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101.

If `vecr.veca=vecr.vecb=vecr.vecc=0 " where "veca,vecb and vecc` are non-coplanar, thenA. `vecrbot(veccxxveca)`B. `vecrbot(vecaxxvecb)`C. `vecrbot(vecbxxvecc)`D. `vecr=vec0`

Answer» Correct Answer - d
Let `vecr ne vec0` then `vecr.veca = vecr.vecb = vecr.vecc =0`
Hence, `veca, vecb and vecc` are coplanar, which is a contradiction.
Therefore, `vecr =vec0`
Discussion
102.

If the vectors `3vecP+vecq, 5vecP - 3vecq and 2 vecp+ vecq, 4 vecp - 2vecq` are pairs of mutually perpendicular vectors, the find the angle between vectors `vecp and vecq`.

Answer» `3 vecp + vecq and 5vecp-3vecq` are perpendicular. thereforem,
`(3vecP +vecq). (5vecp -3vecq)=0`
`15vecp^(2)-3vecq^(2)=4vecp.vecq`
`2vecp +vecq and 4vecp -2vecq` are perpendicular , therefore,
`(2vecp +vecq).(4vecp - 2vecq)=0`
`8vecp^(2)=2vecq^(2)`
`vecq^(2)=4vecp^(2)`
Now, `cos theta= (vecp.vecq)/(|vecp||vecq|)`
Substituting `vecq^(2)=4vecp^(2) "in" (i), 3vecp^(2)=4vecp.vecq`
`cos theta = 3/4.vecp^(2)/(|vecp|2|vecp|)=3/8`
or ` theta=cos^(-1)""3/8`
Discussion
103.

If the vertices A,B,C of a triangle ABC are (1,2,3),(-1,0,0) ,(0,1,2) , respectively, then find `angleABC`.

Answer» The vertices of `triangleABC` are given as A(1,2,3) , B(-1,0,0), and C ( 0,1,2)
Also, it is given that `angle ABC` is the angle between the vectors `vec(BA) and vec(BC)` thus
`vec(BA)={1-(-1)}hati+(2-0)hatj+ (3-0)hatk`
`2hati+ 2hatj + 3hatk`
`vec(BC)={0-(-1)}hati+(1-0)hatj + (2-0)hatk`
`= hati + hatj + 2hatk`
`vec(BA).vec(BC)=(2hati+ 2hatj + 3hatk). (hati + hatj + 2hatk)`
`2xx1+2xx1+3xx2`
`= 2 + 2 +6 =10`
`|vec(BA)|=sqrt(2^(2)+2^(2)+3^(2))=sqrt(4+4+9)=sqrt17`
`|vec(BA)|=sqrt(1+1+2^(2))=sqrt6`
Now, it is known that :
`vec(BA). vec(BC)= |vec(BA)||vec(BC)|cos(angleABC)`
`10 =sqrt17 xx sqrt6 cos(angleABC)`
`or cos(angleABC) = 10/(sqrt17xxsqrt6)`
`or angleABC = cos^(-1)(10/sqrt102)`
Discussion
104.

If `veca` satisfies `vecaxx(hati+2hatj+hatk)=hati-hatk" then " veca` is equal toA. `lambdahati+(2lambda-1)hatj+lambdahatk,lambda in R`B. `lambdahati+(1-2lambda)hatj+lambdahatk,lambda in R`C. `lambdahati+(2lambda+1)hatj+lambdahatk,lambda in R`D. `lambdahati+(1+2lambda)hatj+lambdahatk,lambda in R`

Answer» Correct Answer - c
`vecaxx(hati + 2hatj + hatk) = hati-hatk= (hatjxx (hati + 2hatj +hatk))`
`or (veca- hatj) xx (hati + 2hatj +hatk) = vec0`
`or veca - hatj = lambda(hati + 2hatj +hatk)`
`or veca=lambdahati+ (2lambda+1) hatj + lambdahatk , lambda in R`
Discussion
105.

`P(1,0,-1), Q(2,0,-3),R(-1,2,0)a n dS(,-2,-1),`then find the projectionlength of ` vec P Qon vec R Sdot`

Answer» `vec( PQ)=vec(OQ)-vec(OP)=hati -2hatk`
`vec(RS)=vec(OS)-vec(OR)=4hati - 4hatj-hatk`
Projection of `vec(PQ) on vec(RS)= (vec(QP)vec(RS))/(|vec(RS)|) = 6/sqrt33`
Discussion
106.

If vectors `hati-2xhatj-3yhatk and hati+ 3 x hatj + 2yhatk` are orthogonal to each other, then find the locus of th point (x,y).

Answer» It is given that vectors `hati - 2xhatj - 3yhatk and hati + 3x hatj + 2yhatk` are orthogonal , therefore,
`(hati-2xhatj - 3yhatk).(hati+3xhatj+2yhatk)=0`
` or 1-6x^(2)-6y^(2)=0`
`or bx^(2)+6y^(2)=1` whiche is a circle.
Discussion
107.

Show that `|veca|vecb+|vecb|veca` is perpendicular to `|veca|vecb-|vecb|veca` for any two non zero vectors `veca and vecb.

