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51.

If `veca and vecb` are two vectors, such that `veca.vecblt0 and |veca.vecb|=|vecaxxvecb|` then the angle between angles between the vectors `veca and vecb` isA. `pi`B. `7pi//4`C. `pi//4`D. `3pi//4`

Answer» Correct Answer - d
`|veca.vecb|=|vecaxxvecb|`
`Rightarrow|veca||vecb||costheta|=|veca||vecb||sin theta|`
(where,`theta` is the angle between `veca and vecb`)
`Rightarrow |costheta|= |sintheta|`
`Rightarrow theta= pi/4 or (3pi)/4(as 0 le thetalepi)`
But `veca.vecb lt 0`, therefore, `theta= (3pi)/4`
Discussion
52.

vectors `veca,vecb and vecc` are of the same length and when taken pair-wise they form equal angles. If `veca=hati+hatj and vecb=hatj+hatk` then find vector `vecc`.

Answer» `Let vecc=xhati+yhatj+zhatk`. Then `|veca|=|vecb|=|vecc|Rightarrowx^(2)+y^(2)+z^(2)=2`
it is given that the angles between the vectors taken in pairs are equal, say `theta` . Therefore,
`cos theta=(veca.vecb)/(|veca||vecb|)= (0+1+1)/(sqrt2sqrt2)=1/2`
`Rightarrow (veca.vecc)/(|veca||vecc|)=1/2and (vecb.vecc)/(|vecb||vecc|)=1/2`
`(x+y)/(sqrt2sqrt2)=1/2 and(y+z)/(sqrt2sqrt2)=1/2`
`Rightarrow x+y=1 and y+z=1`
y=1-x and z=1-y=1-(1-x)=x
`x^(2)+y^(2)+z^(2)=2Rightarrowx^(2)+(1-x)^(2)+x^(2)=2`
`Rightarrow (3x+1)(x-1)= Rightarrowx=1,1//3` ,
`y=1-xRightarrow0 "for" x=1andy=4//3for x=-1//3`
`Hence, vecc=hati+0hatj+hatk and vecc=1/3hati+4/3hatj-1/3hatk`
Discussion
53.

Prove that : `vecixx(vecaxxveci)+vecjx(vecaxxvecj)+veckxx(vecaxxveck)=vec(2a)`

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`
Discussion
54.

If `hati xx[(veca-hatj)xxhati]xx[(veca-hatk)xxhatj]+veckxx[(veca-veci)xxhatk]=0` , then find vector `veca`.

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)`
Discussion
55.

If `veca= vecP + vecq, vecP xx vecb = vec0 and vecq. vecb =0` then prove that `(vecbxx(veca xx vecb))/(vecb.vecb)=vecq`

Answer» `veca=vecp+vecq`
` or vecaxxvecb = vecpxxvecb+vecq xxvecb`
`or veca xx vecb =vecq xx vecb " " ( vecp xx vecb = vec0)`
`or vecb xx (veca xx vecb)= vecb xx (vecq xx vecb)`
`(vecb.vecb)vecq - (vecb.vecq) vecb`
` = (vecb.vecb)vecq " " ( vecb. vecq =0)`
`or (vecb xx (veca xx vecb))/(vecb.vecb) = vecq`
Discussion
56.

Lelt two non collinear unit vectors `hata and hatb` form and acute angle. A point P moves so that at any time t the position vector `vec(OP)` (where O is the origin) is given by `hatacost+hatbsint.` When P is farthest from origin O, let M be the length of `vec(OP) and hatu` be the unit vector along `vec(OP)` Then (A) `hatu= (hata+hatb)/(|hata+hatb|) and M=(1+hata.hatb)^(1/2)` (B) `hatu= (hata-hatb)/(|hata-hatb|) and M=(1+hata.hatb)^(1/2)` (C) `hatu= (hata+hatb)/(|hata+hatb|) and M=(1+2hata.hatb)^(1/2)` (D) `hatu= (hata-hatb)/(|hata-hatb|) and M=(1+2hata.hatb)^(1/2)`A. `,hatu = (hata+hatb)/(|hata + hatb|) and M = (1 + hata.hatb)^(1//2)`B. `,hatu = (hata-hatb)/(|hata - hatb|) and M = (1 + hata.hatb)^(1//2)`C. `hatu = (hata+hatb)/(|hata + hatb|) and M = (1 + 2hata.hatb)^(1//2)`D. `,hatu = (hata-hatb)/(|hata - hatb|) and M = (1 + 2hata.hatb)^(1//2)`

Answer» Correct Answer - a
`|vec(OP)|= |hata cos t +hatb sin t |`
`= (cos^(2)t + sin^(2) t + 2cos t tin t hata.hatb) ^(1//2)`
` (1 + 2 cos t tin t hata.hatb)^(1//2)`
` (1+sin 2t hata.hatb)^(1//2)`
`|vec(OP)|_(max)= (1+hata.hatb)^(1//2)"when " t=pi//4`
`hatu = (hata+hatb)/(sqrt2(|hata+hatb|)/sqrt2)= (hata+hatb)/(|hata+hatb|)`
Discussion
57.

If `veca and vecb` are non - zero vectors such that `|veca + vecb| = |veca - 2vecb|` thenA. `2 veca. vecb= |vecb|^(2)`B. ` veca. vecb= |vecb|^(2)`C. least value of `veca . Vecb + 1/(|vecb|^(2) + 2) " is " sqrt2`D. least value of `veca .vecb + 1/(|vecb|^(2) + 2) " is " sqrt2 -1 `

Answer» Correct Answer - a,d
` |veca+vecb|= |veca-2vecb|`
`Rightarrow veca.vecb= (|vecb|^(2))/2`
Also `veca.vecb + 1/(|vecb|^(2) +2)`
`(|vecb|^(2)+2)/2+ 1/(|vecb|^(2)+2)-1`
`ge sqrt2-1 `
Discussion
58.

