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1.

If `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 and [3veca+vecb 3vecb+vecc 3vecc + veca] =lambda|{:(veca.hati,veca.hatj,veca.hatk),(vecb.hati,veca.hatj,hatb.hatk),(vecc.hati,vecc.hatj,vecc.hatk):}| " then find the value of " lambda/4`

Answer» Correct Answer - 7
`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`
L.H.S = ` [3veca +vecb 3vecb+vecc 3vecc +veca]`
`[3veca 3vecb 3vecc] + [vecb vecc veca]`
`3^(3)[veca vecb vecc] + [veca vecb vecc]`
`28 [veca vecb vecc]`
Discussion
2.

`veca, vecb and vecc` are unit vecrtors such that `|veca + vecb+ 3vecc|=4` Angle between `veca and vecb is theta_(1)` , between `vecb and vecc is theta_(2)` and between `veca and vecb` varies `[pi//6, 2pi//3]` . Then the maximum value of `cos theta_(1)+3cos theta_(2)` isA. 3B. 4C. `2sqrt2`D. 6

Answer» Correct Answer - b
`|veca + vecb + 3 vecc |^(2) = 16`
`Rightarrow |veca|^(2) + |vecb|^(2) + 9 |vecc|^(2) + 2cos theta_(1) + 6 cos theta_(2)`
` + 6 cos theta_(3) = 16, theta_(3) ne [ pi//6, 2pi//3]`
`or 2 cos theta_(1)+ 6 cos theta_(2) = 5-6 cos theta_(3)`
` or ( cos theta_(1) + 3 costheta_(2))_(max) = 4`
Discussion
3.

The volume of a tetrahedron fomed by the coterminus edges `veca , vecb and vecc is 3` . Then the volume of the parallelepiped formed by the coterminus edges `veca +vecb, vecb+vecc and vecc + veca` isA. 6B. 18C. 36D. 9

Answer» Correct Answer - c
`3 =-1/6 [veca vecb vecc] `
`or [veca vecb vecc] =m 18`
volume of the required parallelepiped lt brgt `[ veca + vecb vecb + vecc vecc + veca]`
` 2[veca vecb vecc] =36 `
Discussion
4.

Let `vecb=4hati+3hatj and vecc` be two vectors perpendicular to each other in the xy-plane. Find all vetors in te same plane having projection 1 and 2 along `vecb and vecc` respectively.

Answer» Correct Answer - `2hati-hatj`
Let `vecc = alpha hati + beta hatj`
Give n that ` vecb bot vecc`
` vecb.vecc=0`
`Rightarrow (4hati+3hatj) .(alphahati+betahatj)=0`
`or 4alpha + 3beta=0`
`or alpha/3 = beta/(-4) =0`
`or alpha=3 lambda, beta= -4 lambda`
Now let `veca = xhati + yhatj` be the required vectors,porjection of `veca " along " vecb` given
`(veca.vecb)/(|vecb|) = (4x + 3y)/(sqrt(4^(2)+3^(2)))=1`
`or 4x + 3y =5 `
Also projection of `veca ` along `vecc` given
`(veca.vecc)/(|vecc|)=2`
` Rightarrow (alphax +betay)/(sqrt(alpha^(2) +beta^(2)))=2`
`or 3lambda xx - 4 lambday = 10 lambda`
`or 3x - 4y =10`
solving (ii) and (iii) we get x=2,y=-1
therefore, the required vector is `2hati - hatj`
Discussion
5.

Let `veca and vecb` be unit vectors that are perpendicular to each other l. then `[veca+ (veca xx vecb) vecb + (veca xx vecb) veca xx vecb]` will always be equal toA. 1B. 0C. `-1`D. none of these

Answer» Correct Answer - a
`[ veca + (hata xx vecb) vecb + (veca xx vecb) veca xx vecb]`
` (veca + (veca xx vecb)) . (( vecb + (veca xx vecb))xx (veca xx vecb))`
` (veca + (veca xx vecb)) . (vecb xx (veca xx vecb))`
` (veca + (veca xx vecb)) . (veca- (veca.vecb) vecb)`
`veca. veca =1 " " (as veca. vecb =0, veca, (veca xx vecb) =0)`
Discussion
6.

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 + vecc)xxvecb- vecb -veca]`B. `1/2 [(veca - vecc)xxvecb+ vecb +veca]`C. `1/2 [(veca - vecb)xxvecc+ vecb +veca]`D. `1/2[(veca-vecc)xx veca + vecb -veca]`

Answer» Correct Answer - c
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)]`
Discussion
7.

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-vecc) xx vecc-vecb+veca]`B. `1/2[(veca-vecb) xx vecc+vecb-veca]`C. `1/2[veccxx(veca-vecb) + vecb +veca]`D. none of these

Answer» Correct Answer - b
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)]`
Discussion
8.

Let `veca=hati+4hatj+2hatk,vecb=3hati-2hatj+7hatk and veca=2hati-2hatj+4hatk` . Find a vector `vecd` which perpendicular to both `veca and vecb and vecc.vecd=15`.

Answer» Vector `vecd` is perpendicular to vectors `veca=hati+4hatj+2hatk,hatn=3hati-2hati-2hatj+7hatk`
`vecd=lambda|{:(hati,hatj,hatk),(1,4,2),(3,-2,7):}|=lambda(32hati-hatj-14hatk)`
`Also" " vecc.vecd=15`
`lambda(2hati-hatj+4hatk).(32hati-hatj-14hatk)15`
` or 9 lambda =15`
`or lambda = 5/3`
Hence, the required vector `vecd=1/3(160hati-5hatj-70hatk)`
Discussion
9.

The value of a so thast the volume of parallelpiped formed by vectors `hati+ahatj+hatk, hatj+ahatk, ahati+hatk` becomes minimum is (A) `sqrt93)` (B) 2 (C) `1/sqrt(3)` (D) 3

Answer» `V=|{:(1,a,1),(0,1,a),(a,0,1):}|=1-a+a^(3)`
`Rightarrow" " (dV)/(da)=3a^(2)-1`
V is minimum at a `=1/sqrt3`
Discussion
10.

