `vecb` is perpendicular to both `veca and vecc Rightarrow vecb = lamda (veca xx vecc) ` |
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| 37. |
Given two orthogonal vectors `vecA and VecB` each of length unity. Let `vecP` be the vector satisfying the equation `vecP xxvecB=vecA-vecP`. then which of the following statements is false ?A. vectors `vecP, vecA and vecP xx vecB` ar linearly dependent.B. vectors `vecP, vecB and vecP xx vecB` ar linearly independentC. `vecP` is orthogonal to `vecB` and has length `1/sqrt2`.D. none of these |
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Answer» Correct Answer - d `vecPxxvecB=vecA-vecP and |vecA|=|vecB|=1 and vecA.vecB=0`is given Now `vecP xx vecB = vecA-vecP` `(vecPxxvecB) xxvecB= (vecA-vecP)xxvecB` (taking cross product with `vecB` on both sides) `or (vecP.vecB)vecB- (vecB.vec B)vecP=vecAxxvec B -vecP xx vecB` `or (vecP.vecB) vecB-vecP=vecA xx vecB-vecA+vecP` `or 2vecP = vecA-vecA-vecAxxvec B-(vecP.vecB)vecB` `or vecP= (vecA-vecAxxvecB=-(vecP.vecB)vecB)/2` taking dot product with `vecB` on both sides of (i) get `vecP.vecB=vecA.vecB-vecP.vecB`ltbr gt `Rightarrow vecP.vecB=0` `Rightarrow vecP=( vecA + vecBxxvecA)/2` Now `(vecP xx vecB) xx vecB= (vecP.vecB) vecB-(vecB.vecB) vecP=-vecP` `vecP,vecA,vecPxxvecB(=vecA-vecP)` are depenment Also `vecP.vecB=0` ` and |vecP|^(2)=|(vecA-vecAxxvecB)/2|^(2)` `(|vecA|^(2)+|vecAxxvecB|^(2))/4` `(1+1)/4=1/2or |vecP|=1/sqrt2` |
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| 38. |
Given two orthogonal vectors `vecA and VecB` each of length unity. Let `vecP` be the vector satisfying the equation `vecP xxvecB=vecA-vecP`. then `vecP` is equal toA. `vecA/2 + (vecAxxvecB)/2`B. `vecA/2 + (vecBxxvecA)/2`C. `(vecAxxvecB)/2 -vecA/2`D. `vecA xxvecB` |
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Answer» Correct Answer - b `vecPxxvecB=vecA-vecP and |vecA|=|vecB|=1 and vecA.vecB=0`is given Now `vecP xx vecB = vecA-vecP` `(vecPxxvecB) xxvecB= (vecA-vecP)xxvecB` (taking cross product with `vecB` on both sides) `or (vecP.vecB)vecB- (vecB.vec B)vecP=vecAxxvec B -vecP xx vecB` `or (vecP.vecB) vecB-vecP=vecA xx vecB-vecA+vecP` `or 2vecP = vecA-vecA-vecAxxvec B-(vecP.vecB)vecB` `or vecP= (vecA-vecAxxvecB=-(vecP.vecB)vecB)/2` taking dot product with `vecB` on both sides of (i) get `vecP.vecB=vecA.vecB-vecP.vecB`ltbr gt `Rightarrow vecP.vecB=0` `Rightarrow vecP=( vecA + vecBxxvecA)/2` Now `(vecP xx vecB) xx vecB= (vecP.vecB) vecB-(vecB.vecB) vecP=-vecP` `vecP,vecA,vecPxxvecB(=vecA-vecP)` are depenment Also `vecP.vecB=0` ` and |vecP|^(2)=|(vecA-vecAxxvecB)/2|^(2)` `(|vecA|^(2)+|vecAxxvecB|^(2))/4` `(1+1)/4=1/2or |vecP|=1/sqrt2` |
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| 39. |
Vectors `vecA and vecB` satisfying the vector equation `vecA+ vecB = veca, vecA xx vecB =vecb and vecA.veca=1`. Vectors and `vecb` are given vectosrs, areA. `vecA = ((vecaxxvecb)-veca)/(a^(2))`B. `vecB = ((vecbxx veca) + veca (a^(2) - 1))/a^(2)`C. `vecA = ((vecaxxvecb)+veca)/(a^(2))`D. `vecB = ((vecbxx veca) - veca (a^(2) - 1))/a^(2)` |