Let O be an interior point of `DeltaABC` such that `bar(OA)+2bar(OB) + 3bar(OC) = 0`. Then the ratio of a `DeltaABC` to area of `DeltaAOC` is

Answer» Correct Answer - 3
`("Area of" triangleABC)/("Area of" triangleAOC)= (1/2|vecaxxvecb+vecbxxvecc+veccxxveca|)/(1/2 |veca xxvecb|)`
now `veca + 2vecb + 3vecc = vec0`
Cross multiply with `vecb, vecaxxvecb +3vecc xxvecb=vec0`
`Rightarrow vecaxxvecb= (vecb xxvecc)`
`vecaxxvecb = 3/2(veccxxveca) = 3(vecb xxvecc)`
`Let (vecc xxveca)= vecp`
`veca xxvecb = (3vecp)/2, vecbxxvecc= vecp/2`
`Ratio = (|vecaxxvecb=vecb xxvecc +vecc xxveca|)/(|veccxxveca|)`
`(|(3vecp)/2 + vecP/2 +vecp|)/(|vecp|)`
`(3|vecp|)/|vecp|=3`
Discussion
59.

Find the volume of a parallelopiped having three coterminus vectors of equal magnitude `|a|` and equal inclination `theta` with each other.

Answer» Correct Answer - `|veca|^(3)sqrt(1+ 2cos theta ) (1 - cos theta)`
Let `veca, vecb and vecc` be three vectors of magnitude `|veca|` and equal inclination `theta` with each other.
volume of parallelepiped = `( veca. (vecb xx vecc) = [veca vecb vecc]`
`and [veca vecb vecc]^(2)= |{:(veca.veca,veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc),(veca.vecc, vecb.vecc, vecc.vecc):}|`
`|veca|^(2)|{:( 1, costheta, costheta),(costheta,1,costheta),(costheta, costheta, 1) :}|`
` |veca|^(6) ( 2cos^(3) theta- 3cos^(2)theta + 1) `
`|veca|^(6) (1-costheta)^(2) ( 1+ 2costhetaa)`
` or [veca vecb vecc] = |veca|^(3) sqrt(1 + 2cos theta) ( 1- cos theta)`
Discussion
60.

Show that `(veca-vecb)xx(veca+vecb)=2vecaxx vecb` and give a genometrical interpretation of it.

Answer» `(veca - vecb) xx (veca + vecb) `
`= vecaxxveca + vecaxxvecb -vecbxxveca -vecbxxvecb`
`= veca xx veca + veca xx vecb + veca xx vecb -vecb xx vecb`
`= vec0 + 2veca xx vecb - vec0 = 2vea xx vecb`
Geometrically, the vector area of a parallelogram, whose sides are along vectors, `veca and vecb is veca xx vecb`. Also diagonals are long vectors, `veca - vecb and veca + vecb` and the vector area in terms of diagonal vectors is
`1/2 [(veca -vecb) xx (veca + vecb)]`
Discussion
61.

Given that `vecA,vecB,vecC` form triangle such that `vecA=vecB+vecC`. Find a,b,c,d such that area of the triangle is `5sqrt(6)` where `vecA=aveci+bveci+cveck. vecB=dveci+3vecj+3veck and vecC=3veci+vecj-2veck`.

Answer» Correct Answer - a= -8, b=4, c=2, d =-11
Here, `vecA,vecB and vecC` are the vectors representing the sides of triangle ABC, where `vecA= ahati + bhatj +chatk`
`vecB = hati + 3hatj + 4hatk and vecC = 3hati + hatj - 2hatk`
Given that `vecA =vecB +vecC`, Therefore,
`ahati + bhatj + chatk = (d+3) hati + 4hatj + 2hatk`
` Rightarrow a=d +3, b=4,c=2`
`vecB xxvecC= |{:(hati ,hatj ,hatk),(d,3,4),(3,1,-2):}|`
`10 hati + (2d + 12) hatj + (d -9) hatk`
Area of `triangle = 1/2 |vecB xx vecC|`
`1/2 sqrt([100 + (2d + 12)^(2) + (d-9)^(2)])`
`5 sqrt6`
` or sqrt((5d^(2)+30d + 325)) = 10 sqrt6`
`or 5d^(2) + 30d - 275 = 0 or d^(2) + 6 d - 55 =0`
`or (d +11) (d-5) =0
`Rightarrow d=5 or -11 `
where d= 5 a= 8, b= 4 and c=2 and where d = -11 , a = -8 ,b =4 and c= 2
Discussion
62.

given that `veca. vecb = veca.vecc, veca xx vecb= veca xx vecc and veca ` is not a zero vector. Show that `vecb=vecc`.

Answer» we have er`veca. Vecb = veca .vecc.` therefore,
`veca.vecb-veca .vecc = 0 or veca. (vecb -vecc) = 0 `
Therefore, there are three possibilities : (i) ,
(ii) `vecb - vecc = vec0 and (iii) veca` is perpendicu
`vecb - vecc`
Again, `veca xx vecb = veca xx vecc`, therefore, `veca xx vecb - veca xx vecc = vec0`
`or veca xx ( vecb - vecc) = vec0`
Therefore, again there are three posibilities,
`(i) veca= vec0, (ii) vecb - vecc = vec0 and (iii) veca` is parallel to `vecb - vecc`.
now ` veca` is given to be a non-zero vector. therefore, we have the following possibilities left :
`1. vecb -vecc= vec0`
2. `veca` is -perendicular to `vecb - vecc and veca` is parallel to `vecb - vecc`, which is absurd.
Therefore, the only possibility , left is `vecb -vecc = vec0 or vecb = vecc`
Discussion
63.

If `veca=-hati+hatj+hatk andvecb=2hati+0hatj+hatk` then find vector `vecc` satisfying the following conditions, (i) that it is coplaner with `veca and vecb` , (ii) that it is `bot "to" vecb` and (iii) that `veca.vecc=7`.

