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1.

If a, b,c are the pth, qth, and rth terms of a HP, then the vectors ` vecu= a^(-1) hati +b^(-1) hatj +c^(-1) hatk and vecv = ( q -r) hati + ( q -r) hati + ( r-p) hatj + ( p-q) hatk`A. are parallelB. are othogonalC. satisfy `vecu .vecv =1`D. satisfy `|vecu xx vecv| =hati +hatj +hatk`

Answer» Correct Answer - B
Let A be the first term and D be the common difference of the corresponding AP. Then,
` 1/a = A + ( p -1) D, 1/b = A + ( q-1) D and , 1/c = A + ( r-1)D`
` a^(-1) (q-r) + b^(-1) (r-p) +c^(-1) (p-q) =0`
` Rightarrow vecu.vecv =0 Rightarrow vecu bot vecv`
Hence, ` vecu and vecv ` are othogonal vectors.
Discussion
2.

The number of vectors of unit length perpendicular to the vectors ` hata = hati +hatj and vecb = hatj + hatk` isA. 1B. 2C. 4D. infinite

Answer» Correct Answer - B
Required vectors are given by `+-(vecaxxvecb)/(|vecaxxvecb|)`
Hence , there are two vectors.
Discussion
3.

For any vector `vecr, (vecr.hati) ^(2) + (vecr.hatj)^(2) + ( vecr.hatk)^(2) ` is equal toA. 1B. `|vecr|`C. `vecr`D. `|vecr|^(2)`

Answer» Correct Answer - D
Let `vecr= xhati +y hatj +z hatk,` Then ,
` vecr.hati =x, vecr.hatj =y and vecr.hatk =z `
` ( vecr.hati)^(2) + ( vecr.hatj)^(2) + ( vecr.hatk) ^(2) =x^(2) +y^(2) +z^(2) = |vecr|^(2)`
Discussion
4.

A vector of magnitude 4 which is equally inclined to the vectors `hati +hatj,hatj +hatk and hatk +hati`, isA. `4/sqrt3 (hati -hatj -hatk)`B. ` 4/sqrt3 (hati +hatj -hatk)`C. ` 4/sqrt3 (hati +hatj +hatk)`D. `4/sqrt3 (hati +hatj -hatk)`

Answer» Correct Answer - C
Let the required vector be ` vecr = xhati + yhaty + zhatk` , Then
` |vecr| =4 Rightarrow x^(2) +y^(2) +z^(2) =16`
Now, `vecr`, sequally inclined to the vectors `hati +hatj ,hatj + hatk and hati`
` (vecr. (hati + hatj)/(|vecr|sqrt2)= (vecr.(hatj +hatk))/(|vecr|sqrt2) = (vecr.(hatj+hati))/(|vecr|sqrt2)`
` Rightarrow x + y=y +z =z +x =lambda (say) `
` Rightarrow 2(x +y +z) =3 lambda Rightarrow x+y +z = (3 lambda)/2`
Now, ` x + y+ lambda and x +y + z = (3lambda)/2 Rightarrow z = lamnbda/2`
Similarly, we have, ` x=y = lambda/2`
Substituting these values in (i) , we get ` lambda = +- 8/sqrt3`
Hence, ` vecr= -+ 8/(2sqrt3) ( hati +hatj +hatk) = -+ 4/sqrt3 (hati +hatj + hatk)`
Discussion
5.

Let ` vecu = u_(1)hati + u_(2)hatj +u_(3)hatk` be a unit vector in ` R^(3) and vecw = 1/sqrt6 ( hati + hatj + 2hatk)` , Given that there exists a vector `vecv " in " R^(3)` such that ` | vecu xx vecv| =1 and vecw . ( vecu xx vecv) =1` which of the following statements is correct ?A. There is exactly one choice for such ` vecv`B. There are exactly two for such `vecv`C. There are exactly for such `vecv`D. There are infinitely many choices for such `vecv`

Answer» Correct Answer - D
Clearly, `vecw` is a unit vector such that ` |vecu xx vecv|=1 and vecw , (vecu xx vecv) =1 ` Now,
` ` vecw .(vecu xx vecv) =1`
` Rightarrow vecw = vecu xx vecv`
` Rightarrow |vecw| = |vecu xx vecv|`
` Rightarrow 1 = |vecu||vecv| sin theta " where " theta` is the angle betweeen ` vecu and vecv`
` Rightarrow vecc sin theta =1`
Clearly, p can take infinitely many positions on the line at a unit distance from OA. Consequently, ` vec(OP) =vecv` has, infinitely, many chocies.
Discussion
6.