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Answer» Correct Answer - b,c, we have `vecA.vecB =veca` `or vecA .veca + vecB.veca = veca.veca` `or 1+ vecB.veca = a^(2)` `or vecB.veca= a^(2)-1` Also `vecA xx vecB =vecb` `or veca xx (vecA xx vecB ) = veca xx vecb` ` or (veca. vecB)vecA - (veca.vecA) vecB = veca xx vecb` `or (a^(2)-1) vecA -vecB = veca xx vecb` ( using (i) and ` veca. vecA =1`) (ii) ` and vecA + vecB =a` form (ii) and (iii) , we have `vecA= ((vecaxxvecb)+veca)/a^(2)` `vecB=veca-{((vecaxxvecb)+veca)/a^(2)}` ` = ((vecbxxveca) + veca (a^(2) -1))/a^(2)` thus `vecA= ((vecaxxvecb)+veca)/a^(2)` ` and vecB= ((vecbxxveca)+veca (a^(2) -1))/a^(2)` |
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| 40. |
If `|veca|=|vecb|=|veca+vecb|=1` then find the value of `|veca-vecb|` |
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Answer» we have `|veca+vecb|^(2)+|veca-vecb|^(2)=2(|veca|^(2)+|vecb|^(2))` `1+|veca-vecb|^(2)=4or|veca-vecb|sqrt3` |
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| 41. |
If `veca=4hati+6hatj and vecb=3hati+4hatk` find the vector component of `veca alond vecb`. |
| Answer» The components of vector `veca long vecb` is `((veca.vecb)vecb)/(|vecb|^(2))=18/25(3hati+4hatk)` | |
| 42. |
If `vecb and vecc` are any two mutually perpendicular unit vectors and `veca` is any vector, then `(veca.vecb)vecb+(veca.vecc)vecc+(veca.(vecbxxvecc))/(|vecbxxvecc|^2)(vecbxxvecc)=` (A) 0 (B) `veca (C) `veca/2` (D) `2veca` |
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Answer» Correct Answer - `veca` Let `vecalpha, vecbeta,vecgamma` be any three mutually perpendicular non-coplanar, unit vectors and `veca` be any vector, then `veca= (veca.vecalpha)vecalpha+ (veca.vecbeta)+(veca.vecgamma)vecgamma` Here `vecb, vecc` are two mutually perpendicular vectors, therefore, `vecb , vecc and (vecb xx vecc)/(|vecb xx vecc|)` are three mutually Perpendicular non-coplanaar unit vectors. Hence `veca=(veca .vecb)vecb+(veca.vecc)vecc` `+(veca.(vecbxxvecc)/(|vecb xx vecc|))(vecb xx vecc)/(|vecb xx vecc|)` `(veca.vecb)vecb+(veca.vecc)vecc` `+(veca.(vecbxxvecc))/(|vecb xx vecc|^(2))(vecbxxvecc)` |
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| 43. |
A non vector `veca` is parallel to the line of intersection of the plane determined by the vectors `hati,hati+hatj` and thepane determined by the vectors `hati-hatj,hati+hatk` then angle between `veca and hati-2hatj+2hatk` is = (A) `pi/2` (B) `pi/3` (C) `pi/6` (D) `pi/4` |
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Answer» Correct Answer - `pi//4 or 3pi//4` A vector normal to the plane containing vectors ` hati and hati + hatz` is `vecp=|{:(hati,hatj,hatk),(1,0,0),(1,1,0):}|=hatk` A vector normal to the plane containing vectors ` hati- hatj, hati + hatk` is `vecq=|{:(hati,hatj,hatk),(1,-1,0),(1,0,1):}|=-hati-hatj+hatk` vector `veca` is parllel to vector `vecp xx vecq` `vecp xx vecq=|{:(hati,hatj,hatk),(0,0,1),(-1,-1,1):}|=hati-hatj` Therefore, a vector in direction of `veca` is `hati - hatj` Now `theta` is the angle between `hata and hati - 2hatj + 2hatk` then `cos theta=+-(1.1+(-1).(-2))/(sqrt(1+1)sqrt(1+4+4))=+-3/(sqrt(2).3)` `Rightarrow +-1/sqrt2 Rightarrow theta=pi/4or (3pi)/4` |
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| 44. |