Answer» `Let vecc = x x hati + y hatj+zhatk`
then from conditon (i).
`|{:(x,y,z),(-1,1,1),(2,0,1):}|=0 or x+3y-2z=0`
from condition (ii),
2x+z=0
from condition (iii),
-x+y+z=7
solving (i),(ii) and (iii) we get the values of x,y and z , hence vector `vecc,1/2(-3hati+5hatj+6hatk)`
Discussion
64.

if the vectors `vecc, veca = xhati +yhatj + zhatk and vecb = hatj` are such that `veca , vecc and vecb` from a right -handed system, then find `vecc`.

Answer» Since `veca ,vecc and vecb` form a right - handed system,
`vecc = vecb xx veca`
`= hatj xx (xhati + y hatj + z hatk) `
= - x hatk + zhati = z hati - x hatk`
Discussion
65.

Two vectors in space areequal only if they have equal component ina. a given direction b. two given directionsc. three given directions d. in any arbitrary directionA. a given directionB. two given directionsC. three given directionD. in any arbitrary direaction

Answer» Correct Answer - c
if `vecx = vecy Rightarrow hata.vecx = hata.vecy` this equality must hold for any arbitary `hata`
Discussion
66.

If `veca, vecb and vecc` are vectors such that `|veca|=3,|vecb|=4 and |vec|=5 and (veca+vecb)` is perpendicular to vecc,(vecb+vecc) is perpendicular to veca and (vecc+veca)` is perpendicular to `vecb` then `|veca+vecb+vecc|=` (A) 4sqrt(3)` (B) `5sqrt(2)` (C) 2 (D) 12

Answer» Given , `(veca+vecb)=0Rightarrowveca.vecc+vecb.vecc=0`
`(vecb+vecc).veca=0Rightarrowveca.vecb+vecc.veca=0`
`(vecc+veca).vecb=0Rightarrowvecb.vecc+veca.vecb=0`
`2(veca.vecb+vecb.vecc+vecc.veca)=0``Now, |veca+vecb+vecc|^(2)+|vecb|^(2)+|vecc|^(2)+2(veca.vecb+vecb.vecc+vecc.veca)=50`
`|veca+vecb+vecc|=5sqrt2`
Discussion
67.

Let ` veca, vecb , vecc` be three vectors of equal magnitude such that the angle between each pair is ` pi/3` . If ` |veca + vecb + vec| = sqrt6 , " then " |veca|=`A. 2B. `-1`C. 1D. `sqrt6//3`

Answer» Correct Answer - c
`|veca+ vecb +vecc|^(2)=6`
`or |veca|^(2)+ |vecb|^(2)+|vecc|^(2)`
`Rightarrow |veca|=|vecb|=|vecc|and veca.vecb`
` = |veca|vecb|.cos(pi/3)`
`i.e. veca.vecb= 1/2 |veca|^(2)`
` 3 |veca|^(2)+3|veca|^(2)=6`
`Rightarrow |veca|^(2)or |veca|=1`
Discussion
68.

Prove that `[veca+vecb vecb+vecc vecc+veca]=2[veca vecb vecc]`

Answer» `[veca+vecb vecb +vecc " " vecc+veca]=(veca+vecb).((vecb+vecc)xx(vecc+veca))`
`=(veca+vecb).(vecbxxveccxxvecbxxveca+veccxxvecc)`
`=[veca vecbvecc]+[vecb veccveca]=2[veca vecbvecc]`
Discussion
69.

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 the value of `|vecaxxvecb -veca xx vecc|` isA. `1//2`B. 1C. 2D. none of these

Answer» Correct Answer - b
`|vecaxxvecb-vecaxxvecc|^(2)=|vecaxx(vecb-vecc)|^(2)`
`|veca|^(2)|vecb-vecc|^(2)`
`-(veca.(vecb-vecc))^(2)=|vecb-vecc|^(2)`
`= |vecb|^(2)+|vecc|^92)-2|vecb||vecc|cos (pi/3)=1`
Discussion
70.

Let `veca, vecb, vecc` be three unit vectors and `veca.vecb=veca.vecc=0` . If the angle between `vecb and vecc` is `pi/3` then find the value of `|[veca vecb vecc]|`

Answer» `veca.vecb=veca. Vecc=0`
is perpendicular to vectors `vecb and vecc`. Thus
`veca=lambda(vecbxxvecc)`
`|veca|=|lambda(vecbxxvecc)=|lambdasqrt3/2|=1`
` |[veca vecbvecc]|=|veca.(vecbxxvecc)|`
`=|lambda||(vecbxxvecc)|^(2)`
`=2/sqrt3|vecb|^(2)|vecc|^(2)sin^(2)(pi/3)=2/sqrt3xx(sqrt3/2)^(2)=sqrt3/2`
Discussion
71.

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. 1B. 2C. 3D. 4

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`
Discussion
72.

The vectors which is/are coplanar with vectors `hati+hatj+2hatk and hati+2hatj+hatk` and perpendicular to the vector `hati+hatj+hatk ` is /are (A) `hatj-hatk` (B) `-hati+hatj` (C) `hati-hatj` (D) `-hatj+hatk`A. `hatj - hatk`B. `-hati + hatj`C. `hati -hatj`D. `-hatj +hatk`

Answer» Correct Answer - a,d
Any vector in the plane of `veca = hati +hatj + 2hatk and vecb = hati + 2hatj + hatk` is
`vecr=lamda(hati+hatj+2hatk) +mu(hati+2hatj+hatk)`
`= (lamda+mu)hati+(lamda+2mu)hatj= (2lamda+mu)hatk`
Also `vecr` is perpendicular to the vector `hati + hatj +hatk`
`Rightarrow vecr.vecc=0`
` Rightarrow lamda + mu =0`
Possible vectors are `hatj -hatk or -hatj +hatk`
Discussion
73.