IF the force represented by ` 3hati +2hatk` is acting through the point ` 5hati +4hatj -3hatk` , then its moment about th point (1,3,1) isA. ` 14 hati -8 hatj +12hatk`B. ` -14hati +8hatj -12hatk`C. ` -6hati -hatj +9hatk`D. ` 6hati +hatj -9hatk`

Answer» Correct Answer - A
we have, ` vecF= 3hatj +2hatk`
Let P ( 5,4,-3) and O (1,3,1) be the given points. Then ,
` vectr = vec(OP) =4hati +hatj -4hatk`
Moment of ` vecF ` about O is given by
` vecr xx vecF = |{:(hati,hatj,hatk),(4,1,-4),(0,3,2):}|=14ghati -8hatj +12hatk`
Discussion
7.

Forces of magnitudes 5 and 3 units acting in the directions ` 6hati + 2hatj + 3hatk and 3 hati - hati +6hatk` respectively act on a particle which is displaced from the point ( 2,2,-1) to ( 4,3,1) . The work done by the forces, isA. 148 unitB. ` 148/7` unitC. 296 unitsD. none of these

Answer» Correct Answer - B
Let ` vecF` be the resultant force and `vecd` be the displacement vector. Then,
` vecF=5 ((6hati +2hatj +3hatk))/ (sqrt(36+4+9))+3 ((3hati -2hatj +6hatk)) /(sqrt(9 +4+ 36)) = 1/7 (39 hati + 4hatj +33 hatk)`
` and vecd = (4hati + 3hatj +hatk) - (2hati +2hatj -hatk) = 2hati +hatj + 2hatk`
Total work done = ` vecF. vecd`
`Rightarrow ` Total work done = ` 1/7 (39 hati +4hatj +33hatk).(2hati +hatj +2hatk) `
Total work done ` = 1/7 ( 78 +4 +66) = 148/7` units.
Discussion
8.

Unit vectors equally inclined to the vectors ` hati , 1/3 ( -2hati +hatj +2hatk) = +- 4/sqrt3 ( 4hatj +3hatk)` areA. `+- 1/sqrt51 (hati - 5hatj +5hatk)`B. ` +- 1/sqrt51 (hati -5hatj -5hatk)`C. `+- 1/sqrtt51 (hati+ 5hatj + 5hatk)`D. none of these

Answer» Correct Answer - A
Let the require unit be ` vecb = xhati + yhati + yhatj + zhatk`
It is equally inclined to the given units vectors. Therefore,
`(x hati +yhatj +zhatk) .hati = 1/3 ( - 2hati +hatj +2hatk) . (x hati +yhatj +zhatk)`
` -1/5 ( 4hatj +3hatk) . (xhati +yahtj +3hatk)`
` x = 1/3 ( -2x +y +2z) = -1/5 ( 4y +3z)`
` x= 1/3 ( -2 x + y +2z) = -1/5 ( 4y +3z)`
` Rightarrow 5x -y -2x =0 and 5x +4y +3z=0`
` Rightarrow x/1 = y/(-5) = z/5 lambda (say) , Rightarrow x=lambda, y =-5 lambda, z= 5lambda`
It is given that
` vecr =xhati +yhatj +zhatk` is a unit vector.
` |vecr| =1 Rightarrow x^(2) +y^(2) +z^(2) =1 Rightarrow lambda = +- 1/sqrt51`
` vecr = +- 1/sqrt51 (hati -5hatj+5hatk)`
Discussion
9.