If `vecr and vecs` are non-zero constant vectors and the scalar b is chosen such that `|vecr+bvecs|` is minimum, then the value of `|bvecs|^(2)+|vecr+bvecs|^(2)` is equal toA. `2|vecr|^(2)`B. `|vecr|^(2)//2`C. `3|vecr|^(2)`D. `|vecr|^(2)`

Answer» Correct Answer - b
For minimum value `|vecr+ bvecs|=0`
Let `vecr and vecs` are anti - parallel so `bvecs =- vecr`
`|bvecs|^(2) + |vecr + bvecs|^(2) = |-vecr|^(2) + |vecr-vecr|^(2) = |vecr|^(2) `
Discussion
11.

If `P`is any arbitrary point onthe circumcirlce of the equllateral trangle of side length `l`units, then `| vec P A|^2+| vec P B|^2+| vec P C|^2`is always equal to`2l^2`b. `2sqrt(3)l^2`c. `l^2`d. `3l^2`A. `2l^(2)`B. `2sqrt3l^(2)`C. `l^(2)`D. `3l^(2)`

Answer» Correct Answer - a
Let P.V. of A,B and C be `vecp, veca, vecb and vecc`
respectively, and `O(vec0)` be the circumcentre of equilateral traingle ABC. Then
`|vecP| = |vecb| = |veca|= |vecc| = l/ sqrt3`
Now `|vec(PA)|^(2) = |veca - vecp|^(2)= |veca|^(2) + |vecp|^(2) - 2vecp`
` and |vecPC|^(2) = |vecc|^(2) + |vecp|^(2) - 2vecp. vecc`
` Rightarrow sum|vec(PA)|^(2) = 6. l^(2)/3 - 2vecp . (veca + vecb + vecc)`
` 2l^(2) (as veca + vecb + vecc//3 = vec0)`
Discussion
12.

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-hatj+5hatk`C. `2hati+3hatj+3hatk`D. `2hati+hatj+5hatk`

Answer» Correct Answer - b
Let the required vector be `vecr`. Then
`vecr = x_(1) vecb + x_(2) vecc and vecr .veca = sqrt(2/3) " " (|veca|)=2`
Now ` vecr.veca = x_(1) veca. Vecb + x_(2) veca.vecc`
` Rightarrow 2 = x_(1) ( 2-2-1) + x_(2) ( 2-1-2)`
` Rightarrow x_(1) + x_(2) = -2`
` Rightarrow vecr = x_(1) ( hati + 2hatj -hatk) + x_(2) (hati +hatj - 2hatk)`
`= hati (x_(1) + x_(2) +hatj ( 2x_(1) + x_(2) ) -hatk ( 2x_(2) +x_(1))`
` 2hati + hatj (x_(1) -2) -hatk(-4-x_(1))`,
Discussion
13.

If `vecaxxvecb=veccxxvecd and vecaxxvecc=vecbxxvecd` show that `(veca-vecd)` is parallel to `(vecb-vecc). It is given that `vec!=vecd and vecb!=vecc.

Answer» `{:("we have ",vecaxxvecb=veccxxvecd),(and,vecaxxvecc=vecbxxvecd):}]`
`veca-vecd "will be parallel to" vecb-vecc`
`if (veca-vecd)xx(vecb-vecc)=vec0`
` if(veca-vecd)xx(vecb-vecc)=vec0`
`i.e. if vecaxxvecb-vecaxxvecc-vecd xxvecb+vecd xxvecc=vec0`
`if (vecaxxvecb+vecd xx vecc)-(vecaxxvecc+vecd xxvecb)=vec0`
`if (veca xx vecb-veccxxvecd)-(vecaxxvecc-vecb xxvecd)=vec0`
`if vec0-vec0=vec0`
`vec0=vec0` which is ture
Hence the result.
Discussion
14.

The position vectors of points A,B and C are `hati+hatj,hati + 5hatj -hatk and 2hati + 3hatj + 5hatk`, respectively the greatest angle of triangle ABC isA. `120^(@)`B. `90^(@)`C. `cos^(-1)(3//4)`D. none of these

Answer» Correct Answer - b
Since `vec(OA) = hati +hatj + hatk`
` vec(OB) = hati + 5hatj -hatk`
`vec(OC) = 2hati + 3hatj + 5hatk`
` a = BC |vec(BC)|= |vec(OC) - vec(OB)|`
`|hati - 2hatj + 6hatk| = sqrt41`
`b = CA= |vec(CA)|= |vec(OA) -vec(OC)|`
` = | -hati - 2hatj - 4hatk| = sqrt21`
`and c= AB= |vec(AB)| = |vec(OB)-vec(OA)|`
`|0 hati + 4hatj - 2hatk| =sqrt20`
Since `a gt b gt c`, A is the greatest angle l. therefore,
` cos A = (b^(2) + c^(2)-a^(2))/2bc) = (21 + 20 -41)/(2. sqrt21 . sqrt20) = 0`
`angleA = 90^(@)`
Discussion
15.

If `veca and vecb` be two non-collinear unit vectors such that `vecaxx(vecaxxvecb)=1/2vecb` then find the angle between `veca and vecb`.

Answer» `vecaxx (vecaxxvecb) =1/2 vecb Rightarrow(veca.vecb)veca-(veca.veca) vecb= 1/2vecb`
`Rightarrrow veca.vecb = 0 , veca. Veca = -1/2 ` ( which is not possible).
thus, the given information is inconsistent.
Discussion
16.