The value of x for which the angle between ` veca = 2x^(2) hati + 4x hatj =hatk +hatk and vecb = 7hati -2hatj =x hatk` , is obtuse and the angle between ` vecb` and the z-axis is acute and less than `pi//6`, areA. `altxlt1//2`B. `1//2ltxlt15`C. `xlt1//2orxlt0`D. none of these

Answer» Correct Answer - b
The angle between `veca and vecb` is obtuse , therefore,
`veca.vecb lt 0`
`Rightaarrow 14x^(2) - 8x + x lt 0`
`or 7x ( 2x - 1) lt 0`
`or 0 lt x lt 1//2`
The angle between `vecb` and the z-axis is acute and less than `pi//6` therefore,
`(vecb.veck)/(|vecb||veck|)gtcospi//6 ( thetalt pi//6 Rightarrowcos thetagtcos pi//6)`
`Rightarrow x/(sqrt(x^(2)+53) gt sqrt3/2`
`or 4x^(2) gt 3x^(2) + 159`
`or x^(2) gt 159`
`Rightarrow x gt sqrt159 or x gt - sqrt159`
clearly, (i) and (ii) cannot hold together.
Discussion
74.

Find the area of the parallelogram whsoe adjacent sides are given by the vectors `veca=hati-hatj+3hatk dn vecb=2hati-7hatj+hatk``

Answer» The area of the parallelgram whose adjacent sides are `veca and vecb is |veca xx vecb|`
Adjacent sides are given as
`veca=hati-hatj+3hatkand vecb=2hati-7hatk+hatk`
`vecaxxvecb=|{:(hati,hatj,hatk),(1,-1,3),(2,-7,1):}|=hati(-1+21)-hatj(1-6)+hatk(-7+2)=20hati=5hatj-5hatk`
`|vecaxxvecb|=sqrt(20^(2)+5^(2)+(-5)^(2))=sqrt(400+25+25)=15sqrt2`
Hence, the area of the given parllelogram is `15sqrt2` square units.
Discussion
75.

If `veca,vecbandvecc` are unit vectors such that `veca+vecb+vecc=0`, then the value of `veca.vecb+vecb.vecc+vecc.veca` is

Answer» squaring `(veca+vecb+vecc)= vec0` we get
`|veca|^(2)+|vecb|^(2)=|vecc|^(2)+2 veca.vecb+ 2vecb.vecc + 2vecc.veca=0`
`2(veca.vecb+vecb.vecc+vecc.veca)=-3`n
`veca.vecb+vecb.vecc+vecc.veca= -3/2`
Discussion
76.

Vector `1/3 (2hati - 2hatj +hatk) ` isA. a unit vectorB. makes an angle `pi//3` with vector `(2hati- 4hatj + 3hatk) `C. parallel to vector `( -hati + hatj - 1/2 hatk)`D. perpendicular to vector `3hati + 2hatj - 2hatk`

Answer» Correct Answer - a,c,d
`veca=1/3(2hati-2hatj+hatk)`
`|veca|^(2)=1/9(4+4+1)=1or |veca|=1`
Let `vecb = 2hati - 4hatj + 3hatj`, then
` cos theta = (veca.vecb)/(|veca||vecb|) = 5/sqrt29 Rightarrow vecc ||veca`
Let `vecd = 3hati + 2hatj + 2hatk, " then " veca. vecd =0 Rightarrrow veca bot vecd`
Discussion
77.

find the projection of the vector `hati+3hatj+7hatk` on the vector `7hati-hatj+8 hatk`

Answer» Let `veca=hati+3hatj+7hatj+7hatkand vecb=7hatj-hatj+8hatk`
Now, projection of vector `veca on vecb` is given by
`1/(|vecb|)(veca.vecb)=1/(sqrt(7^(2)+(-1)^(2)+8^(2))){(7)+3(-1)+7(8)}`
`(7-3=56)/(sqrt(49+1+64))=60/(sqrt114)`
Discussion
78.

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` is

Answer» Correct Answer - 7
Let the vertices be, A ,B , C , D and O be the origin.
`vecOA=hati -6hatj+10hatk,vecOB=hati-3hatj +7hatk`,
`vecOC= -5hati-hatj+lambdahatk,vecOD=7hati -4hatj+7hatk`
`vecAB=vecOB-vecOA= -2hati+3hatj-3hatk`
`vecAC=vecOC-vecOA= -4hati + 5hatj + (lambda-10)hatk`
`vecAC=vecOC -vecOA=4hati+5hatj+(lamda-10)hatk`
`vecAD=vecOD-vecOA = 6hati +2hatj-3hatk`
volume of tetrahedron
`1/6[vecAB vecAC vecAD]=1/6|{:(-2,3,-3),(4,5,lamda-10),(6,2,-3):}|`
`1/6 {-2(-15-2lambda+20)-3(-12-6lambda+60)-3(8-30)}`
`1/6 {4lambda- 10 -144 + 18 lambda+66}`
`= 1/6 (22lambda - 88) =11`
`or 2lambda -8 =6`
`or 2lambda -8 =6`
`or lambda=7`
Discussion
79.

If `veca and vecb` are vectors in space given by `veca= (hati-2hatj)/sqrt5and vecb= (2hati + hatj + 3hatk)/sqrt14` then find the value of `(2veca + vecb) . [(vecaxxvecb) xx (veca- 2vecb)]`

Answer» Correct Answer - 5
`E= (2veca + vecb).[|veca|^(2) vecb-(veca.vecb)veca-2(veca.vecb)vecb +2|vecb|^(2)veca]`
`veca.vecb= (2-2)/sqrt70=0`
`|veca|=1`
`|b|=1`
`E= (2veca+vecb).[ 2|vecb|^(2)veca + |veca|^(2)vecb]`
`4|veca|^(2)|vecb|^(2)+|veca|^(2)(veca.vecb)`
`+2|vecb|^(2)(vecb.veca)+|veca|^(2)|vecb|^(2)`
`5|veca|^(2)|vecb|=5`
Discussion
80.

if `vecr.veci=vecr.vecj=vecr.veck and|vecr|= 3, "then find vector" vecr`.