A force of 39 kg. wt is acting at a point p ( -4,2,5) in the direaction ` 12hati -4hatj -3hatk` . The moment of this force about a line through the origin having the direction of ` 2hati -2hatj + hatk` isA. 76 unitsB. - 76 unitsC. ` 42hati + 144 hatj -24 hatk`D. none of these

Answer» Correct Answer - B
we have,
Force, ` (vecF) = 39 ((12hati -4hatj -3hatk))/(sqrt (144 +16+9) ) = 36 hati -12 hatj -9hatk`
Let `vecM` be the moment of ` vecF` about a point O ( origin) on the given line . Then
` vecM = vecr xx vecF, " where " vecr = vec(OP) = - 4hati +2 hatj +5hatk`
`Rightarrow vecM=|{:(hati,hatj,hatk),(-4,2,5),(36,-12,-9):}|=42hati+144hatj-24hatk`
So, the moment of ` vecF` about a line having the direction `veca =2hati -2hatj +hatk` is given by
` vecM l.hata = (42 hati +144 hatj -24hatk) . (( 2hati -2hatj +hatk))/(sqrt( 4+4+1)) = -76 ` units
Discussion
10.

If ` veca, vecb` are unit vectors such that ` |veca +vecb|=1 and |veca -vecb|=sqrt3`, " then " |3veca +2vecb|=`A. 7B. 4C. ` sqrt7`D. `sqrt19`

Answer» Correct Answer - C
Let `theta` the angle between `veca and vecb` . Then,
` tan "" theta/2= (|veca -vecb|)/(|veca +vecb|) Rightarrow tan "" theta/2 = sqrt3 Rightarrow theta =120^(@)`
` veca.vecb = |veca||vecb| cos theta =cos 120^(@) = -1/2`
Now ,
` |3 veca = 2vecb|^(2) = 9 |veca|^(2) + 4 |vecb|^(2) + 12(veca .vecb)`
` Rightarrow |3veca + 2veca|^(2) = 9 +4+12 xx ( - 1/2) =7`
` Rightarrow |3veca +2vecb|= sqrt7`
Discussion
11.

Let `veca , vecb , vecc` be three unit vectors such that ` |veca + vecb +vecc| =1 and veca bot vecb , " if " vecc` makes angles ` delta beta " with " veca, vecb` respectively, then ` cos delta + cos beta ` is equal toA. 1B. -1C. `3/2`D. 0

Answer» Correct Answer - B
we have ` veca bot vecb` , therefore, ` veca.vecb =0`
Also, `vecc. Veca = cos delta and vecc. Vecb =cos beta` .
now , ` |veca + vecb + vecc|= 1`
`Rightarrow |veca + vecb + vecc|^(2) =1`
` Rightarrow |veca|^(2) + |vecb|^(2) + |vecc|^(2) + 2 (veca.vecb + vecb.vecc + vecc.veca) =1`
` Rightarrow 1+1+1+2 ( cos delta + cos beta) =1`
` Rightarrow cos delta + cos beta = -1`
Discussion
12.

If the vectors `veca = ( c log_(2) x ) hatk ` make an obtuse angle for any ` x ne ( 0 , oo) ` then c belongs toA. ` ( - oo, 0)`B. ` ( - oo , -4//3)`C. ` ( -4//3,0)`D. ` (-4//3, oo)`

Answer» Correct Answer - C
For the vectors ` veca and vecb` to be inclined at an obtuse angle we must have
` veca. Vecb lt o) " for all " x ne ( 0, oo)`
` Rightarrow 2(log_(2) x) ^(2) - 12 + 6 c ( log_(2) x ) lt 0 " for all " x ne ( 0, oo)`
` Rightarrow 2y^(2) + 6 cy -12 lt 0 " for all " y ne R, " where " y = log_(2) x `
` Rightarrow c lt 0 and 36 c^(2) + 48 c lt 0`
` Rightarrow c lt 0 and c ( 3c + 4) lt 0`
` c lt 0 and - 4/3 lt c lt 0`
` Rightarrow c in ( -4//3 , 0)`
Discussion
13.