Answer» `(|veca|vecb+ |vecb|veca).(|veca|vecb- |vecb|veca)`
=`|veca|^(2)vecb.vecb- |veca||vecb|^(2)vecb.veca`
`+ |vecb||veca|veca.vecb-|veca||vecb|^(2)veca.veca`
`= |veca|^(2)|vecb|^(2)-|vecb|^(2)|veca|^(2)`
=0
Hence, `|veca|vecb + |vecb|veca and |veca|vecb- |vecb|veca` are perpendicular to each other .
Discussion
108.

If `A ,B ,C ,D`are four distinct point inspace such that `A B`is not perpendicular to `C D`and satisfies ` vec A Bdot vec C D=k(| vec A D|^2+| vec B C|^2-| vec A C|^2=| vec B D|^2),`then find the value of `kdot`

Answer» Let A be the origin, and the position vectors of B,C
and D be `vecb,vecc,vecd`
`vec(AB).vec(CD)=k(|vec(AD)|^(2)+|vec(BC)|^(2)-|vec(AC)|^(2)-|vec(BD)|^(2))`
or ` (vecb).(vecd - vecc)`
`k = [(vecd)^(2)+(vecc-vecb)^(2)-(vecc)^(2)- (vecd-vecb)^(2)]`
`or vecb.vecd-vecc=k (-2vecb.vecc+ 2vecb .,vecd)`
or k 1/2
Discussion
109.

If `|veca|=3, |vecb|=4` and the angle between `veca and vecb is 120^(@)` . Then find the value of `|4 veca + 3vecb|`

Answer» `|4veca+3.vecb|=sqrt((4veca+3vecb).(4veca+3vecb))`
`sqrt(16|veca|^(2)+9|vecb|^(2)+24veca.vecb)`
`sqrt(144 + 144 + 24 xx 3xx4 xx(-1/2))`
12
Discussion
110.

A vector `vecd` is equally inclined to three vectors `veca=hati-hatj+hatk,vecb=2hati+hatj and vecc=3hatj-2hatk.` Let `vecx,vecy and vecz` be three vectors in the plane of `veca,vecb;vecb,vec;vecc,veca,` respectively. ThenA. `vecz.vecd =0`B. `vecx.vecd=1`C. `vecy.vecd= 32`D. `vecr.vecd= 0 , " where " vecr= lambdavecx+ muvecy+ gammavecz`

Answer» Correct Answer - a,d
` veca =hati -hatj +hatk`
` vecb = 2hati +hatj`
`and vecc = 3hatj -2hatk`
since, `[veca vecb vecc]= |{:(1,-1,1),(2,1,0),(0,3,-2):}|=0`
therefore, `veca,vecb and vecc` are coplanar vectors, furthere, sicne `vecd` is equally inclined to `veca, vecb anv vecc` we have
`vecd. veca=vecd.vecb=vecd.vecc=0`
` vecd.vecx=vecd.vecy =vecd.vecz=0`,
`vecd.vecr =0`
Discussion
111.

A vector `vecd` is equally inclined to three vectors `veca=hati-hatj+hatk,vecb=2hati+hatj and vecc=3hatj-2hatk.` Let `vecx,vecy and vecz` be three vectors in the plane of `veca,vecb;vecb,vec;vecc,veca,` respectively. ThenA. `vecx.vecd=-1`B. `vecy.vecd= 1`C. vecz.vecd=0`D. vecr.vecd=0, " where " vecr=lambda vecx + mu vecy +deltavecz`

Answer» Correct Answer - c.d
Since `[veca vecb vecc] =0, veca, vecb and vecc` are coplanar vectors.
Further, since, `vecd` is equally inclined to `veca ,vecb and vecc,`
`vecd.veca =vecd.vecb = vecd.vecc=0`
`vecd.vecx =vecd.vecy = vecd.vecz =0`
`vecd.vecr =0`
Discussion
112.

If `veca=hati+hatj + hatk and vecb = hati - 2 hatj+hatk` then find the vector `vecc` such that `veca.vecc =2 and veca xx vecc=vecb`.

Answer» `vecaxx vecb = vecaxx(vecaxxvecc) = (veca.vecc)veca- (veca.veca)vecc`
`2veca - 3vecc`
But `veca xx vecb = 1|{:(hati,hatj , hatk),(1,1,1),(1,-2, 1):}|= 3hati-3hatk`
Hence, `3vecc= 2veca - (3hati - 3hatk)`
`= (2hati + 2hatj =2hatk)-(3hati - 3hatk)`
`= - hati + 2hatj + 5hatk`
`or vecc = 1/3 (-hati = 2hatj + 5hatk) `
Discussion
113.

Let `veca and vecb` be unit vectors such that `|veca+vecb|=sqrt3`. Then find the value of `(2veca+5vecb).(3veca+vecb+vecaxxvecb)`

Answer» `(2veca+5vecb).(3veca+vecb+vecaxxvecb)=6veca.veca+17veca.vecb+5vecb.vecb=11+17veca.vecb`
`(veca.(vecaxxvecb)=vecb.(vecaxxvecb)=0,as veca and vecb` are perpendicular to `veca xx vecb`)
`|veca+vecb|=sqrt3`
`|veca+vecb|^(2)=3`
`|veca|^(2)+|vecb|^(2)+2veca.vecb=3`
`veca.vecb=1/2`
`Rightarrow (2veca+5vecb).(3veca+vecb+veca xx vecb)11+17/2=39/2`
Discussion
114.