Answer» let `vecr= x hati+ yhatj+ zhatk`. Since `vecr.hati= vecr.Hatj = vecr.hatk`
x=y=z
Also `|vecr|= sqrt(x^(2)+y^(2)+z^(2))=3`
`x= sqrt3`
Hence, the required vector, `vecr= +-sqrt3(hati+hatj+hatk)`
Discussion
81.

Statement 1: Distance of point D( 1,0,-1) from the plane of points A( 1,-2,0) , B ( 3, 1,2) and C( -1,1,-1) is `8/sqrt229` Statement 2: volume of tetrahedron formed by the points A,B, C and D is `sqrt229/ 2`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.

Answer» Correct Answer - d
`vec(AD)=2hatj -hatk ,vec(BD) =-2hati -hatj -3hatk and vec(CD)=2hati-hatj`
volume of tetrahedron = `1/6 [vec(AD) vec(BD) vec(CD)]`
`=1/6||{:(0,2,-1),(-2,-1,-3),(2,-1,0):}||=8/3`
Also, area of the triangle ABC = `1/2 |vec(AB) xx vec(AC)|`
`1/2||{:(hati,hatj,hatk),(2,3,2),(-2,3,-1):}||`
` 1/2 |-9hati - 2hatj +12 hatk|`
`sqrt229/2`
Then ` 8/3 = 1/3 xx` (Distance of D from base ABC) `xx` (Area of triangle ABC)
Distance of D from base ABC` = 16//sqrt229`
Discussion
82.

If `veca and vecb` are two non collinear vectors `and `vecuveca0(veca.vecb)vecb and vecv=vecaxxvecb` then `|vecv|` is (A) `|vecu|` (B) `|vecu|+|vecu.vecb|` (C) `|vecu|+|vecu.veca|` (D) none of theseA. `|vecu|`B. `|vecu|+ |vecu. Veca|`C. `|vecu| + |vecu.vecb|`D. `|vecu|+ vecu. (veca + vecb)`

Answer» Correct Answer - a,c
we have
`vecv= vecaxxvecb= |veca||vecb| sin theta hatn = sin theta hatn`
where `veca and vecb` are unit vectors. Therefore,
`|vecv|= sin theta`
Now, `vecu = veca - (veca.vecb)vecb`
`= veca -vecb cos theta ( " where " veca. Vecb = cos theta)`
`|vecu|^(2) = | veca-vecb cos theta|^(2)`
` 1 + cos^(2) theta -2 cos theta . cos theta`
` =1 - cos^(2) theta = sin^(2) theta = |v|^(2)`
` Rightarrow |vecu|= |vecv|`
Also , `vecu . vecb = veca. vecb - (veca.vecb) (vecb.vecb)`
` = veca.vecb-veca.vecb=0`
`|vecu.vecb|=0`
`|vecv|=|vecu|+ |vecu.vecb|` is also correct.
Discussion
83.

Let `vecu, vecv and vecw` be three unit vectors such that `vecu + vecv + vecw = veca, vecuxx (vecv xx vecw)= vecb, (vecu xx vecv) xx vecw= vecc, vec a.vecu=3//2, veca.vecv=7//4 and |veca|=2` Vector `vecu` isA. ` 2 veca - 3vecc`B. `3vecb - 4c`C. `-4vecc`D. `veca+vecb + 2vecc`

Answer» Correct Answer - c
taking dot product of `vecu + vecv + vecw =veca "with" vecu,` we have
`1+vecu.vecv+vecu.vecw=veca.vecu=3/2or vecu.vecv+vecu.vecw=1/2 (i)`
similarly,l taking dot product with `vecv`.we have
`vecu .vecv + vecw.vecv = 3/4`
Also , `veca. vecu+ veca.vecv +veca.vecw=veca.veca =4 `
`Rightarrow veca.vecw =4 - (3/2 + 7/4) = 3/4`
Again, taking dot product with `vecw`, we have
`vecu.vecw+vecu.vecw=3/4-1=-1/4`
Adding (i), (ii) and (iii) , we have
`2(vecu.vecv+vecw+vecv.vecw)=1`
`or vecu.vecv+vecu.vecw+vecv.vecw=1/2`
Subtracting (i), (ii) and (iii) form (iv) , we have
`vecv.vecw=0,vecu.vecw -1/4 and vecu.vecv=3/4`
Now , the equations `vecu xx (vecv xx vecw) = vecb and (vecuxx vecv) xx vecw =vecc` can be written as
`(vecu.vecw)vecv- (vecu.vecv) vecw=vecb`
`and (vecu.vecw)vecv- (vecv.vecw)vecu=vecc`
`Rightarrow -1/4vecv-3/4vecw=vecb,-1/4vecv=vecc,i.e, vecv= -4vecc`
`Rightarrow vecc-3/4vecw=vecb Rightarrow vecw=4/3 (vecc-vecb)`
`and vecu=veca-vecv-vecw=veca+4vecc-4/3vecc+4/3vecb`
`veca+4/3vecb+ 8/3vecc`
Discussion
84.