If ` veca , vecb` are unit vectors such that the vector ` veca + 3vecb ` is peependicular to ` 7 veca - vecb and veca -4vecb` is prependicular to ` 7 veca -2vecb` then the angle between ` veca and vecb` isA. ` pi//6`B. ` pi//4`C. ` pi//3`D. ` pi//2`

Answer» Correct Answer - C
Let ` theta` be the angle between `veca and vecb` is
we have,
` ( veca + 3vecb) bot ( 7 veca - 5vecb)`
` Rightarrow ( veca + 3vecb) . ( 7 veca -5vecb) =0 `
` Rightarrow 7|veca|^(2) + 16(veca .vecb) -15 |vecb|^(2) =0`
` Rightarrow 15-30 cos theta =0 Rightarrow cos theta = 1/2 Rightarrow 1/2 Rightarrow theta = pi/3`
And,
` (veca - 4 vecb ) bot ( 7veca - 2vecb)`
` Rightarrow (veca -4vecb) . ( 7 veca - 2 vecb) =0`
` Rightarrow 7| veca|^(2) + 8|vecb|^(2) -30 (veca .vecb) =0`
` Rightarrow 15-30 cos theta =0 Rightarrow = 1/2 Rightarrow = pi /3`
Discussion
14.

If ` veca , vecb` are unit vectors such that ` |vec a+vecb|=-1 " then " |2veca -3vecb| =`A. 19B. `sqrt19`C. `sqrt13`D. 4

Answer» Correct Answer - B
we have,
`|veca +vecb|=1`
` Rightarrow |veca|^(2) +|vecb|^(2) + 2(veca .vecb) =1 `
` Rightarrow 1+1 +2 (veca.vecb) =1 Rightarrow veca.vecb= -1/2 `
Now,
` |2 veca -3 vecb|^(2) =4 |veca|^(2) +9|vecb|^(2) -12(veca.vecb)`
` Rightarrow |2veca -3vecb|^(2) =4+9 =19`
` Rightarrow |2veca -3vecb|= sqrt19`
Discussion
15.

The moment of the couple consisting of the force through the point `2hati -3hatj -hatk` isA. 5B. `5sqrt5`C. `sqrt5`D. 25

Answer» Correct Answer - B
Let A and B be the points having position vectors
` hati - hatj + hatk and 2hati -3hatj -hatk` respectively, then
`vecr = vec(BA) = (hati -hatj +hatk) - ( 2hati -3hatj -hatk) = - hati + 2hatj + 2hatk`
Let `vecM` b the moment of the couple. Then,
` vecM = vecr xx vecF |{:(hati, hatj ,hatk),(-1,2,2),( 3,2,-1):}| = -6 hati +5hatj -8hatk`
Moment fo the couple = ` |vecM|= sqrt(36+25+ 64) = 5sqrt5`
Discussion
16.

If two out to the three vectors , `veca, vecb , vecc` are unit vectors such that ` veca + vecb + vecc =0 and 2(veca.vecb + vecb .vecc + vecc.veca) +3=0` then the length of the third vector isA. 3B. 2C. 1D. 0

Answer» Correct Answer - C
Let ` |veca|-1 and |vecb| =1`
we have,
` veca + vecb+ vecc = vec0`
` Rightarrow |veca|^(2) + |vecb|^(2) + 2 (veca.vecb +vecb.vecc + vecc.veca)=0`
` Rightarrow 1+1 + |vecc|^(2) -3=0 Rightarrow |vecc| =1`
Discussion
17.