Vectors `3veca-5vecb and 2veca + vecb` are mutually perpendicular. If `veca + 4 vecb and vecb - veca` are also mutually perpendicular, then the cosine of the angle between `veca nad vecb` isA. `19/(5sqrt43)`B. `19/(3sqrt43)`C. `19/(sqrt45)`D. `19/(6sqrt43)`

Answer» Correct Answer - a
`(3veca-5vecb).(2veca+vecb)=0`
`or 6|veca|^(2)-5|vecb|^(2)= 7 veca.vecb`
`Also , (veca+4vecb).(vecb-veca)=0`
`or -|veca|^(2)+ 4|vecb|^(2)= 3 veca.vecb`
`or 6/7 |veca|^(2)-5/7|vecb|^(2)= -1/3 |veca|^(2)+ 4/3|vecb|^(2)`
` or 25|veca|^(2) = 43 |vecb|^(2)`
` Rightarrow 3 veca. vecb= - |veca|^92) + 4|vecb|^(2)= 57/25 |vecb|^(2)`
`or 3 |veca||vecb| cos theta = 57/25 |vecb|^(2)`
`or 3sqrt(43/25) |vecb|^(2) cos theta = 57/25 |vecb|^(2)`
` or cos theta = 19/(5sqrt43)`
Discussion
115.

If the vectors `A ,B ,C`of a triangle `A B C`are `(1,2,3),(-1,0,0),(0,1,2),`respectively then find `/_A B Cdot`

Answer» The vertices of `triangleABC` are given as A(1,2,3) , B(-1,0,0), and C ( 0,1,2)
Also, it is given that `angle ABC` is the angle between the vectors `vec(BA) and vec(BC)` thus
`vec(BA)={1-(-1)}hati+(2-0)hatj+ (3-0)hatk`
`2hati+ 2hatj + 3hatk`
`vec(BC)={0-(-1)}hati+(1-0)hatj + (2-0)hatk`
`= hati + hatj + 2hatk`
`vec(BA).vec(BC)=(2hati+ 2hatj + 3hatk). (hati + hatj + 2hatk)`
`2xx1+2xx1+3xx2`
`= 2 + 2 +6 =10`
`|vec(BA)|=sqrt(2^(2)+2^(2)+3^(2))=sqrt(4+4+9)=sqrt17`
`|vec(BA)|=sqrt(1+1+2^(2))=sqrt6`
Now, it is known that :
`vec(BA). vec(BC)= |vec(BA)||vec(BC)|cos(angleABC)`
`10 =sqrt17 xx sqrt6 cos(angleABC)`
`or cos(angleABC) = 10/(sqrt17xxsqrt6)`
`or angleABC = cos^(-1)(10/sqrt102)`
Discussion
116.

If in a right-angledtriangle `A B C ,`the hypotenuse `A B=p ,t h e n vec A BdotA C+ vec B Cdot vec B A+ vec C Adot vec C B`is equal to`2p^2`b. `(p^2)/2`c. `p^2`d. none of theseA. `2p^(2)`B. `p^(2)/2`C. `p^(2)`D. none of these

Answer» Correct Answer - c
we have
` vec(AB). Vec(AC) + vec(BC) + vec(BA) + vec(CA) . Vec(CB)`
` AB ( AC cos theta + BC sin theta)`
` AB ( AC cos theta + BC sin theta)`
`AB (((AC)^(2))/(AB)) + ((BC)^(2))/(AB)`
` AC^(2) + BC^(2) = AB^(2) + p^(2)`
Discussion
117.

`A , B , C , D`are any four points, provethat ` vec A Bdot vec C D+ vec B Cdot vec A D+ vec C Adot vec B D=0.`

Answer» `vec(AD)=vec(AB)+vec(BC)+vec(CD)=veca+vecb+vecc`
`vec(AC)=vec(AB)+vec(BC)=veca+vecborvec(CA)=-(veca+vecb)`
`vec(BD)=vec(BC)=vec(CD)=vecb+vecc`
`vec(AB).vec(CD)+vec(BC).vec(AD)+vec(CA).vec(BD)`
`veca.vecc+vecb.(veca+vecb+vecc)+(=-veca-vecb).(vecb+vecc)`
`= veca.vecc+vecb.veca+vecb.vecb+vecb.vecc-veca.vecb-veca.vecc`
`-vecb.vecb-vecb.vecc=0`
Discussion
118.

A vector of magnitude `sqrt2` coplanar with the vectors `veca=hati+hatj+2hatk and vecb = hati + hatj + hatk,` and perpendicular to the vector `vecc = hati + hatj +hatk` isA. `-hatj+hatk`B. `hati and hatk`C. `hati - hatk`D. hati- hatj`

Answer» Correct Answer - a
A vector coplanar with `veca and vecb` and perpendicular to `vecc is lambda (veca xx vecb) xx vecc)`
but `lambda(veca xx vecb xx vecc) = lambda [ (veca . Vecc) vecb- (vecb .vecc) veca]`
` lambda [ 4vecb - 4 veca]`
`4 lambda [ vecj -hatk]`
Hence, the required vector vector, is `hatj - hatk or -hatj + hatk`
Discussion
119.