Let `vecu, vecv and vecw` be three unit vectors such that `vecu + vecv + vecw = veca, vecuxx (vecv xx vecw)= vecb, (vecu xx vecv) xx vecw= vecc, vec a.vecu=3//2, veca.vecv=7//4 and |veca|=2` Vector `vecu` isA. `veca-2/3vecb+vecc`B. `veca+4/3vecb+ 8/3vecc`C. `2veca-vecb + 1/3 vecc`D. `4/3veca-vecb+2/3vecc`

Answer» Correct Answer - b
taking dot product of `vecu + vecv + vecw =veca "with" vecu,` we have
`1+vecu.vecv+vecu.vecw=veca.vecu=3/2or vecu.vecv+vecu.vecw=1/2 (i)`
similarly,l taking dot product with `vecv`.we have
`vecu .vecv + vecw.vecv = 3/4`
Also , `veca. vecu+ veca.vecv +veca.vecw=veca.veca =4 `
`Rightarrow veca.vecw =4 - (3/2 + 7/4) = 3/4`
Again, taking dot product with `vecw`, we have
`vecu.vecw+vecu.vecw=3/4-1=-1/4`
Adding (i), (ii) and (iii) , we have
`2(vecu.vecv+vecw+vecv.vecw)=1`
`or vecu.vecv+vecu.vecw+vecv.vecw=1/2`
Subtracting (i), (ii) and (iii) form (iv) , we have
`vecv.vecw=0,vecu.vecw -1/4 and vecu.vecv=3/4`
Now , the equations `vecu xx (vecv xx vecw) = vecb and (vecuxx vecv) xx vecw =vecc` can be written as
`(vecu.vecw)vecv- (vecu.vecv) vecw=vecb`
`and (vecu.vecw)vecv- (vecv.vecw)vecu=vecc`
`Rightarrow -1/4vecv-3/4vecw=vecb,-1/4vecv=vecc,i.e, vecv= -4vecc`
`Rightarrow vecc-3/4vecw=vecb Rightarrow vecw=4/3 (vecc-vecb)`
`and vecu=veca-vecv-vecw=veca+4vecc-4/3vecc+4/3vecb`
`veca+4/3vecb+ 8/3vecc`
Discussion
85.

`vecb and vecc` are non- collinear if `veca xx (vecb xx vecc) + (veca .vecb) vecb = ( 4-2x- sin y) vecb + ( x^(2) -1) vecc andd (vec. vecc) veca =veca ` thenA. x =1B. x = -1C. `y = (4 n+1) pi/2, n in I `D. `y ( 2n + 1) pi/2, n in I`

Answer» Correct Answer - a,c
` veca xx (vecb xx vecc) + (veca .vecb)vecb`
`= ( 4 - 2 x - sin y) vecb + ( x^(2) -1) vecc`
` or (veca .vecc0 vecb - (veca. Vecb) vecc + ( veca . Vecb) vecb`
` = ( 4-2x - sin y) vecb + (x^(2) -1) vecc`
now `(vecc . Vecc) veca= vecc`. therefore,
` (vecc.vecc) (veca.vecc) = (vecc.vecc) or veca. vecc =1 `
` Rightarrow 1+ veca . vecb = 4 -2x-sin y, x^(2) -1 = - (veca. vecb)`
` or 1 = 4 -2x -sin y + x^(2) -1`
but ` sin y le 1 Rightarrow x =1 , sin y =1`
` Rightarrow y = (4 n +1) pi/2, n in I`
Discussion
86.

Statement1: A component of vector `vecb = 4hati + 2hatj + 3hatk` in the direction perpendicular to the direction of vector `veca = hati + hatj +hatk is hati - hatj` Statement 2: A component of vector in the direction of `veca = hati + hatj + hatk is 2hati + 2hatj + 2hatk`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.

Answer» Correct Answer - c
Component of vector `vecb = 4hati + 2hatj +3hatk ` in the direction of `veca=hati +hatj +hatk is (veca.vecb)/ (|veca|) veca/ (|veca|) or 3hati + 3hatj + 3hatk `. Then componant in the direaction perpendicular to the direction of `veca =hati+hatj + hatk "is " vecb -3hati + 3hatj +3hatk= hati -hatj`
Discussion
87.

Let `veca=a_(1)hati+a_(2)hatj+a_(3)hatk,vecb=b_(2)hatj+b_(3)hatk and vecc=c_(1)hati+c_(2)hatj+c_(3)hatk` gve 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 prove that `|{:(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3)):}|p=1/4 (a_(1)^(2)+a_(2)^(2)+a_(3)^(2))(b_(1)^(2)+b_(2)^(2)+b_(3)^(2))`

Answer» `|{:(a_(1) , a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3)):}|^(2)= [veca vecb vecc]^(2)`
` = ((veca xx vecb) .vecc)^(2)`
` ( ab sin theta vecc. vecc)^(2)` .
`(a^(2)b^(2))/4`
` 1/4 (a_(1)^(2) + a_(2)^(2)+a_(3)^(2))(b_(1)^(2) =b_(2)^(2) + b_(3)^(2)) `
Discussion
88.

If `veca, vecb and vecc` are non - zero vectors such that `veca.vecb= veca.vecc`,.the find the goemetrical relation between the vectors.

Answer» ` veca.vecb = veca.vecc`
`veca. Vecb-veca . Vecc =vec0`
`veca. (vecb.vecc)=vec0`
Either `vecb=-vecc=vec0 or veca bot (vecb-vecc)`
`Rightarrow vecb = vecc or veca (vecb- vecc)`
Discussion
89.

If the unit vectors ` veca and vecb ` are inclined of an angle ` 2 theta` such that ` |veca -vecb| lt 1 and 0 le theta le pi` then ` theta` in the intervalA. `[0, pi//6)`B. `(5 pi//6, pi]`C. `[pi//6, pi//2]`D. `(pi//2, 5pi//6]`

Answer» Correct Answer - a,b
we have `|veca -vecb|^(2) = |veca|^(2) + |vecb|^(2) -2 (veca. Vecb)`
`or |vec-vecb|^(2) = |veca |^(2) + |vecb|^(2) -2 |veca||vecb|cos2theta`
` |veca -vecb|^(2) = 2 - 2cos theta " " (|veca|=|vecb|=1)`
` = 4sin^(2) theta `
` or |veca -vecb| =2 |sin theta|`
Now, `|veca- vecb| gt 1`
` Rightarrow 2|sin theta|lt 1`
` or |sin theta| lt 1/2 `
` Rightarrrow theta in [ 0, pi//6) or theta in ( 5 pi//6, pi]`
Discussion
90.