Given three vectors e`veca, vecb and vecc` two of which are non-collinear. Futrther if `(veca + vecb)` is collinear with `vecc, (vecb +vecc)` is collinear with `veca, |veca|=|vecb|=|vecc|=sqrt2` find the value of `veca. Vecb + vecb.vecc+vecc.veca`A. 3B. `-3`C. 0D. cannot of these

Answer» Correct Answer - b
` veca + vecb = lambdavecc`
` and vecb + vecc = mu veca`
` ( lambda vecc - veca) = mu veca ("putting" vecb = lambda vecc - veca)`
` Rightarrow ( lambda = mu = -1)`
` Rightarrow veca + vecb + vecc =0`
` or |veca|^(2) + |vecb|^(2) + |vecc|^(2)`
` + 2 (veca .vecb + vecb .vecc + vecc. veca) =0`
` or veca. vecb + vecb . vecc + vecc. veca = -3`
Discussion
17.

Let `veca, vecb and vecc` be non-zero vectors such that no two are collinear and `(vecaxxvecb)xxvecc=1/3 |vecb||vecc|veca` if `theta` is the acute angle between vectors `vecb and vecc` then find value of `sin theta`.

Answer» we have `(veca xx vecb ) xx vecc = 1/3 |vecb||vecc|veca`
`or (veca .vecc) vecb - (vecb .vecc)veca = 1/3 |vecb||vecc|veca`
`or (veca. Vecc) vecb- {(vecb.vecc) + 1/3 |vecb|vecc|} veca =vec0`
`Rightarrow veca.vecc =0 and vecb.vecc + 1/3 |vecb|vecc|=0`
(`veca and vecb` are non-collinear)
or `|vecb||vecc|cos theta+ 1/3 |vecb |vecc| =0`
`or cos theta = -1//3`
`Rightarrow sin theta = sqrt(8/9) = ( 2sqrt2)/3`
Discussion
18.

If `veca xx vecb = vecb xx vecc ne 0 ` where `veca , vecb and vecc` are coplanar vectors, then for some scalar k prove that `veca+vecc = kvecb`.

Answer» since `veca xx vecb = vecb xx vecc ne vec0` , we have
`veca xx vecb - vecb xx vecc = vec0`
` or veca xx vecb + vecc xx vecb = vec0`
`or ( veca + vecc) xx vecb = vec0 `
Hence, `veca + vecb` is parallel to `vecb` . Thus ,
`veca + vecc = k vecb`
Discussion
19.

If `veca = 2vecj+3vecj-veck, vecb =-veci+2vecj-4veck and vecc=veci + vecj + veck`, then find the value of `(veca xx vecb).(vecaxxvecc)`

Answer» `vecaxxvecb= |{:(hati, hatj,hatk),(2,3,-4),(-1,2,-4):}|= - 10 veci + 9 vecj + 7 veck`
`vecaxxvecb= |{:(hati, hatj,hatk),(2,3,-1),(1,1,1):}|= - 4 veci - 3 vecj -veck` `Rightarrow (vecaxxvecb) . (vecaxxvecc) = - 40 - 27 - 7 = -74`
Discussion
20.

If `veca,vecb, vecc` are unit coplanar vectors then the scalar triple product `[2veca-vecb 2vecb-c vec2c-veca]` is equal to (A) `0` (B) `1` (C) `-sqrt(3)` (D) `sqrt(3)`

Answer» Correct Answer - a
`veca vecb and vecc` are unit coplanar vectors, `2veca- vecb , 2vecb - 2vecc and 2vecc - veca` are also coplanar, vectors ], being a linear combination of `veca , vecb and vecc` thus ` [ 2 veca - vecb 2vecb - vecc 2vecc - veca] =0`
Discussion
21.

Find `veca.vecb if |veca|2, |vecb|=5,a and |vecaxxvecb|=8`

Answer» `veca.vecb= |veca||vecb|costheta,but |vecaxxvecb|=|veca||vecb|sintheta`
` or sin theta (|veca xx vecb|)/(|veca||vecb|)= 4/5 Rightarrow 4/5 Rightarrow theta= 3/5`
thereforem , `veca.vecb= 2xx5xx 3/5 = 6`
Discussion
22.

If vectors `veca and vecb` are non collinear then `veca/(|veca|)+vecb/(|vecb|)` is (A) a unit vector `in the plane of `veca and vecb` (B) in the plane of `veca and vecb` (C) equally inclined ot vecas and vecb` (D) `perpendiculat to `veca xx vecb`A. a unit vectorB. in the plane of `veca and vecb`C. equally inclined to `veca and vecb`D. perpendicular to `veca xx vecb`

Answer» Correct Answer - b,c,d
Obviosusly, ` veca/ (|veca| ) + vecb/(|vecb|)` is a vector in the palane of `veca and vecb` and hence perpendicular to `veca xx vecb` . It is also equally inclined to `veca and vecb` as it is along bisector.
Discussion
23.

If `veca, vecb and vecc` are three non-coplanar non-zero vectors, then prove that `(veca.veca) vecb xx vecc + (veca.vecb) vecc xx veca + (veca.vecc)veca xx vecb = [vecb vecc veca] veca`

Answer» As `veca ,vecb and vecc` are non- coplanar, `vecb xx veca,veccxxveca and vecaxxvecb` are also non-coplanar, So, any vector con be expressend as a linear combination of these vectors.
`veca=lambdavecbxxvecc+mu vecc xx veca+ v vecaxxvecb`
`veca.veca=lambda[vecbveccveca],veca.vecb=mu [vecc vecavecb], veca.vecc=v[vecavecbvecc]`
`veca=((veca.veca)vecbxxvecc)/([vecb veccveca])+((veca.vecb)veccxxveca)/([veccvecavecb])+((veca.vecc)vecaxxvecb)/([vecavecbvecc])`
Discussion
24.

If `veca xx vecb = vecb xx vecc ne 0 ` where `veca , vecb and vecc` are coplanar vectors, then for some scalar k prove that `veca+vecc = kbvecb`.