Let the position vectors ofthe points `Pa n dQ`be `4 hat i+ hat j+lambda hat ka n d2 hat i- hat j+lambda hat k ,`respectively. Vector ` hat i- hat j+6 hat k`is perpendicular to theplane containing the origin and the points`Pa n dQ`. Then `lambda`equals`1//2`b. `1//2`c. `1`d. none of theseA. `-1//2`B. `1//2`C. 1D. none of these

Answer» Correct Answer - a
A vector perpendicular to the plane of O , P and Q is
` vec(OP) xx vec(OQ) `
Now, `vec(OP) xx vec(OQ) = |{:(hati,hatj,hatk),(4,1,lambda),(2,-1,lambda):}|`
` = 2lambda hati - 2hatj - 6hatk`
therefore, `hatim -hatj + 6hatk` is parallel to
`2lambda hati - 2lambdahatj - 6hatk`
Hence, `1/(2lambda) = (-1)/(-2lambda) = 6/(-6)`
`lambda = -1/2`
Discussion
120.

If `veca and vecb` are any two vectors of magnitudes 1and 2. respectively, and `(1-3veca.vecb)^(2)+|2veca+vecb+3(vecaxxvecb)|^(2)=47` then the angle between `veca and vecb ` isA. `pi//3`B. `pi-cos^(-1) (1//4)`C. `(2pi)/3`D. `cos^(-1) (1//4)`

Answer» Correct Answer - c
` 1 + 9 (veca .vecb)^(2) - 6( veca .vecb) + 4 |veca|^(2)`
`+ |vecb|^(2) + 9 |veca xx vecb|^(2) + 4veca.vecb = 47`
`or 1 + 4+4 +36 - 4 cos theta = 47`
`or cos theta = -1/2`
Hence, the angle between `veca and vecb is (2pi)/3`
Discussion
121.

Let `P`be a point interior to theacute triangle `A B Cdot`If `P A+P B+P C`is a null vector, thenw.r.t traingel `A B C ,`point `P`is itsa. centroid b. orthocentrec. incentre d. circumcentreA. centroidB. orthocentreC. incentreD. circumcentre

Answer» Correct Answer - a
`veca - vecp + vecb - vecp + vecc -vecp =0`
`or vecp = (veca + vecb + vecc)/3`
Hence, P is centroid .
Discussion
122.

If `|vecaxxvecb|^(2)+(veca.vecb)^(2)=144and|veca|=4,"then "|vecb|` is equal to ……..

Answer» Since `(veca xx vecb)^(2) + (veca. Vecb)^(2) = 144` if the angle between `veca and vecb is theta`. Then
`|veca|^(2)|vecb|^(2) sin ^(2)theta + |veca|^(2).|vecb|^(2)cos^(2)theta=144`
`or |veca|^(2)|vecb|^(2) = 144`
`or |veca||vecb|= 12`
`or 4 |vecb|=12`
`or |vecb|= 3`
Discussion
123.

`veca and vecc` are unit vectors and `|vecb|=4` the angle between `veca and vecb is cos ^(-1)(1//4) and vecb - 2vecc=lambdaveca` the value of `lambda` isA. 3,-4B. 1/4,3/4C. `-3,4`D. `-1//4,3/4`

Answer» Correct Answer - a
` vecb - 2vecc = lambda veca`
` or vecb = 2vecc + lambda veca`
` or |vecb|^(2) = |2vecc + lambdaveca|^(2)`
`Rightarrow 16 = 4 |vecc|^(2) + lambda^(2)|veca|^(2) + 4lambda veca.vecc`
` = 4 + lambda^(2) + 4lambda 1/4`
`or lambda^(2) + lambda - 12 =0`
`or lambda = 3, -4`
Discussion
124.

`veca and vecc` are unit vectors and `|vecb|=4` the angle between `veca and vecc` `is cos ^(-1)(1//4) and vecb - 2vecc=lambdaveca` the value of `lambda` isA. 3,-4B. 1/4,3/4C. `-3,4`D. `-1//4,3/4`

Answer» Correct Answer - a
` vecb - 2vecc = lambda veca`
` or vecb = 2vecc + lambda veca`
` or |vecb|^(2) = |2vecc + lambdaveca|^(2)`
`Rightarrow 16 = 4 |vecc|^(2) + lambda^(2)|veca|^(2) + 4lambda veca.vecc`
` = 4 + lambda^(2) + 4lambda 1/4`
`or lambda^(2) + lambda - 12 =0`
`or lambda = 3, -4`
Discussion
125.

if the volume of a parallelpiped whose adjacent egges are `veca=2hati+3hatj+4hatk,vecb=hati+alphahatj+2hatk, vecc=veci+2hatj + alphahatk is 15` then find of `alpha if (alpha gt 0) `

Answer» `|{:(2,3,4),(1 , alpha,2),(1,2,alpha):}|=15`
`or 2(alpha^(2)-4)+3(2-alpha)+ 4(2-alpha) =15`
`or 2 alpha^(2)-8 + 6 -3alpha=8 - 4alpha = 15`
`or 2alpha^(2) - 9alpha=2alpha-9=0`
` or ( alpha+1) (2 alpha-9) =0`
`or alpha = -1 , 9//2`
Discussion
126.

Given `|veca|=|vecb|=1 and |veca + vecb|= sqrt3 "if" vecc` is a vector such that `vecc -veca - 2vecb = 3(veca xx vecb) ` then find the value of `vecc . Vecb`.