`veca and vecb` are two unit vectors that are mutually perpendicular. A unit vector that if equally inclined to `veca, vecb and veca xxvecb` is equal toA. `1/sqrt2(veca+vecb+vecaxxvecb)`B. `1/2(vecaxxvecb+veca+vecb)`C. `1/sqrt3(veca+vecb+vecaxxvecb)`D. `1/3(veca+vecb+vecaxxvecb)`

Answer» Correct Answer - a
Let the required vector `vecr` be such that
` vecr =x_(1) veca + x_(2) vecb = x_(3) veca xx vecb`
we must have `vecr. Veca = vecr .vecb = vecr . (veca xx vecb) `
` vecr,veca, vecb and veca xx vecb` are unit vectors and `vecr` is equally inclined to `veca ,vecb and veca xx vecb)`
Now ` vecr. veca = x_(1) , vecr. vecb = x_(2) vecr. (veca xx vecb) = x_(3)`
`Rightarrow vecr = lambda ( veca + vecb + (veca xx vecb))`
Also , `vecr .vecr=1`
` Rightarrow lambda^(2) (veca + vecb + veca xx vecb) . (veca + vecb + (veca xx vecb)) =1 `
`or lambda^(2) (|veca|^(2) + |vecb|^(2) + |veca xx vecb|^(2) ) =1`
` or lambda^(2) = 1/3`
`or lambda = +- 1/sqrt3`
` Rightarrow vecr = +- 1/sqrt3 (veca = vecb + vecaxx vecb) `
Discussion
91.

` vec a , vec b ,a n d vec c`are three unit vectors andevery two are inclined to each other at an angel `cos^(-1)(3//5)dot`If ` vec axx vec b=p vec a+q vec b+r vec c ,w h e r ep ,q ,r`are scalars, then find thevalue of `qdot`

Answer» `veca xx vecb = pveca =qvecb + rvecc`
taking dot product with `veca, vecb and vecc`, we get
`0 = P + 3/5 q = 3/5 r`
`0 = 3/5 p+q+ 3/5 r`
`[veca vecb vecc] = 3/5p + 3/5 q + r `
`Also, [veca vecb vecc] ^(2)=|{:(veca.veca,veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc),(vecc.veca,vecc.vecb,vecc.vecc):}|`
`=[{:(1,3//5,3//5),(3//5, 1, 3//5),(3//5,3//5,1):}]= 44/125` solving (i), (ii) and (iii) for q, we get
`5/11 [veca vecb vecc] = - 2/3 q `
` 5/11 xx sqrt(44/125) = - 2/3 q`
` q = - 3 /sqrt55`
Discussion
92.

Find the angel between the following pairs of vectors `3hati+2hatj-6hatk, 4hati-3hatj+hatkhati-2hatj+3hatk, 3hati-2hatj+hatk`

Answer» the given vectors are
`|oversettoa|= sqrt(1^(2)+(-2)^(2)+3^(2))=sqrt(1+4+9)= sqrt14`
`|vecb|= sqrt(3^(2)+(-2)^(2)+1^(2))=sqrt(9+4+1)= sqrt14`
now, `oversettoa.oversetto= (hati-2hatj+3hatk).(3hati-2hatj+hatk)`
= 1.3+ (-2)(-2)+ 3.1
= 3+4+3
= 10
now , `oversettoa.oversetb= |oversettoa||oversettob|costheta.`
`10 = sqrt14sqrt14costheta`.
` cos theta= 10/14`
`theta= cos^(-1)(5/7)`
Discussion
93.

Let `veca, vecb` and `vecc` be three vectors having magnitudes 1,1 and 2 resectively. If `vecaxx(vecaxxvecc)+vecb=vec0` then the acute angel between `veca` and `vecc` is

Answer» Correct Answer - `pi//6`
`veca xx (vecaxxvecc) +vecb=vec0`
`or (veca.vecc)veca-(veca.veca)vecc+vecb=vec0`
`or 2 cos theta. veca-vecc+vecb=vec0`
(using `|veca|=1,|vecb|=1,|vecc|=2`)
`or (2costheta veca-vecc)^(2)=(-vecb)^(2)`
`or 4cos^(2)theta.|veca|^(2)+|vecc|^(2)`
`-2.2 cos theta.veca.vecc=|vecb|^(2)`
`or 4cos^(2)theta+4-8costheta.costheta=1`
`or cos^(2)theta-8cos^(2)theta+4=1`
`or 4 cos^(2) theta=3`
`or cos theta = +- sqrt3//2`
for `theta` to be acute , `cos theta= sqrt3/2 Rightarrow theta= pi/6`
Discussion
94.

If `veca and vecb` are two unit vectors inclined at an angle `pi//3 then { veca xx (vecb+veca xx vecb)} .vecb` is equal toA. `(-3)/4`B. `1/4`C. `3/4`D. `1/2`

Answer» Correct Answer - a
` { veca xx (vecb + veca xx vecb)} .vecb`
` { veca xx vecb + veca xx (veca xx vecb)} .vecb`
`= [veca vecb vecb] = {( veca.vecb) veca - (veca .veca) (vecb.vecb)`
` cos^(2) pi/3 -1 = -3/4`
Discussion
95.