Answer» since `veca xx vecb = vecb xx vecc ne vec0` , we have
`veca xx vecb - vecb xx vecc = vec0`
` or veca xx vecb + vecc xx vecb = vec0`
`or ( veca + vecc) xx vecb = vec0 `
Hence, `veca + vecb` is parallel to `vecb` . Thus ,
`veca + vecc = k vecb`
Discussion
25.

If `veca and vecb` are unequal unit vectors such that `(veca - vecb) xx[ (vecb + veca) xx (2 veca + vecb)] = veca+vecb` then angle `theta " between " veca and vecb` is

Answer» Correct Answer - b,d
`(vecaxxvecb)xx[(vecb+veca)xx (2veca+ vecb)] = vecb+veca`
`or {(veca-vecb).(2veca+vecb) } (vecb +veca) `
`-{(veca-vecb). (vecb+veca)}(2veca + vecb)= vecb+veca`
`or (2-veca.vecb- 1) (vecb +veca) = veca.vecb=1`
`Rightarrow "either " vecb +veca =vec0 or 1 - veca.vecb=1`
`Rightarrow "either " vecb = -veca or veca.vecb=0`
`Rightarrow "either " theta= pi" "theta = pi//2`
Discussion
26.

If `(vecaxxvecb)xx(vecbxxvecc)=vecb, where veca,vecb and vecc` are non zero vectors then (A) `veca,vecb and vecc can be coplanar (B) `veca,vecb and vecc` must be coplanar (C) `veca,vecb and vecc cannot be coplanar (D) none of theseA. `veca,vecb and vecv` can be coplanarB. `veca, vecb and vecc` must be coplanarC. `veca,vecb and vecc` cannot be coplanarD. none of these

Answer» Correct Answer - c
`(veca xx vecb) xx (vecb xx vecc) = vecb`
` or [veca vecb vecc] vecb = vecb`
` or [veca vecb vecc] =1 `
therefore, `veca , vecb and vecc` cannot be coplanar.
Discussion
27.

If `veca and vecb` are two unit vectors perpenicualar to each other and `vecc= lambda_(1)veca+ lambda_2 vecb+ lambda_(3)(vecaxx vecb),` then which of the following is (are) true ?A. `lambda_(1)= veca.vecc`B. `lambda_(2)= |vecb xx vecc|`C. `lambda_(3)= |(vecaxx vecb|xxvecc|`D. `lambda_(1)veca + lambda_(2) vecb + lambda_(3) (vecaxxvecb)`

Answer» Correct Answer - a,d
Given `vecc= lambda_(1)veca+lambda_(2)vecb + lambda_(3)(vecaxxvecb)`
`and veca. vecc= lambda_(1),vecc.vecb=lambda_(2)`
` and vecc. (veca xx vecb) |veca xx vecb|^(2) lambda_(3)`
` (1.1 sin 90^(@))^(2) lambda_(3) = lambda_(3)`
Hence, ` lambda_(1), + lambda_(2), + lambda_(3) = (veca + vecb + veca xx vecb) . vecc`
Discussion
28.

Unit vectors `veca and vecb` ar perpendicular , and unit vector `vecc` is inclined at an angle `theta` to both `veca and vecb . If alpha veca + beta vecb + gamma (veca xx vecb)` then.A. `alpha = beta `B. `gamma^(2) = 1- 2alpha^(2)`C. `gamma^(2) =-cos 2 theta`D. `beta^(2) = (1+ cos 2theta)/2`

Answer» Correct Answer - a,b,c,d
Since, `veca, vecb, and vecc` are unit vectors inclined at an angle `theta` we have
`|veca|=|vecb|=1 and cos theta = veca.vecc = vecb.vecc`
Now, ` vecc = alpha veca + beta vecb + gamma (veca xx vecb)`
` Rightarrow veca.vecc=alpha(veca. veca)+beta(veca.vecb) +gamma{veca.(vecaxx vecb)}`
`Rightarrow cos theta=alpha|veca|^(2) " " (therefore veca. vecb=0, veca. (vecaxxvecb) =0) = alpha`
similarly, by taking dot product on both sides of (i)
by `vecb " we get " beta =cos tehta`
Again ` vecc= alpha veca + beta vecb + gamma (veca xx vecb) `
` Rightarrow |vecc|^(2) = |alphaveca + beta vecb + gamma (veca xx vecb)`
`=alpha^(2) |veca|^(2) +beta^(2) |vecb|^(2) +gamma^(2)|vecaxxvecb|^(2)`
` + 2alphabeta(veca .vecb) + 2alphagamma{veca. (vecaxxvecb)}`
`+ 2 betagamma(vecb.{vecaxxvecb})`
` Rightarrow 1 = alpha^(2) +beta^(2) +gamma^(2) |vecaxxvecb|^(2)`
` = 2alpha^(2) + gamma^(2) {|veca|^(2) |vecb|^(2) sin^(2) pi//2}`
`= 2alpha^(2) +gamma^(2) or alpha^(2)(1-gamma^(2))/2`
But `alpha = beta = cos theta`
`1=2alpha^(2) +gamma^(2) Rightarrow1-2cos^(2) theta-cos 2theta`
` beta^(2) (1-gamma^(2))/2 = (1+cos2theta)/2`
Discussion
29.

If `veca , vecb and vecc` are non- coplanar vectors and `veca xx vecc` is perpendicular to `veca xx (vecb xx vecc)` , then the value of `[ veca xx ( vecb xx vecc)] xx vecc` is equal toA. ` [veca vecb vecc] vecc`B. `[ veca vecb vecc] vecb`C. ` vec0`D. `[veca vecb vecc] veca`

Answer» Correct Answer - c
Given that `veca, vecb and vecc` are non-coplanar, thus
` [veca vecb vecc] ne 0`
again ` veca xx (vecb xx vecc) . (veca xx vecc) =0`
`or (veca .vecc) =0`
Hence, ` veca and vecc` are perpendicular,
` veca xx (vecb xx vecc) = (veca .vecc) vecb - (veca .vecb) vecc`
` or [veca xx (vecb xx vecc)] xx vecc = vec0`
Discussion
30.