Answer» We, have, `|veca + vecb|=sqrt3`
` or |veca + vecb|^(2) =3`
`or |veca|^(2) + |vecb|^(2) + 2(veca .vecb) =3`
`or veca. Vecb = 1//2` (i)
now ` vecc - veca - 2vecb = 3(veca xx vecb) `
`or (vecc - veca - 2vecb) .vecb = 3{(veca xx vecb) .vecb}`
`or vecc . vecb - 1/2 - 2xx 1 =0`
or `vecc.vecb = 5//2`
Discussion
127.

If `vecl,vecm,vecn` are three non coplanar vectors prove that `[vec` vecm vecn](vecaxxvecb) =|(vec1.veca, vec1.vecb, vec1),(vecm.veca, vecm.vecb, vecm),(vecn.veca, vecn.vecb, vecn)|`

Answer» Let `vecl =l_(1)hati + l_(2)hatj +l_(3)hatk, vecm = m_(1)hati + m_(2)hatj + m_(3)hatk`
` vecn=n_(1)hati+n_(2)hatj +n_(3)hatk, veca = a_(1)hati +a_(2)hatj +a_(3) hatk`
`vecb =b_(1)hati , b_(2)hatj+b_(3)hatk,` therefore,
`vecl.veca =l_(1)a_(1)+ l_(2)a_(2) +l_(3)a_(3) = sum l_(1) a_(1) `
similarly `vecl.vecb = suml_(1)b_(1)`.etc.
`now, [vecl vecm vecn] (vecaxxvecb) = |{:(l_(1),l_(2),l_(3)),(m_(1),m_(2),m_(3)),(n_(1),n_(2),n_(3)):}|xx|{:(hati,hatj ,hatk),(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)):}|`
`|{:(suml_(1)hati,suml_(1)a_(1),suml_(1)b_(1)),(summ_(1)hati,summ_(1)a_(1),summ_(1)b_(1)),(sumn_(1)hati, sumn_(1)a_(1),sumn_(1)b_(1)):}|`
`=|{:(vecl,vecl.veca,vecl.vecb),(vecm,vecm.veca,vecm.vecb),(vecn,vecn.veca,vecn.vecb):}|=|{:(vecl.veca,vecl.vecb,vecl),(vecm.veca,vecm.vecb,vecm),(vecn.veca,vecn.vecb,vecn):}|`
Discussion
128.

If `veca , vecb , vecc and vecd` are four non-coplanar unit vectors such that `vecd` makes equal angles with all the three vectors `veca, vecb, vecc` then prove that `[vecd vecavecb]=[vecd veccvecb]=[vecd veccveca]`

Answer» Since `vecd` makes equalw angles with the vectors `veca1 , vecb and vecc`, we have,
`d= (mu(veca + vecb + vecc))/3`
(`vecd` passes through the centroid of the triangle with position vectors, `veca , vecb and vecc`)
Again `[veca vecb vecc]vecd = [ vecd vecb vecc] + [vecd vecc vecd] vecb`
`+ [vecd veca vecb]vecc`
From (i) and (ii) , we get `[veca vecb vecc] = [vecd vecc veca] = [ vecd veca vecb] `
Discussion
129.

Points `veca , vecb vecc and vecd` are coplanar and `(sin alpha)veca + (2sin 2beta) vecb + (3sin 3gamma) vecc - vecd= vec0` . Then the least value of `sin^(2) alpha + sin^(2) 2beta + sin^(2) 3gamma` isA. `1//14`B. 14C. 6D. `1//sqrt6`

Answer» Correct Answer - a
Points `veca , vecb , vecc and vecd` are coplanar, therefore,
`sin alpha + 2 sin 2 beta + 3 sin 3 gamma =1 `
` now |sin alpha + 2sin 2 beta + 2sin 3 gamma|`
`le sqrt(1+4+9). sqrt(sin^(2)alpha +sin^(2)2beta + sin^(2) 3gamma)`
`or sin ^(2) alpha + sin^(2) 2beta + sin ^(2) 3 gamma le 1/14`
Discussion
130.

prove that `(veca.hati)(vecaxxhati)+(veca.hatj)(vecaxxhatj)+(veca.hatk)(vecaxxhatk)=vec0`

Answer» Let `veca = a_(1) hati + a_(2)hatj + a_(3) hatk`1, therefore,
`veca. Hati = (a_(1) hati = a_(2) and veca . Hatk = a_(3)`
`and vecaxxhati = (a_(1)hati+a_(2)hat j + a_(3) hatk) xx hati = a_(2)hatk + a_(3) hatj` ,
`(veca . hati ) (vecaxx hati) + (veca . hatj) (veca xx hatj) + (veca. hatk) (veca xx veck)`
`-a_(1)a_(2)hatk + a_(1)a_(3)hatj + a_(1)a_(2) hatk +a_(3)a_(2)hati`
`+a_(3)a_(2) hati -a_(3)a_(1)hati`
`vec0`
Discussion
131.

