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1.

The volume of the parallelepiped whose coterminous edges are represented by the vectors `2vecb xx vecc, 3vecc xx veca` and `4veca xx vecb` where `veca=(1+sintheta)hati+costhetahatj+sin2thetahatk` , `vecb=sin(theta+(2pi)/(3))hati+cos(theta+(2pi)/(3))hatj+sin(2theta+(4pi)/(3))hatk`, `vecc=sin(theta-(2pi)/(3))hati+cos(theta-(2pi)/(3))hatj + sin(2theta-(4pi)/(3))hatk` is 18 cubic units, then the values of `theta`, in the interval `(0,pi/2)`, is/areA. `pi/9`B. `(2pi)/(9)`C. `pi/3`D. `(4pi)/9`

Answer» Correct Answer - A::B::D
Volume `=|[2vecb xx vecc 3 vecb]|=18`
`rArr 24[vecavecbvecc]^(2)=18`
`rArr [|vecavecbvecc|]=sqrt(3)/2`
Now, `[vecavecbvecc]= |{:(1+sintheta,costheta,sin2theta),(sintheta+(2pi)/(3),costheta+(2pi)/(3),sin2theta+(4pi)/(3)),(sintheta-(2pi)/(3), costheta-(2pi)/(3), sin2theta-(4pi)/(3)):}|`
Applying `R_(1) to R_(1) + R_(2)+R_(3)` and expanding
`|[vecavecbvecc]|=sqrt(3)|cos3theta|=sqrt(3)/2`
`rArr cos3theta=+-1/2 rArr 3theta=pi/3,(2pi)/(3), (4pi)/(3)`
`rArr theta=pi/9, (2pi)/9, (4pi)/9`
Discussion
2.

Let `(hatp xx vecq) xx (hatp.vecq)vecq` `=(x^(2)+y^(2))vecq + (14-4x-6y)vecp` Where `hatp` and `hatq` are two non-collinear vectors `vecp` is unit vector and x,y are scalars. Then the value of `(x+y)` isA. 4B. 5C. 6D. 7

Answer» Correct Answer - B
`(vecp xx vecq) xx hatp+(hatp.vecc)vecq=(x^(2)+y^(2))vecq+(14-4x-6y)hatp`
`rArr (hatp.hatp)vecq+(1-4x-6y)hatp`
Since `hatp` and `vecq` are non-zero non-collinear.
We can compare coefficients of vectors `hatp` and `hatq`.
`therefore 1+hatp.vecq=x^(2)+y^(2)`.............(i)
And `hatp.vecq=4x+6y-14`............(ii)
From (i) and (ii), we get
`rArr x^(2) + y^(2)-4x-6y+13=0`
`rArr (x-2)^(2)+(y-3)^(2)=0`
x=2 and y=3
Discussion
3.

If `veca, vecb, vecc` are non coplanar vectors and `vecp, vecq, vecr` are reciprocal vectors, then `(lveca+mvecb+nvecc).(lvecp+mvecq+nvecr)` is equal toA. `l^(2)+m^(2)+n^(2)`B. lm+mn+nlC. 0D. None of these

Answer» Correct Answer - A
The vectors reciprocal to `veca,vecb,vecc`, are given by `vecp=(vecb xx vecc)/[vecavecbvecc], vecq= (vecc xx veca)/[vecavecbvecc], vecr = (veca xx vecb)/([vecavecbvecc])`such that `veca.vecp=1, veca.vecq=veca.vecr=0`,
`vecb.vecq=1, vecb.vecp=vecb.vecr=0`
`vecc.vecr=1, vecc. vecp=vecc.vecq=0`
This gives `(lveca+mvecb+nvecc).(lvecp+mvecq+nvecr)`
`=l^(2)+m^(2)+n^(2)`.
Discussion
4.