If `veca,vecb and vecc` are three non-coplannar vectors, then prove that `(|hataxx(hatbxxhatc)|)/sinA=(|hatbxx(hatcxxhata)|)/sinB=(|hatcxx(hataxxhatb)|)/sin C = (prod|hata xx(hatbxx hatc)|)/(|sum sinalpha cosbeta cosgamma hatn_(1)|)`

Answer» Since `veca,vecb and vecc` are non - coplanar, vectors `vecaxxvecb,vecb xxveccandveccxxveca` are also non-coplanar. Let
`vecd=l(vecbxxvecc)+vecm(veccxxveca)+vecn(vecaxxvecb)`
now multiplying both sides of (i) scalarly by `veca` we have
`veca.vecd=lveca.(vecbxxvecc)+mveca.(veccxxveca)+nveca.(vecaxxvecb)=l[vecavecc veca]([veca vecc veca]=0=[veca veca vecb])`
`l=(veca.vecd)//[veca vecb vecc]`
putting these values oif l,nm and n and (i) , we get the required relation.
Discussion
31.

Let `hata,vecb and vecc` be the non-coplanar unit vectors. The angle between `hatb and hatc is alpha "between" hatc and hata is beta and "between" hata and hatb is gamma`. If `A(hatacos alpha),B(hatbcosbeta) and C(hatc cosgamma),` then show that in triangle ABC, `(|hataxx(hatbxxhatca)|)/(sinA)=(|hatbxx(hatcxxhata)|)/sinB = (|hatcxx(hataxxhatb)|)/sinC=(prod|hataxx(hat xx hatc|))/(sumsin alpha-cosbeta. cos gammahatn_(1))` where `hatn_(1)=(hatbxxhatc)/(|hatbxxhatc|),hatn_(2)=(hatcxxhata)/(|hatcxxhata|)and hatn_(3)=(hataxxhatb)/(|hataxxhatb|)`

Answer» From the sine rule, we get
`(AB)/(sin C)=(AC)/(sinB)=(BC)/(sinA)= ((AB)(BC)(CA))/(2DeltaABC)`
`BC=|vec(BC)|=|hatc cos gamma=-hatbcosbeta|=|(hata.hatb)hatc-(hatc.hata)hatb|=|(hataxx(hatbxxhatc))|`
`AC = |vec(AC)|=|hatbxx(hatcxxhata)|and AB = |vec(AB)|=hatcxx(hataxx hatb)|`
`DeltaABC=1/2|vec(BC)xxvec(BA)|`
`=1/2 |(hatc cosgamma-hatb cos beta)xx(hata cosalpha-hatbcosbeta)|`
`=1/2 |(hatc xxhata)cosalpha cosgamma+(hatbxxhatc)cosalphacosbeta+(hata xx hatb)cos beta cos alpha|`
`2DeltaABC=|sumhatn_(1)sinalphacosbeta cosgamma|`
`(|hataxx(hatbxxhatc)|)/sinA=(|hatbxx(hatcxxhata)|)/sinB=(|hatcxx(hataxxhatb)|)/sin C = (prod|hata xx(hatbxx hatc)|)/(|sum sinalpha cosbeta cosgamma hatn_(1)|)`
Discussion
32.

Value of `[vec a xx vec b,vec a xx vecc,vec d]` is always equal toA. `(veca.vecd) [vecavecbvecc]`B. `(veca.vecc)[veca vecb vecd]C. `(veca.vecb)[veca vecb vecd]`D. none of these

Answer» Correct Answer - a
` [veca xx vecb veca xx vecc vecd]`
`= (veca xx vecb) .(( veca xx vecc) xx vecd)`
` (veca xx vecb) . ((veca.vecd)vecc= (vecc.vecd) veca)`
` (veca.vecd) [veca vecb vecc]`
Discussion
33.

If `4veca+5vecb+9vecc=0 " then " (vecaxxvecb)xx[(vecbxxvecc)xx(veccxxveca)]` is equal toA. a vector perpendicular to the plane of `veca, vecb and vecc`B. a scalar quantityC. `vec0`D. none of these

Answer» Correct Answer - c
`4 veca + 5 vecb + 9 vecc =0`
`Rightarrow "vcectors" veca,vecb and vecc` are coplanar.
`Rightarrow vecb xx vecc and vecc xx veca` are collinear.
`Rightarrow (vecb xx vecc) xx (vecc xx veca) = vec0`
Discussion
34.

Let `vecr, veca, vecb` and `vecc` be four non-zero vectors such that `vecr.veca=0, |vecrxxvecb|=|vecr||vecb|,|vecrxxvecc|=|vecr||vecc|` then `[(veca, vecb, vecc)]=`A. |a||b||c|B. `-|a||b||c|`C. 0D. none of these

Answer» Correct Answer - c
`vecr.veca = 0, |vecr xx vecb| = |vecr||vecb| and |vecr xx vecc|`
`|vecr||vecc|`
` [ veca vecb vecc] =0`
Discussion
35.

If `veca and vecb` are two vectors and angle between them is `theta` , thenA. `|vecaxxvecb|^(2)+ (veca.vecb)^(2)= |veca|^(2)|vecb|^(2)`B. `|vecaxxvecb|^(2)+ (veca.vecb)^(2), if theta= pi//4`C. `veca xx vecb = (veca. Vecb) hatn` ( where `hatn` is a normal unit vector ) `if theta f= pi//4`D. `(veca xx vecb ) . (veca + vecb) =0`

Answer» Correct Answer - a,b,c,d
`vecaxx vecb= |veca||vecb| sin theta hatn `
`or |vecaxx vecb|=|veca||vecb|sintheta`
`or sin theta (|vecaxxvecb|)/(|veca||vecb|)`
`veca. vecb= |veca||vecb|cos theta`
`Rightarrow cos theta= (|veca.vecb|)/(|veca||vecb|)`
From (i) and (ii) .
`sin^(2)theta + cos ^(2) theta=1`
`if theta= pi//4 "then" sintheta=costheta= 1//sqrt2.` Therefore,
`|vecaxxvecb|= (|veca||vecb|)/sqrt2 and veca.vecb= (|veca||vecb|)/sqrt2`
`|vecaxxvecb|= veca.vecb`
`vecaxxvecb= |veca||vecb|sinthetahatn = (|veca||vecb|)/sqrt2hatn`
`(veca.vecb)hatn`
Discussion
36.