Let `veca, vecb, and vecc` be three non- coplanar vectors and `vecd` be a non -zero , which is perpendicular to `(veca + vecb + vecc). Now vecd = (veca xx vecb) sin x + (vecb xx vecc) cos y + 2 (vecc xx veca) `. ThenA. `(vecd. (veca + vecc))/([veca vecb vecc])=2`B. `(vecd. (veca + vecc))/([veca vecb vecc])=-2`C. minimum value of `x^(2) + y^(2) is pi^(2)//4`D. minimum value of `x^(2) + y^(2) is 5 pi^(2)//4`

Answer» Correct Answer - b,d
`vecd. veca = [veca vecb vecc] cos y=-vecd. (vecb+vecc)`
` or cos y = (vecd. (vecb +vecc))/([veca.vecb.vecc]) `
similarly, ` sin x = (vecd.(veca+vecb))/([veca vecb vecc]) and (vecd.(veca+vecc))/([veca vecb vecc])=-2`
`sin x + cos y + 2 = 0`
` or sin x + cos y =-2 `
` or sin x =-1 , cos y = -1 `
since we want the minimum value of `x^(2) + y^(2) x = -pi//2 , y = pi`
,Therefore, the minimum value of `x^(2) + y^(2) is 5 pi^(2)//4`
Discussion
132.

Let `veca=hati+hatj and vecb=2hati-hatk.` Then the point of intersection of the lines `vecrxxveca=vecbxxveca and vecrxxvecb=vecaxxvecb` is (A) `(3,-1,10` (B) `(3,1,-1)` (C) `(-3,1,1)` (D) `(-3,-1,-10`A. `hati-hatj+hatk`B. `3hati-hatj+hatk`C. `3hati+hatj-hatk`D. `hati-hatj-hatk`

Answer» Correct Answer - c
`vecrxxveca =vecbxxveca or (vecr-vecb) xxveca=0`
`vecrxxvecb = vecaxx vecb or (vecr-veca) xx vecb=0`
`if vecr=xhati + yhatj +zhatk`then
`|{:(hati,hatj,hatk),(x-2,y,z+1),(1,1,0):}|=0and|{:(hati,hatj,hatk),(x-1,y-1,z),(2,0,-1):}|=0`
`Rightarrow z+1=0,x-y=2`
`and y-1=0,x-1+2z=0`
`Rightarrow x=3,y=1,z=-1`
Discussion
133.

prove that `(veca.(vecbxxhati)hati(veca.(vecbxxhatj))hatj+ (veca.(vecbxxhatk))hatk=vecaxxvecb`

Answer» `veca.(vecbxx hati)hati = ((vecaxxvecb).hati)hati`
`if veca xx vecb = xhati + yhatj + zhatk , then (veca xx vecb). Hati = x `
similarly, `(veca. (vecb xx vecj)) hatj = y and (veca . (vecb xx veck)) =z `
`Rightarrow (veca.(vecb.hati)) hati+ (veca . vecb xx vecj)) hatj + (veca .(vecb xx veck)) hatk`
= xhati + y hatj = zhatk = veca xx vecb`
Discussion
134.

If `veca = (hati + hatj +hatk), veca. vecb= 1 and vecaxxvecb = hatj -hatk , " then " vecb` isA. `hati - hatj + hatk`B. `2hati - hatk`C. `hati`D. `2hati`

Answer» Correct Answer - c
`(veca xx vecb) xxveca = (veca.veca)vecb- (veca.vecb)veca`
`(hatj-hatk)xx (hati +hatj +hatk) = (sqrt3)^(2)vecb - (hati +hatj +hatk)`
`or 3vecb = 3hati or vecb = hati`
Discussion
135.

if `veca , vecb and vecc` are three non-zero, non- coplanar vectors and `vecb_(1)=vecb-(vecb.veca)/(|veca|^(2))veca,vecb_(2)=vecb+(vecb.veca)/(|veca|^(2))veca,vecc_(1)=vecc-(vecc.veca)/(|veca|^(2))veca+ (vecb.vecc)/(|vecc|^(2))vecb_(1),vecc_(2)=vecc-(vecc.veca)/(|veca|^(2)) veca-(vecbvecc)/(|vecb_(1)|^(2))vecb_(1),vecc_(3)=vecc- (vecc.veca)/(|vecc|^(2))veca + (vecb.vecc)/(|vecc|^(2))vecb_(1), vecc_(4)=vecc - (vecc.veca)/(|vecc|^(2))veca= (vecb.vecc)/(|vecb|^(2))vecb_(1)`, then the set of orthogonal vectors isA. `(veca,vecb_(1),vecc_(3))`B. `(vecca,vecb_(1),vecc_(2))`C. `(veca, vecb_(1),vecc_(1))`D. `(veca,vecb_(2),vecc_(2))`

Answer» Correct Answer - c
we observe that
`veca.vecb_(1)=veca.vecv- ((vecb.veca)/(|veca|^(2)))veca. veca=veca.vecb- veca.vecb=0`
`veca.vecc_(2)=veca(vecc-(vecc.veca)/(|veca|^(2))veca- (vecc.vecb_(1))/(|vecb_(1)|^(2))vecb_(1))`
`veca.vecc- (veca.vecc)/(|veca|^(2))|veca|^(2)-(vecc.vecb_(1))/(|vecb_(1)|^(2))(veca.vecb_(1))`
`veca.vecc-veca.vecc-0 " " (therefore veca.vecb_(1)=0)`
`and vecb_(1).vecc_(2)=vecb_(1)(vecc- (vecc.veca)/(|veca|^(2))veca- (vecc.vecb_(1))/(|vecb_(1)|^(2))vecb_(1))`
`vecb_(1).vecc-((vecc.veca)(vecb_(1).veca))/(|veca|^(2))- (vecc.vecb_(1))/(|vecb_(1)|^(2))vecb_(1).vecb_(1)`
`vecb_(1).vecc-0-vecb_(1).vecc` (using `vecb_(1).veca =0)`
Discussion
136.