If `veca, vecb, vecc` are three non-zero vectors, then which of the following statement(s) is/are true?A. `veca xx (vecb xx vecc), vecb xx (vecc xx veca), (vecc xx veca), vecc xx (veca xx vecb)` form a right handed systemB. `vecc, (veca xx vecb) xx vecc, veca xx vecb` from a right handed systemC. `veca.vecb+vecb.vecc+vecc.veca lt 0` if `veca+vecb+vecc=vec0`D. `((veca xx vecb).(vecb xx vecc))/((vecb xx vec).(veca xx vecc))=-1` if `veca + vecb + vecc=0`

Answer» Correct Answer - B::C::D
a) `veca xx (vecb xx vecc) + vecb xx (vecc xx veca) + vecc xx (veca xx vecb)=vec0`
`rArr` Vectors are coplanar, so do not form RHS.
b) `(veca xx vecb) xx vecc, veca, veca xx vecb, vecc` in that order from RHS.
`rArr vecc, (veca xx vecb) xx vecc, veca xx vecb` also form RHS as they are in same cycle order.
c) `veca + vecb+vecc=0 rArr (veca +vecb+vecc)=0`
`rArr veca^(-2)+vecb^(-2)=-2a(veca.vecb+vecb.vecc+vecc.veca)`
Hence, `veca.vecb+vecb.vecc+vecc.veca lt 0`
d) `veca+vecb+vecc=0 rArr veca xx vecb=vecb xx vecc = vecc xx veca`
Using this we get result.
Discussion
5.

Vectors `veca, vecb, vecc` are three unit vectors and `vecc` is equally inclined to both `veca` and `vecb`. Let `veca xx (vecb xx vecc) + vecb xx (vecc xx veca)` `=(4+x^(2))vecb-(4xcos^(2)theta)veca`, then `veca` and `vecb` are non-collinear vectors, `x gt 0`A. `x=2`B. `theta=0^(@)`C. `theta=x`D. `x=4`

Answer» Correct Answer - A::B::C
`|veca|=|vecb|=|vecc|=1`
and `veca.vecc=vecb.vecc`………….(i)
`veca xx (vecb xx vecc) +vecb xx (vecc xx veca)`
`=(vecc. vecc)vecb-(veca.vecb)vecc+(veca.vecb)vecc-(vecb.vecc)veca`
`=(veca.vecc)vecb-(vecb.vecc)veca`
`rArr veca.vecc+x^(2)` and `vecb.vecc=4xcos^(2)theta`. ........(ii)
From (i) and (ii), `4+x^(2)=4xcos^(2)theta`
`rArr 4+x^(2)=4x-4sin^(2)theta`
`rArr (x-2)^(2)=-4xsin^(2)theta`
`rArr x=2` or `sintheta=0`.
Discussion
6.

`veca=2veci+vecj+veck`, `vecb=b_(1)hati+b_(2)hatj+b_(3)hatk`, `veca xx vecb=5hati+2hatj-12hatk, veca.vecb=11`, then `b_(1)+b_(2)+b_(3)=`A. 3B. 5C. 7D. 9

Answer» Correct Answer - B
`vecb= ((vecaxxvecb)xx veca+(veca.vecb)veca)/(|veca|^(2))`
`=6hati-3hatj+2hatk`
`rArr vecb_(1)+vecb_(2)+vecb_(3)=6-3+2=5`
Discussion
7.

If `veca` and `vecb` are unequal unit vectors such that `(veca -vecb) xx [(vecb+veca) xx (2veca+vecb)]=veca+vecb`, then angle `theta` between `veca` and `vecb` can beA. `pi/2`B. 0C. `pi`D. `pi/4`

Answer» Correct Answer - A::C
`(veca-vecb) xx [(vecb + veca) xx (2veca + vecb)]=vecb + veca`
`rArr {(veca -vecb).(2veca+ vecb)}(vecb+veca)-{(veca-vecb).(vecb+veca)}(2veca+vecb)=vecb+veca`
`rArr vecb+veca=vec0` or `1-veca.vecb=1`
`rArr vecb=-veca`or `veca.vecb=0`
`rArr theta=pi` or `theta=pi/2`
Discussion
8.

if `veca=hati+hatj+2hatk, vecb=hati+2hatj+2hatk` and `|vecc|=1` Such that `[veca xx vecb vecb xx vecc vecc xx veca]` has maximum value, then the value of `|(veca xx vecb) xx vecc|^(2)` is

Answer» Correct Answer - A
`[veca xx vecbvecb xx veccvecc xx veca]`
`=[vecavecbvecc]^(2) = |vecc. (veca xx vecb)|^(2) cos^(2)theta`
The maximum value of `[vecavecbvecc]` is possible only when `vecc` is parallel to `veca xx vecb`.
`rArr theta=0` or `pi`
Hence, `|(veca xx vecb) xx vecc|^(2) =|veca xx vecb|^(2)|vecc|^(2)sin^(2)theta=0`
Discussion
9.