If ` veca. Vecb and vecc` are unit vectors satisfying ` |veca -vecb|^(2) +|vecb -vecc| ^(2) |vecc -veca| =9 , " then " |2 veca + 5 vecb + 5 vecc| ` is equal to

Answer» Correct Answer - D
we know that
` |veca + vecb + vecc|^(2) = |veca|^(2) +|vecb|^(2) +|vecc|^(2) + 2 ( veca .vecb + vecb.vecc + vecc.veca)`
` and |veca -vecb|^(2) + |vecb -vecc|^(2) +|vecc -veca|^(2) `
` = 2 (|veca|^(2) +|vecb|^(2) +|vecc|^(2) ) -2 (veca.vecb + vecb.vecc + vecc.veca)`
` |veca -vecb|^(2) + |vecb - vecc|^(2) + |vecc -veca|^(2) `
` = 3 { |veca|^(2) +|vecb|^(2) |vecc|^(2) } - |veca +vecb +vecc|^(2)`
` Rightarrow 9 = 3xx 3 - |veca + vecb + vecc|^(2) `
` Rightarrow |veca + vecb + vecc|^(2) =0`
` Rightarrow veca + vecb + vecc = vec0`
` Rightarrow vecb + vecc =- veca`
` therefore | 2 veca + 5 vecb + 5vecc| = | 2 vecb +5( -veca)| = 3|veca|=3`
Discussion
18.

If ` veca and vecb` are unit vectors inclined to x-axis at angle ` 30^(@) and 120^(@)` then ` |veca +vecb|` equalsA. `sqrt(2//3)`B. `sqrt2`C. `sqrt3`D. 2

Answer» Correct Answer - B
Clearly, `veca and vecb` are at angle angle.
` |veca +vecb|^(2) =|vecb|^(2) +2|veca||vecb| cos 90^(@)`
` Rightarrow | veca +vecb|^(2) = 1+ 1+0=2 Rightarrow |veca + vecb| = sqrt2`
Discussion
19.

The vactors ` veca = 3hati -2hati + 2hatk and vecb =- hati -2hatk` are the adjacent sides of a parallelogram. Then , the acute angle between ` veca and vecb` isA. `pi//4`B. ` pi//3`C. ` 3pi//4`D. `2pi//3`

Answer» Correct Answer - A
The diagonals of the parallelogram are given by
` vecalpha =veca =vecb and vecbeta = +- (veca -vecb)`
i.e ` vecalpha = 2hati -2hatj and vecbeta = +- ( 4hati -2hatj +4hatk)`
Let ` theta` be the angle between the diagonals. Then ,
` cos theta = ( vecalpha .vecbeta)/(|vecalpha||vecbeta|)`
` Rightarrow cos theta= 1/sqrt2 or, cos theta = - 1/sqrt2 Rightarrow theta = pi//4, or , theta = 3pi//4`
Discussion
20.

Let ` veca , vecb,vecc` be three vectors such that ` veca bot ( vecb + vecc), vecb bot ( vecc + veca) and vecc bot ( veca + vecb) , " if " |veca| =1 , |vecb| =2 `, ` |vecc| =3 , " then " | veca + vecb + vecc|` is,A. ` sqrt6`B. 14C. ` sqrt14`D. none of these

Answer» Correct Answer - C
we have,
` veca bot ( vecb + vecc), vecb bot ( vecc + veca) and vecc bot ( veca + vecb) `
` Rightarrow veca.vecb + veca.vecc =0, vecb.veca =0, vecc.veca + vecc.vecb =0`
` Rightarrow veca.vecb = vecb.vecc= vecc.veca =0`
` |veca + vecb + vecc|^(2) = |veca|^(2) + 2 (veca.vecb + vecb.vecc + vecc. veca)`
` Rightarrow |veca + vecb + vecc|^(2) = 1+ 4+9 =14`
` Rightarrow |veca + vecb+ vecc| = sqrt 14`
Discussion
21.

The length of the longer diagonal of the parallelogram constructed on ` 5veca + 2vecb and veca - 3vecb, ` if it is given that ` |veca|=2sqrt2, |vecb|=3 and veca. Vecb= pi/4` isA. 15B. `sqrt3`C. `sqrt593`D. `sqrt369`

Answer» Correct Answer - C
The diagonals of the parallelogram are
` vecalpha = 5veca + 2vecb + veca -3vecb = 6veca - vecb and beta = +- ( 4 veca + 5vecb)`
Now,
` |vecalpha|= | 6 veca - vecb|`
` Rightarrow |vecalpha| = sqrt(36 |veca|^(2) + |vecb|^(2) -12 (veca .vecb))`
`|vecalpha|=sqrt(36xx8+9-12xx 2sqrt2xx3xx1/sqrt2) =15`
and,
` |vecbeta|=|4veca +5vecb|`
`|vecbeta|=sqrt(16|veca|^(2)+25|vecb|^(2)+40(veca.vecb))`
`|vecbeta|=sqrt(16xx8+25xx9 +40xx2sqrt2xx3xx1/sqrt2)=sqrt593`
Clearly, `|vecbeta| gt |vecalpha|`
Hence, the length of the longer diagonal is `sqrt593`
Discussion
22.