The unit vector which is orthogonal to the vector `5hati + 2hatj + 6hatk ` and is coplanar with vectors `2hati + hatj + hatk and hati - hatj + hatk `isA. `(2hati - 6hatj + hatk)/sqrt41`B. `(2hati-3hatj)/sqrt13`C. `(3 hati -hatk)/sqrt10`D. `(4hati + 3hatj - 3hatk)/sqrt34`

Answer» Correct Answer - c
Any vector coplanar, to `veca and vecb ` can be written as
`vecr=muveca + lamdavecb`
`vecr = (mu + 2lamda) hati + ( -mu+lamda) hatj + ( mu + lamda) hatk`
since `vecr` is othorgonal to `5hati + 2hatj + 6hatk`
` Rightarrow 5 ( mu + 2lambda) +2 ( - mu+ lamda) + 6 ( mu + lambda) = 0`
`or 9mu + 18 lamda =0`
` or lamda =-1/2 mu `
` vecr = lamda ( 3hatj -hatk)`
Since `hatr` is a unit vector s, `hatr = (3hati -hatk) / sqrt10`
Discussion
137.

Let `veca = 2i + j + k, and b = i+ j ` if c is a vector such that `veca .vecc = |vecc|, |vecc -veca| = 2sqrt2` and the angle between `veca xx vecb and vec is 30^(@)` , then `|(veca xx vecb)|xx vecc|` is equal toA. `2//3`B. `3//2`C. 2D. 3

Answer» Correct Answer - b
`|(veca xxvecb)xxvecc|=|vecaxxvecb||vecc|sin30^(@)`
`=1/2 (vecp +vecq+vecr)a^(2)`
`or vecx=1/2 (vecp +vecq +vecr)`
we have `veca = 2hati+hatj -2hatk and vecb = hati+hatj`
`Rightarrow vecaxx vecb = 2hati -2hatj+hatk`
`or |veca xx vecb|=sqrt9=3`
Also given `|vecc-veca|^(2)=8`
`or |vecc|^(2)=|veca|^(2)-2veca.vecc=8`
Given `|aveca|=3 and veca. vecc =|vecc|` , using these we get
`|vecc|^(2) -2|vecc|+1=0`
`or (|vecc|-1)^(2)=0`
`or |vecc|=1`
Substituting values of `|veca xx vecb|and |vecc|` in (i), we get
`|(veca xx vecb)xxvecc|=1/2xx 3xx 1= 3/2`
Discussion
138.

Sholve the simultasneous vector equations for `vecx aedn vecy: vecx+veccxxvecy=veca and vecy+veccxxvecx=vecb, vec!=0A. `vecx=(vecbxxvecc+veca+(vecc.veca)vecc)/(1+ vecc.vecc)`B. `vecx=(veccxxvecb+vecb+(vecc.veca)vecc)/(1+vecc.vecc)`C. `vecy=(vecaxxvecc+vecb+(vecc.vecb)vecc)/(1+vecc.vecc)`D. none of these

Answer» Correct Answer - b
`vecx +vecc xx vecy = veca`
`vecy +vecc xx vecx =vecb`
taking cross with `vecc` , we have
`vecc xx vecy + vecc xx (vecc xx vecx) =vecc xx vecb`
` Rightarrow (veca - vecx) + (vecc .vecx)vecc- (vecc.vecc)vecx = vecc xx vecb`
Also `vecx + vecc xx vecy= veca`
` Rightarrow vecc. vecx + vecc.(vecc xx vecy) =vecc. veca`
` or vecc . vecx + 0 = vecc.vcea`
` Rightarrow veca- vecx + (vcec - veca) vecc -(vecc-vecc)vecx = vecc xx vecb`
`or vecx ( 1 + (vecc.vecc)) = vecb xx vecc + veca + (vecc .veca). vecc`
`or vecc= (vecb xx vecc + veca + (vecc . vecb) vecc)/ (1+vecc.vecc)`
similarly on taking cross product of Eq (i) , we find
` vecy = (veca xx vecc + vecb + (vecc. veca) vecc)/ (1 + vecc .vecc)`
Discussion
139.

The condition for equations `vecrxxveca = vecb and vecr xx vecc = vecd` to be consistent isA. `vecb.vecc=veca.vecd`B. `veca.vecb=vecc.vecd`C. `vecb.vecc+veca.vecd=0`D. `veca.vecb+vecc.vecd=0`

Answer» Correct Answer - c
` vecr xx veca = vecb `
` or vecd xx ( vecr xx veca)= vecd xx vecb`
`or (veca .vecd) vecr - (vecd.vecr) veca= vecd xx vecb`
` vecr xx vecc = vecd`
` or vecb xx ( vecr xx vecc) = vecb xx vecd`
` or (vecb .vecc) vecr - (vecb.vecr) vecc = vecb xx vecd`
Adding (i) and (ii) , we get
` ( veca. vecd + vecb . vecc) vecr- (vecd.vecr) veca - (vecb.vecr) vecc `
Now `vecr.vecd = 0 and vecb.vecr=0 as vecd and vecr` as `vecb and vecr` are mutually perpendicular.
Hence ` ( vecb.vecc + veca.vecd)vecr = vec0`
Discussion
140.