If `vecca xx (vec b xx vecc)` is perpendicular to `(veca xx vecb ) xx vecc`, we may haveA. `(veca.vecb)|vecb|^(2)= (veca.vecb)(vecb.vecc)`B. `veca.vecb=0`C. `veca.vecc=0`D. `vecb.vecc=0`

Answer» Correct Answer - a,c
`vecaxx(vecbxxvecc)= (veca.vecc)vecb-(veca.vecb)vecb]and (vecaxx vecb) xxvecc=-(vecc. vecb) veca + (veca.vecc)vecb`
we have been givn
`(veca xx (vecbxx vecc)). ((vecaxxvecb)xxvecc)=0`
` or (veca . vecc)^(2)|vecb|^(2)- (veca.vecc)(vecb.vecc) (veca.vecb)`
` - (veca.vecb) (veca.vecc)(vecb.vecc)+ (veca.vecb)(vecb.vecc)(vecc.veca)=0`
`or (veca.vecc)^(2)|vecb|^(2)= (veca.vecc)(veca.vecb)(vecb.vecc)`
`or (veca.vecc)((veca.vecc)(vecb.vecb)- (veca.vecb) (vecb.vecc))=0`
`veca.veca=0 or (veca.vecc) |vecb|^(2) = (veca.vecb)(vecb.vecc)`
Discussion
37.

`a_(1), a_(2),a_(3) in R - {0} and + a_(1)+ a_(2)cos2x+ a_(3)sin^(2)x=0` " for all " x in R` thenA. vectors `veca=a_(1) hati+ a_(2) hatj + a_(3) hatk and vecb = 4hati + 2hatj + hatk` are perpendicular to each otherB. vectors `veca=a_(1) hati+ a_(2) hatj + a_(3) hatk and vecb = hati + hatj + 2hatk` are parallel to each each otherC. if vector `veca=a_(1) hati+ a_(2) hatj + a_(3) hatk` is of length `sqrt6` units, then on of the ordered trippplet `(a_(1), a_(2),a_(3)) = (1, -1,-2)`D. if `2a_(1) + 3 a_(2) + 6 a_(3) + 6a_(3) = 26 , " then " |veca hati + a_(2) hatj + a_(3) hatk | is 2sqrt6`

Answer» Correct Answer - a,b,c,d
`a_(1)+a_(2)cos2x+a_(3)sin^(2)x=0AAx inR`
`or (a_(1)+ a_(2)) + sin ^(2)x (a_(3)- 2a_(2)=0`
`Rightarrow a_(1)+a_(2)=0 and a_(3)-2a_(2)=0`
` a_(1)/(-1)=a_(2)/1=a_(3)/2 = lambda(ne0)`
`Rightarrow a_(1) =lamda, a_(2)-lamda, a_(3)= 2lambda`
Discussion
38.

If `vecp=(vecbxxvecc)/([(veca,vecb,vecc)]),vecq=(veccxxveca)/([(veca,vecb,vecc)]),vecr=(vecaxxvecb)/([(veca,vecb,vecb)])` where `veca,vecb,vecc` are three non-coplanar vectors, then the value of the expression `(veca+vecb+vecc).(vecp+vecq+vecr)` isA. `x [veca vecb vecc] + ([vecp vecqvecr])/x ` has least value 2B. `x^(2) [veca vecb vecc]^(2) + ([vecp vecqvecr])/x^(2) ` has least value `(3//2^(2//3))`C. `[vecp vecq vecr] gt 0 `D. none of these

Answer» Correct Answer - a,c
we have ` [vecp vecq vecr] = 1/ ([ veca vecb vecc])` therefore,
` [vecp vecq vecr] gt 0`
a. ` x gt 0, x [veca vecb vecc] + ([vecp vecq vecr])/x ge 2`
( using ` A.M. ge G.M.`)
b . Similarly, use `A.M. ge G.M`
Discussion
39.

Let `vecr` be a unit vector satisfying `vecr xx veca = vecb, " where " |veca|= sqrt3 and |vecb| = sqrt2`A. `vecr= 2/3(veca+ veca xx vecb)`B. `vecr= 1/3(veca+ veca xx vecb)`C. `vecr= 2/3(veca- veca xx vecb)`D. `vecr= 1/3(-veca+ veca xx vecb)`

Answer» Correct Answer - b,d
`veca xx (vecr xx veca) = vecaxxvecb`
` 3 vecr - (veca.vecr) veca = veca xx vecb`
Also, `|vecrxxveca|= |vecb|`
`Rightarrow sin^(2)theta= 2/3`
`or (1-cos^(2)theta) = 2/3`
`or 1/3 = cos^(2)theta`
`Rightarrow veca.vecr = +- 1`
`Rightarrow 3 vecr +- veca = veca xx vecb1`
`or vecr = 1/3(vecaxx vecb +- veca)`
Discussion
40.