Prove that vectors` vec u=(a l+a_1l_1) hat i+(a m+a_1m_1) hat j+(a n+a_1n_1) hat k`` vec v=(b l+b_1l_1) hat i+(b m+b_1m_1) hat j+(b n+b_1n_1) hat k`` vec w=(b l+b_1l_1) hat i+(b m+b_1m_1) hat j+(b n+b_1n_1) hat k`

Answer» `[vecuvecv vecw]=|{:(al=a_(1)l_(1),am+a_(1)m_(1),an+a_(1)n_(1)),(bl+b_(1)l_(1),bn+b_(1)m_(1),bn+b_(1)n_(1)),(cl+c_(1)l_(1),cm+c_(1)m_(1),cn+c_(1)n_(1)):}|`
`=|{:(a,a_(1),0),(b,b_(1),0),(c,c_(1),0):}||{:(l,l_(1),0),(m,m_(1),0),(n,n_(1),0):}|=0`
Therefore, the given vectors are coplanar.
Discussion
96.

Position vector ` hat k`is rotated about the originby angle `135^0`in such a way that theplane made by it bisects the angel between ` hat ia n d hatjdot`Then its new position is`+-( hat i)/(sqrt(2))+-( hat j)/(sqrt(2))`b. `+-( hat i)/2+-( hat j)/2-( hat k)/(sqrt(2))`c. `( hat i)/(sqrt(2))-( hat k)/(sqrt(2))`d. none of theseA. `+-hati/sqrt2+-hatj/sqrt2`B. `+-hati/2+-hatj/2-hatk/sqrt2`C. `hati/sqrt2-hatk/sqrt2`D. none of these

Answer» Correct Answer - d
Let `vecr` be the new position . Then `vecr= lambdahatk + mu ( hati + hatj) `
` Also, vecr.hatk = - 1/sqrt2 Rightarrow lambda= - 1/sqrt2`
` Also lambda^(2)+ 2mu^(2) = 1 Rightarrow 2mu^92) = 1/2 or mu = +- 1/2`
` vecr = +- 1/2 (hati + hatj) =- hatk/sqrt2`
Discussion
97.

Let `veca=hati-hatj, vecb=hatj-hatk, vecc=hatk-hati. If hatd` is a unit vector such that `veca.hatd=0=[vecb vecc vecd]` then `hatd` equalsA. `+- (hati + hatj - 2hatk)/sqrt6`B. `+- (hati + hatj - hatk)/sqrt3`C. `+- (hati + hatj + hatk)/sqrt3`D. `+- hatk`

Answer» Correct Answer - a
Let ` vecd= xhati + yhatj + zhatk`
where `x^(2) + y^(2) +z^(2)=1`
( `vced ` being a unit vector)
`veca .vced=0`
` Rightarrow x-y =0 or x=y`
`[vecb vecc vecd]=0`
`Rightarrow |{:(0,1,-1),(-1,0,1),(x,y,z):}|=0`
or x+y +z=0
or 2x + z=0
or z= -2x
From (i), (ii), and (iii) we have
`x^(2) +x^(2)+4x^(2)=1`
`x = +- 1/sqrt6`
`vecd=+-(1/sqrt6hati1/sqrt6hatj-2/sqrt6hatk)`
`= +- ((hati+hatj -2hatk)/sqrt6)`
Discussion
98.

A vector of magnitude 10along the normal to the curve `3x^2+8x y+2y^2-3=0`at its point `P(1,0)`can be`6 hat i+8 hat j`b. `-8 hat i+3 hat j`c. `6 hat i-8 hat j`d. `8 hat i+6 hat j`A. `6hati + 8hatj`B. `-8 hati + 3hatj`C. `6hati - 8 hatj`D. `8 hati + 6 hatj`

Answer» Correct Answer - a
Differentiate the curve
`6x + 8 ( xy_(1) + y) + 4 yy_(1) =0`
`m_(T)at (1,0) is 6 + 8 (y_(1)(0)) =0`
` y_(1) (0) = -3/4`
`m_(N)= 4/3`
Unit vector =` +- (3hati + 4hatj) / 5`
Again normal of magnitude ` 10 =+- (6hati + 8hatj)`
Discussion
99.

Let the pair of vector `veca,vecb` and `vecc,veccd` each determine a plane. Then the planes are parallel ifA. `(vecaxxvecc)xx(vecbxxvecd)=vec0`B. `(vecaxxvecc).(vecbxxvecd)=vec0`C. `(vecaxxvecc)xx(veccxxvecd)=vec0`D. `(vecaxxvecc).(veccxxvecd)=vec0`

Answer» Correct Answer - c
`veca xx vecb` is a vector perpendicular to the plane containing `veca and vecb` similarly, `vecc xx vecd` is a vector perpendicular to the plane containing , `vecc and vecd` thus , the two planes will be parallel if their normal i.e., `veca xx vecb and vecc xx vecd` are parallel. thus
`(veca xx vecb) xx (vecc xx vecd)) = vec0`
Discussion
100.

If `vecr.veca=vecr.vecb=vecr.vecc=1/2` for some non zero vector `vecr and veca,vecb,vecc` are non coplanar, then the area of the triangle whose vertices are `A(veca),B(vecb) and C(vecc)` isA. `|[veca vecb vecc]|`B. `|vecr|`C. `|[veca vecb vecc]vecr|`D. none of these

Answer» Correct Answer - c
Any vector, `vecr` can be represented in terms of three non- coplanar vectors, `veca, vecb and vecc` as
`vecr = x (veca xx vecb) + y (vecb xx vecc) +z (vecc xx veca) ` (i)
taking dot product with `veca , vecb and vecc` respectively, we have.
`x=(vecr.vecc)/([vecavecb vecc]),y = (vecr.veca)/([veca vecb vecc])and z= (vecr.vecb)/([veca vecb vecc])`
From (i) we have
`[veca vecb vecc] vecr= 1/2 (veca xx vecb + vecb+vecc +veccxxveca)`
Area of `triangleABC`
` 1/2 |veca xx vecb+vecb xx vecc +vecc +vecc xx veca|`
`|[veca vecb vecc] vecr|`
Discussion