Let `(veca (x) = (sin x) hati+ (cos x) hatj and vecb(x) = (cos 2x) hati + (sin 2x) hatj` be two variable vectors `( x in R)`. Then `veca (x) and vecb (x) ` areA. collinear for unique value of xB. perpendicular for infinte values of x.C. zero vectors for unique value of xD. none of these

Answer» Correct Answer - b
if `veca ( x) and vecb are bot, "then" veca. Vecb =0`
`Rightarrow sin x cos 2x + cos x sin 2x =0`
`sin (3x) =0 = sin 0 `
` 3x = npi 0or x = (npi)/3`
therefore, the two vectors are `bot` for infinite values.
Discussion
141.

If `veca=2hati + hatj + hatk,vecb=hati + 2hatj + 2hatk ,vecc= hati + hatj + 2hatk and [vecavecbveci] hati+ [vecavecb vecj] hatj+ [veca vecb hatk] k` is equal to

Answer» Correct Answer - b
Let `veca xx vecb = x hati + y hatj + zhatk`. Therefore,
,` [veca vecb hati] = (veca xx vecb) . Hati =x`
`[veca vecb hatj] = (veca xx vecb ) . Hatj = y `
`[veca vecb veck] = (veca xx vecb) . Hatk =z`
Hence, ` [ veca vecb hati] hati + [ veca vecb hatj] hatj`
` [ veca vecb hatk] hatk = xhati = yhatj + zhatk + veca xx vecb `
Discussion
142.

If `veca=2hati + hatj+ hatk, vecb= hati+ 2hatj + 2hatk,vecc = hati+ hatj + 2hatk and (1 + alpha) hati+ beta(1+ alpha)hatj+gamma(1+alpha)(1 + beta) hatk= vecaxx(vecb xx vecc) , " then " alpha, beta and gamma " are "`A. `-2,-4,-2/3`B. `2,-4,2/3`C. `-2,4,2/3`D. `2,4,-2/3`

Answer» Correct Answer - a
` veca xx (vecb xx vecc) = (veca .vecc) vecb -(veca.vecb) vecc`
` = 5 (hati + 2hatj + 2hatk) - 6 (hati + hatj + 2hatk) `
`Rightarrow ( 1 + alpha) hati + beta (1 + alpha) hatj + gamma ( 1+ alpha) (1+beta) hatk`
` = - hati + 4hatj - 2hatk`
` Rightarrow 1 + alpha = -1 , beta = -4 and gamma (n -1) (-3) = (-2)`
` Rightarrow gamma = -2/3`
Discussion
143.

If `veca .vecb =beta and veca xx vecb = vecc ," then " vecb` isA. `((betaveca-vecaxxvecc))/(|veca|^(2))`B. `((betaveca+vecaxxvecc))/(|veca|^(2))`C. `((betavecc + vecaxxvecc))/(|veca|^(2))`D. `((betavecc + vecaxxvecc))/(|veca|^(2))`

Answer» Correct Answer - a
` veca xx vecb =vecc`
`or veca xx (veca xx vecb) =veca xx vecc `
`or (vec.vecb) veca - |veca|^(2) vecb = veca xx vecc`
`or vecvb = (betaveca -veca xx vecc)/(|veca|^(2) " " (veca .vecb = beta)`
Discussion
144.

Then for any arbitary vector `veca, (((veca xx vecb) + (veca xx vecb))xx (vecb xx vecc)) (vecb -vecc)` is always equal to

Answer» Correct Answer - d
`((veca xxvecb) + (veca xx vecc)) xx (vecb xx vecc)`
`(veca xx vecb) xx (vecb xx vecc) + (veca xx vecc) xx (vecb xx vecc)`
` ((veca xx vecb). Vecc) vecb - ((veca xx vecb) .vecb) vecc`
` +( (veca xx vecc).vecc) vecb - ((vecaxx vecc) .vecb)vecc`
` [veca vecb vecc] (vecb + vecc)`
`Rightarrow (( veca xx vecb ) + (veca xx vecc)) xx (vecb xx vecc)) . (vecb - vecc)`
` [veca vecb vecc] )(vecb + vecc).(vecb -vecc)`
` = [veca vecb vecc] (|vecb|^(2) - |vecc|^(2)) =0`
Discussion
145.

Let `vecV = 2hati +hatj - hatk and vecW= hati + 3hatk . if vecU` is a unit vector, then the maximum value of the scalar triple product `[ vecU vecV vecW]` isA. `-1`B. `sqrt10 + sqrt6`C. `sqrt59`D. `sqrt60`

Answer» Correct Answer - c
Given that `vecV = 2hati +hatj -hatk and vecW =hati + 3hatk and vecU` is a unit vector
` |vecU|=1`
Now `|vecU vecV vecW] = vecU.(vecV xx vecW)`
`= vecU . (2hati +hatj -hatk) xx ( hati + 3hatk)`
`vecU . (3hati -7hatj - hatk)`
` sqrt(3^(2)+7^(2)+ 1^(2)) cos theta`
Which is maximum when `cos theta =1`
therefore, maximum value of [`vecU vecV vecW] - sqrt59`
Discussion