If vector ` vec b=(t a nalpha,-12sqrt(sinalpha//2))a n d vec c=(t a nalpha, t a nalpha-3/(sqrt(sinalpha//2)))`are orthogonal and vector ` vec a=(13,sin2alpha)`makes an obtuse angle withthe z-axis, then the value of `alpha`is`alpha=(4n+1)pi+tan^(-1)2`b. `alpha=(4n+1)pi-tan^(-1)2`c. `alpha=(4n+2)pi+tan^(-1)2`d. `alpha=(4n+2)pi-tan^(-1)2`A. `alpha= ( 4n+1 ) pi + tan^(-1) 2`B. `alpha= ( 4n+1 ) pi - tan^(-1) 2`C. `alpha= ( 4n+2 ) pi + tan^(-1) 2`D. `alpha= ( 4n+2 ) pi - tan^(-1) 2`

Answer» Correct Answer - b,d
Since `veca = 1,3, sin 2 alpha) ` makes on abtuse angle with the z-axis its z-component is negtive, thus,
` -1 le sin 2 alpha lt 0`
But `vecb.vecc=0`
` tan^(2) alpha - tan alpha -6 =0`
`(tan alpha -3) (tan alpha + 2) =0`
` Rightarrow tan alpha 3, -2`
Now, `tan alpha =3,` therefore,
`sin2 alpha = (2 tanalpha)/(1+tan^(2)alpha)= 6/(1+9)= 3/5`
( not possible as `sin 2 alpha lt 0`)
Now , if ` tan alpha = -2`
`Rightarrow sin2 alpha= (2tan alpha)/(1+ tan^(2)alpha)= (-4)/(1+_5) = (-4)/5`
`tan 2 alpha gt0`
Hence, ` 2alpha` is the third quadrant , Also , `sqrt(sin alpha//2)` is meaningful. if `0 lt sin alpha//2 1, ` then
`alpha= (4 n + 1) pi-tan^(-1)2`
`and alpha= (4n+2) pi - tan^(-1) 2`
Discussion
41.

Let ` overset(to)(a) =a_(1) hat(i) + a_(2) hat(j) + a_(3) hat(k) , overset(to)(a) = b_(1) hat(i) +b_(2) hat(j) +b_(3) hat(k) " and " overset(to)(a) = c_(1) hat(i) +c_(2) hat(j) + c_(3) hat(k)` be three non- zero vectors such that `overset(to)(c )` is a unit vectors perpendicular to both the vectors `overset(to)(c )` and `overset(to)(b)`. If the angle between `overset(to)(a) " and " overset(to)(n)` is `(pi)/(6)` then `|{:(a_(1),,a_(2),,a_(3)),(b_(1),,b_(2),,b_(3)),(c_(1),,c_(2),,c_(3)):}|` is equal to

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))`
Discussion
42.

The scalars l and m such that `lveca + m vecb =vecc, " where " veca, vecb and vecc` are given vectors, are equal toA. `l= ((veccxxvecb).(vecaxxvecb))/((vecaxxvecb)^(2))`B. `l = ((vecc xx veca). (vecb xxveca))/((vecbxxveca))`C. `m = ((veccxx veca). (vecbxxveca))/((vecb xx veca)^(2))`D. `m = ((veccxx veca). (vecbxxveca))/((vecb xx veca))`

Answer» Correct Answer - a,c
Here, ` (l veca +mvecb) xx vecb =vecc xx vecb `
`or lveca xx vecb =vcec xx vecb`
`or l(veca xx vecb)^(2) = (vecc xx vecb) / (veca xx vecb)`
`or l= ((vecc xx vecb) . (veca xx vecb))/ ((veca xx vecb)^(2))`
similarly, `m = ((veccxxveca). (vecb xx veca))/((vecbxxveca)^(2))`
Discussion
43.

If the vectors `veca, vecb, vecc` are non -coplanar and `l,m,n` are distinct scalars such that `[(lveca+mvecb+nvecc, lvecb+mvecc+nveca,lvecc+mveca+nvecb)]=0` thenA. ` l + m + n=0`B. roots of the equation `lx^(2) + mx + n =0` are equalC. `l^(2)+m^(2) + n^(2) =0`D. `l^(3) + m^(2) + n^(3) = 3 lmn `

Answer» Correct Answer - a,b,d
`{:(vecV_(1)=lveca+mvecb+nvecc),(vecV_(2)=n veca+lvecb + mvecc),(vecV_(3) = mveca + nvecb+lvecc):}}when,veca, vecb and vecc " are non- coplanar".`
Therefore,
`[vecV_(1)vecV_(2)vecV_(3)]= |{:(l,m,n),(n,l,m),(m,n,l):}|=0`
`or (l + m +n) [(l-m)^(2) + (m-n)^(2)+ (n-l)^(2)=0`
` or l + m+n =0`
obviously, `lx^(2) + mx + n =0` is satisfied by x =1 due to (i).
` l^(3) +m^(3) + n^(3) = 3lmn `
`Rightarrow (l+m +n) (l^(2) + m^(2) +n^(2)-lm -mn -ln) =0`
which is true
Discussion
44.

If `vecr=x_(1)(vecaxx vecb) + x_(2) (vecb xxveca) + x_(3)(vecc xxvecd) and 4[veca vecb vecc] =1 " then " x_(1) + x_(2) + x_(3)` is equal toA. `1/2vecr.(veca + vecb + vecc)`B. `1/4vecr.(veca + vecb + vecc)`C. `2vecr.(veca + vecb + vecc)`D. `4vecr.(veca + vecb + vecc)`

Answer» Correct Answer - d
`vecr= x_(1)(vecaxx vecb) + x_(2)(vecb xx vecc)+x_(3)(vecc xxveca)`
`Rightarrow vecr.veca=x_(2)[veca vecb vecc] . vecr.vecb=x_(3)[vecb vecc veca]`
` and vecr.vecc= x_(1)[vecc veca vecb] =x_(1) [vecb vecc veca]`
` Rightarrow x_(1)+x_(2)+x_(3)=4vecr. (veca +vecb+vecc)`
Discussion
45.