Angle between vectors ` veca and vecb " where " veca,vecb and vecc` are unit vectors satisfying ` veca + vecb + sqrt3 vecc = vec0` isA. ` pi/6`B. `pi/4`C. `pi/3`D. ` pi/2`

Answer» Correct Answer - C
we have,
` veca + vecb + sqrt3 vecc = vec0`
` Rightarrow veca | vecb = sqrt3 vecc`
` Rightarrow |veca + vecb| = sqrt3 |vecc|`
` Rightarrow |veca + vecb|^(2) =3 |vecc|^(2) `
` Rightarrow |veca|^(2) +|vecb|^(2) + 2|veca||vecb| cos theta =3 |veca|^(2) `
where ` theta ` is the angle between ` veca and vecb` .
` Rightarrow 1 +1 + 2 cos theta =3 Rightarrow = 1/2 Rightarrow theta = pi/3`
Discussion
23.

If ` veca +vecb +vecc =vec0, |veca| =3 , |vecb|=5 and |vecc| =7` , then the angle between ` veca and vecb` isA. `pi/2`B. ` pi/4`C. ` pi/6`D. ` pi/3`

Answer» Correct Answer - D
we have,
` veca + vecb +vecc= vec0`
` Rightarrow vecc = - ( veca + vecb)`
` Rightarrow |vecc| = |-(veca + vecb)|`
` Rightarrow |vecb|^(2) = |veca + vecb|^(2)`
` Rightarrow |vecc|^(2) =|veca|^(2) + |vecb|^(2) + 2( veca.vecb)`
` Rightarrow |veca|^(2) = |veca|^(2) +|vecb|^(2) + 2 |veca||vecb| cos theta`
where ` theta` is angle between ` veca and vecb`.
` Rightarrow 49 = 9 + 25 + 30 cos theta`
` Rightarrow 15= 30 cos theta Rightarrow cos theta = 1/2 Rightarrow theta = pi/3`
Discussion
24.

If ` veca,vecb, vecc` are three vectors such that ` veca + vecb +vecc =vec0, |veca| =1 |vecb| =2, | vecc| =3` , then ` veca.vecb + vecb .vecc +vecc + vecc.veca ` is equal toA. 1B. 0C. -7D. 7

Answer» Correct Answer - C
we have,
` veca + vecb + vecc = vec0`
` Rightarrow |veca + vecb + vecc | = 0`
` Rightarrow |veca + vecb + vecc|^(2) =0`
` Rightarrow |veca|^(2) + |vecb|^(2) + 2 ( veca .vecb + vecb .vecc + vecc. veca) =0`
`Rightarrow 1+4+ 9+2 ( veca .vecb +vecb .vecc + vecc .veca) = 0`
` Rightarrow veca. vecb + vecb.vecc + vecc.veca = - 7 `
Discussion
25.

If `|veca|=3,|vecb|= 5 and |vecc|=4 and veca+ vecb + vecc =vec0` then the value of `( veca. Vecb + vecb.vecc)` is equal tio

Answer» Correct Answer - B
we have,
` veca +vecb + vecc= vec0`
` Rightarrow ( veca + vecb + vecc) , vecb = vec0 .vecb` [ Taking dot product with ` vecb`]
` veca.vecb + vecb.vecb + vecc.vecb =0`
` Rightarrow veca. Vecb + |vecb|^(2) + vecb.vecc =0 Rightarrow veca. vecb + vecb.vecc=-25`
Discussion
26.