`veca=2hati+hatj+2hatk, vecb=hati-hatj+hatk` and non zero vector `vecc` are such that `(veca xx vecb) xx vecc = veca xx (vecb xx vecc)`. Then vector `vecc` may be given asA. `4hati+2hatj+4hatk`B. `4hati-2hatj+4hatk`C. `hati+hatj+hatk`D. `hati-4hatj+hatk`

Answer» Correct Answer - A
`(veca xx vecb) xx vecc = veca xx (vecb xx vecc)`
`rArr (veca.vecc)vecb-(vecb.vecc)veca`
`=(veca.vecc)vecc-(veca.vecb)vecc`
`rArr veca` and `vecc` are parallel.
`therefore vecc` may be equal to `4hat(i)+2hatj+4hatk`
Discussion
10.

If the angles between the vectors `veca` and `vecb,vecb` and `vecc,vecc` an `veca` are respectively `pi/6,pi/4` and `pi/3`, then the angle the vector `veca` makes with the plane containing `vecb`and `vecc`, isA. `cos^(-1)sqrt(1-sqrt(2//3))`B. `cos^(-1)sqrt(2-sqrt(3//2))`C. `cos^(-1)sqrt(sqrt(3//2)-1)`D. `cos^(-1)sqrt(sqrt(2//3))`

Answer» Correct Answer - B
Let `theta` be angle between `veca` and plane containing `vecb` and `vecc`.
`therefore 90^(@)-theta`= angle b/w `veca` and `vecb xx vecc`
`|hata xx (hatb xx hatc)|^(2)=|(hata.hatc)hatb-(hata.hatb)hatc|^(2)`
`=sin^(2)(90^(@)-theta)sin^(2)pi/4 = |(cospi/3)hatb-(cospi/6)hatc|^(2)`
`rArr cos^(2)theta xx 1/2 =1/4 +3/4-2 xx 1/2 xx sqrt(3)/2 xx 1/sqrt(2)`
`=1-sqrt(3)/2sqrt(2)`
`rArr cos^(2)theta=2-sqrt(3//2)`
Discussion
11.

Volume of parallelogram whose adjacent sides are given by `veca, vecb, vecb xx vecc` is,A. 18B. 54C. 12D. 36

Answer» Correct Answer - D
`vecb xx vecc = |{:(hati,hatj,hatk),(1,-1,1),(4,2,4):}|=-6hati+6hatk`
`therefore [vecavecb xx vecc]=|{:(2,1,2),(1,-1,1),(-6,0,6):}|=-36`
`therefore` Volume =36
Discussion
12.

A vector along the bisector of angle between the vectors `vecb` and `vecc` is,A. `(2+sqrt(3))hati + (1-sqrt(3))hatj+(2+sqrt(3))hatk`B. `(2+sqrt(3))hati+(1-sqrt(3))hatj-(2+sqrt(3))hatk`C. `(2+sqrt(3))hati-(1-sqrt(3))hatj-(2+sqrt(3))hatk`D. `(2+sqrt(3))hati-(1-sqrt(3))hatj+(2+sqrt(3))hatk`

Answer» Correct Answer - A
Vector along the bisectors is
`(hati-hatj+hatk)/sqrt(3) + (4hati+2hatj+4hatk)/6`
`1/3(2+sqrt(3))hati+(1-sqrt(3))hatj+(2+sqrt(3))hatk`
`therefore` Vectors is `(2+sqrt(3))hati+(1-sqrt(3))hatj+(2+sqrt(3))hatk`
Discussion