If the vectors `veca` and `vecb` are perpendicular to each other then a vector `vecv` in terms of `veca` and `vecb` satisfying the equations `vecv.veca=0, vecv.vecb=1` and `[(vecv, veca, vecb)]=1` isA. `(vecb)/(|vecb|^(2))+ (vecaxx vecb)/(|vecaxxvecb|^(2))`B. `(vecb)/(|vecb|)+ (vecaxx vecb)/(|vecaxxvecb|^(2))`C. `(vecb)/(|vecb|)+ (vecaxx vecb)/(|vecaxxvecb|)`D. none of these

Answer» Correct Answer - a
Let `vecc = xveca + yvecb +z veca xx vecb`
Given : `veca. Vecb = 0, veca. Vecb = 1 , [ vecv veca vecb] =1 `
`Rightarrow vecv.veca=xveca.veca = x|veca|^(2)`
` ( veca. Vecb =0, veca.veca xx vecb =0)`
` Rightarrow x =0`
Again, ` vecv. (veca xx vecb) = z (veca xx vecb)^(2)`
` Rightarrow 1=z (veca xx vecb)^(2) or z= 1/(|veca xx vecb|^(2))`
Hence, `vecv= 1/|vecb|^(2) vecb+ 1 / (|veca xx vecb|^(2)) veca xx vecb`
Discussion
46.

If `vecd=vecaxxvecb+vecbxxvecc+veccxxveca` is a on zero vector and `|(vecd.vecc)(vecaxxvecb)+(vecd.veca)(vecbxxvecc)+(vecd.vecb)(veccxxveca)|=0` then (A) `|veca|+|vecb|+|vecc|=|vecd|` (B) `|veca|=|vecb|=|vecc|` (C) `veca,vecb,vecc` are coplanar (D) `veca+vecc=vec(2b)`A. `|veca|=|vecb|=|vecc|`B. `|veca|+|vecb|+|vecc|=|vecd|`C. `veca, vecb and vecc` are coplanarD. none of these

Answer» Correct Answer - c
`vecd.vecc=vecd.vecb=vecd .vecc= [veca vecb vecc]`
then `| (vecd.vecc) + (vecd.vecb) (vecc xx veca) =0 `
`or [veca vecb vecc] |veca xx vecb +vecb xx vecc +vecc xx veca)|=0`
`or [veca vecb vecc] = 0 " " ( vecd` is non -zero)
Hence. ` veca,vecb vecc` are coplanar.
Discussion
47.

If `a`is real constant `A ,Ba n dC`are variable angles and `sqrt(a^2-4)tanA+atanBsqrt(a^2+4)tanc=6a ,`then the least vale of `tan^2A+tan^2b+tan^2Ci s``6`b. `10`c. `12`d. `3`A. 6B. 10C. 12D. 3

Answer» Correct Answer - d
The given relation can be rewritten as the vector expression
` (sqrt(a^(2)-4) hati + ahatj + sqrt(a^(2) + 4hatk)`
` (tan A hati tanB hatj + tan C hatk) = 6a`
`( sqrt( a^(2) -4 =a^(2) =a^(2) +4))`
`(sqrt(tan^(2)A + tan^(2) B+ tan^(2) C)). (cos theta) = 6a`
` ( veca. vecb = |veca||vecb|cos theta)`
` sqrt3 a sqrt(tan^(2) A + tan^(2) B + tan^(2)C). (cos theta) = 6a`
`tan^(2) A + tan^(2)B + tan^(2)C = 12 sec^(2) theta ge 12`
` (sec^(2) theta ge 1)`
the least value of `tan^(2) A + tan^(2)B + tan^(2)B+ tan^(2)C is 12`.
Discussion
48.

Let ` vecf(t)=[t] hat i+(t-[t]) hat j+[t+1] hat k , w h e r e[dot]`denotes the greatest integer function. Then thevectors ` vecf(5/4)a n df(t),0A. parallel to each otherB. perpendicular to each otherC. inclined at `cos^(-1) 2/(sqrt7(1-t^(2))`D. inclined at `cos ^(-1) (8+t)/(9sqrt(1+t^(2)))`

Answer» Correct Answer - d
`vecf""(5/4)= [5/4] hati+ (5/4-[5/4])hatj + [5/4=1]hatk`
`hati+ (5/4-1)hatj + 2hatk`
`hati + 1/4 hatj + 2hatk`
when `0 lt t 1 ,vecf "" (t) = 0 hati + { t-0} hatj +hatk`
` hatj +hatk`
` vecf "" (5/4) .vecf"" (t) = 2+t/4`
so, `costheta= 4/(|vect+1/4hatj + 2hatk||thatj +hatk|)`
`(2+t/4)/(sqrt(1+1/16=4)sqrt(1+t^(2)))`
` ( 8 +t)/(9sqrt(1 + t^(2)))`
Discussion
49.

Two adjacent sides of a parallelogram ABCD are `2hati+4hatj -5 hatkand hati+2hatj+3hatk`. Then the value of `|vec(AC)xxvec(BD)|` isA. `20sqrt5`B. `22sqrt5`C. `24sqrt5`D. `26sqrt5`

Answer» Correct Answer - b
`|vec(AC) xx vec(BD)|=2 |vec(AB)xxvec(AD)|`
`2|{:(hati,hatj,hatk),(2,4,-5),(1,2,3):}|`
`|2[hati(12+10)-hatj (6+5)+hatk (4-4)]|`
`|2[22hati -11 hatj]|`
`22 |[2hati - hatj]|`
`22xx sqrt5`
Discussion
50.

If `veca and vecb ` are any two vectors of `magnitude` `2 and 3` respectively such that `|2(vecaxxvecb)|+|3(veca.vecb)|=k` then the maximum value of k isA. `sqrt13`B. `2sqrt13`C. `6sqrt13`D. `10sqrt13`

Answer» Correct Answer - c
`k= |2(veca xx vecv) |+ |3 (veca .vecb) |`
` = 12 sin theta + 18 cos theta`
` Rightarrow` maximum value of k = `sqrt(12^(2) + 18^(2)) = 6sqrt13`
Discussion