Let `vec(r)` is a positive vector of a variable pont in cartesian OXY plane such that `vecr.(10hatj-8hati-vecr)=40` and `p_1=max{|vecr+2hati-3hatj|^2},p_2=min{|vecr+2hati-3hatj|^2}`. A tangent line is drawn to the curve `y=8/x^2` at the point A with abscissa 2. The drawn line cuts x-axis at a point BA. 9B. `2sqrt2-1`C. `6sqrt6+3`D. `9-4sqrt2` |
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Answer» Correct Answer - d Let `vecx= x hati + yhatj` `x^(2) + y^(2) + 8x - 10y + 40 =0` , which is a circle centre C(-4,5) , radius r = 1 `p_(1)= max {(x+2)^(2)+ (y-3)^(2)}` `P_(2) = min {(x+2)^(2)+ (y-3)^(2)}` Let P be (-2,3). Then `CP = sqrt2,r=1` ` P_(2)= (2sqrt2-1)^(2)` `P_(1) = (2sqrt2+1)^(2)` `P_(1) + p_(2) =18` Slope = AB = `((dy)/(dx))_(2,2)=-2` Equation of AB, 2x+y=6 `vec(OA)=2hati=2hatj,vec(OB)= 3hati` `vec(AB)=hati-2hatj` `vec(AB).vec(OB)= (hati-2hatj) (3hati)=3` |
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| 45. |
Let `vec(r)` is a positive vector of a variable pont in cartesian OXY plane such that `vecr.(10hatj-8hati-vecr)=40` and `p_1=max{|vecr+2hati-3hatj|^2},p_2=min{|vecr+2hati-3hatj|^2}`. A tangent line is drawn to the curve `y=8/x^2` at the point A with abscissa 2. The drawn line cuts x-axis at a point BA. 2B. 10C. 18D. 5 |
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Answer» Correct Answer - c Let `vecx= x hati + yhatj` `x^(2) + y^(2) + 8x - 10y + 40 =0` , which is a circle centre C(-4,5) , radius r = 1 `p_(1)= max {(x+2)^(2)+ (y-3)^(2)}` `P_(2) = min {(x+2)^(2)+ (y-3)^(2)}` Let P be (-2,3). Then `CP = sqrt2,r=1` ` P_(2)= (2sqrt2-1)^(2)` `P_(1) = (2sqrt2+1)^(2)` `P_(1) + p_(2) =18` Slope = AB = `((dy)/(dx))_(2,2)=-2` Equation of AB, 2x+y=6 `vec(OA)=2hati=2hatj,vec(OB)= 3hati` `vec(AB)=hati-2hatj` `vec(AB).vec(OB)= (hati-2hatj) (3hati)=3` |
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| 46. |
If `veca, vecb, vecc` are unit vectors such that `veca. Vecb =0 = veca.vecc` and the angle between `vecb and vecc is pi/3` , then find the value of `|veca xx vecb -veca xx vecc|` |
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Answer» Correct Answer - 1 `veca.vecb = 0 Rightarrow veca bot vecb` `veca .vecc = 0 Rightarrow veca bot vecc` ` Rightarrow veca bot vecb -vecc` `|veca xx vecb -veca xx vecc|=|veca xx (vecb-vecc)|` `= |veca||vecb-vecc|=|vecb -vecc|` Now `|vecb -vecc|^(2) = |vecb|^(2)+|vecc|^(2) -2 |vecb||vecc|cos "" pi/3` ` = 2 -2 x xx 1/2=1 ` `|vecb -vecc|=1` |
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| 47. |
Let `vecx, vecy and vecz` be three vectors each of magnitude `sqrt2` and the angle between each pair of them is `pi/3 if veca` is a non-zero vector perpendicular to `vecx and vecy xx vecz and vecb` is a non-zero vector perpendicular to `vecy and vecz xx vecx`, 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 Accoding to the question ` vecx. Vecz = vecx .vecy =vecy.vecz=sqrt2.sqrt2. cos " " pi/3=1 ` Given `veca` is perpendicular to `vecx and vecy xx vecz` `veca = lamda_(1)((vecx.vecz) vecy- (vecx .vecy) vecz)` ` Rightarrow veca = lamda _(1) (vecy -vecz)` Now, `veca. vecy =lamda_(1) (vecy. vecy.vecz) = lamda_(1)(2-1)` ` Rightarrow lamda_(1) = veca. vecy` From (1) and )(2), `veca. (veca.vecy)(vecy-vecz)` Similarly, `vecb= (vecb.vecz) (vecz-vecx)` Now `veca.vecb= (veca.vecy) (vecb.vecz) [(vecy-vecz).(vecz-vecx)]` `= (veca.vecy) (vecb.vecz) [ 1-2+1]` ` =- (veca.vecy) (vecb.vecb.vecz)` |
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| 48. |
Let `vec a ` be vector parallel to line of intersection of planes `P_1 and P_2` through origin. If `P_1`is parallel to the vectors `2 bar j + 3 bar k and 4 bar j - 3 bar k` and `P_2` is parallel to `bar j - bar k` and ` 3 bar I + 3 bar j `, then the angle between `vec a` and `2 bar i +bar j - 2 bar k` is :A. `pi//2`B. `pi//4`C. `pi//6`D. `3pi//4` |
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Answer» Correct Answer - b,d Normal to plane `P_(1)` is `vecn_(1)= (2hatj+3hatk)xx)(4hatj-3hatk)=-18hati` Normal to plane `P_(2)` is Therefore, `vecn_(2)= (hatj-hatk)xx (3hati +3hatj)=3hati -3hatj-3hatk` `vecA` is parallel to `+-(vecn_(1)xx vecn_(2))=+- (-54hatj+54hatk)` Now , the angle between `vecA nad 2hati +hatj - 2hatk` is given by `cos thet=+-((-54hatj+54hatk).(2hati+hatj-2hatk))/(54sqrt2 .3)` `=+-1/sqrt2` `theta= pi//4 or 3pi//4` |
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| 49. |
Let a three- dimensional vector `vecV` satissgy the condition , `2vecV + vecV xx ( hati + 2hatj ) = 2hati + hatk . If 3|vecV| = sqrtm` . Then find the value of m. |
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Answer» Correct Answer - 6 `2vecV+vecVxx(hati+2hatj) = (2hati+hatk)` `or 2vecV. (hati+2hatj) +0=(2hati+hatk). (hati+2hatj)` `or 2vecVr. (hati+2hatj)=2` `or |vecV. (hati+2hatj)^(2)|=1` `or |vecV|^(2).|hati +2hatj|^(2) cos^(2)theta=1` (`theta` is the angle between `vecV and hati+2hatj)` `or |vecV|^(2)5(1-sin^(2)theta)=1` `or |vecV|^(2) 5 sin^(2)theta =5|vecV|^(2)-1` from Eq. (i), we have `|2vecV+vecVxx(hati+2hatj)|^(2)=|2hati+hatk|^(2)` `or 4|vecV|^(2)+|vecVxx(hati+2hatj)|^(2)=5` `or 4|vecV|^(2)+|vecV|^(2).|hati + 2hatj|^(2) sin^(2) theta=5` `or 4|vecV|^(2)+5|vecV|^(2)sin^(2)theta=5` `or 4|vecV|^(2)+5|vecV|Y^(2)-1=5` ` 9|vecV|^(2)=6` `or 3|vecV|=sqrt6` ` = sqrt6 = sqrtm` m=6 |
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| 50. |
Let `tianglePQR` be a triangle . Let `veca=overline(QR),vecb = overline(RP) and vecc= overline(PQ).if |veca|=12, |vecb|=4sqrt3and vecb.vecc= 24` then which of the following is (are) true ?A. `|vecc|^(2)/2-|veca|=12`B. `|vecc|^(2)/2-|veca|=30`C. `|vecaxxvecvb + veccxxveca|= 48sqrt3`D. `veca.vecb=-72` |
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Answer» Correct Answer - a,c,d `veca + vecb+vecc =0` `Rightarrow vecb + vecc= -veca` `Rightarrow |vecb|^(2) +|vecc|^(2) + 2vecb.vecc= |veca|^(2)` ` Rightarrow 48 + |vec|^(2) + 48 = 144 ` ` Rightarrow |vecc|^(2)=48` `|vecc|= 4sqrt3` ` (|vecc|)^(2))/2+|veca|=36` Further, `veca+vecb=-vecc` `Rightarrow |veca|^(2)+|vecb|^(2)+2veca.vecb=|vecc|^(2)` `Rightarrow 144 + 48 + 2 veca. vecb= 48` `veca. vecb = -72` `veca.vecb + vecc=0` `Rightarrow veca xx vecb +veca xx vecc=0` `|veca xx vecb +vecc xx veca|` `2|veca xx vecb|` `=2 sqrt(a^(2)b^(2)-(veca.vecb)^(2))` `2sqrt((144)(48)-(72)^(2))=48sqrt3` |
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