If the area of parallelogram whose diagonals coincide with the following pair of vectors is `5sqrt3`,then vectors areA. ` 3hati + 2hati -hatk ,3hati -hatj+4hatk`B. ` 3/2hati +1/2hatj -hatk , 2hati -6hatj +8hatk`C. ` 3hati + hatj -2hatk,hati + 3hatj +4hatk`D. none of these

Answer» Correct Answer - B
If `veca , vecb` are diagonals of a parallelogram, then its area is `1/2 |veca xx vecb|`
Clearly, ` 1/2 |veca xx vecb|= 5sqrt3` is sattisfied by the pair of vectors given in option (b) `
Discussion
27.

If the vectors ` veca.veca = xhati + y hatj + zhatk and vecb =hatj` are such that ` veca, vecb ,and vecb` from a right handed system , then ` vecc ` idA. ` xhati -xhatk`B. `vec0`C. `y hatj`D. `-zhat + x hatk`

Answer» Correct Answer - A
Since ` veca , vecb, vecb` form a right handed system
` veca xx vecc = vecb, vecc xx veca and vecb xx veca = vecc`
Now,
` vecc = vecb xx vecb xx veca Rightarrow vecc= |{:( hati,hatj,hatk),(0,1,0),(x,y,z):}| =zhati -xhatk`
Discussion
28.

if ` veca , vecb ,vecc ` are three vectors such that ` veca +vecb + vecc = vec0` thenA. ` veca .vecb= vecb .vecc= vecc.veca`B. ` veca xx vecb = vecb xx vecc = vecc xx veca`C. ` veca xx vecb = vecb xx vecc = veca xx vecc`D. ` vecb xx veca = vecb xx vecc = vecc xx veca`

Answer» Correct Answer - B
we have,
` veca +vecb +vecc= vec0`
` Rightarrow veca xx (veca + vecb + vecc ) = veca xx vec0` [ Taking cross product with `veca` ]
` Rightarrow veca +veca + veca xx vecb + vecc xx veca =0`
` Rightarrow vec0 + veca xx vecb -vecc xx veca =0`
` Rightarrow veca xx vecb = vecc xx veca`
Taking cross product of (i) with ` vecb, " we get " veca xx vecb = vecb xx vecc`
` veca xx vecb = vecc xx veca`
Discussion
29.

Let ` veca , vecb , vecc` be unit vectors such that ` veca .vecb =0=veca.vecc` . If the angle between `vecb and vecc " is " pi/6 , " then " veca ` equalsA. ` +- 2(vecb xx vecc)`B. ` 2(vecb xx vecc)`C. ` +- 1/2(vecb xx vecc)`D. ` - 1/2 (vecb xx vecc)`

Answer» Correct Answer - A
we have ,
` veca.vecb= veca.vecc = vec0`
` Rightarrow veca bot vecb, veca bot vecc`
` Rightarrow veca || (vecb xx vecc)`
` Rightarrow veca = +- (vecb xx vecc)/(|vecb xx vecc|) =+- (|vecb xx vecc|)/(|vecb||vecc| sin pi/3) = +- 2 ( vecb xx vecc)`
Discussion
30.

Let `triangle PQR` be a triangle. Let ` veca = vec(QR) , vecb = vec(RP) and vecc= vec(PQ) . " if " |veca| = 12, |vecb| = 4sqrt3 and vecb , vecc = 24` , then which of the following is ( are ) true ?A. `1/2|vecc|^(2) -|veca| =12`B. `1/2|vecc|^(2) + |veca| =30`C. ` |veca xx vecb + vecc xx veca| = 48sqrt3`D. `veca.vecb= -72`

Answer» Correct Answer - A::C::D
`veca= vec(QR) ,vecb = vec(RP) and vecc= vec(PQ)`
` Rightarrow veca + vecb + vecc = vec(QR) + vec(RP) + vec(PQ)`
` Rightarrow veca + vecb + vecc = vec(OQ) `
` Rightarrow veca + vecb + vecc= vec0`
` Rightarrow vecb + vecc = -veca`
` Rightarrow vecb + vecc = |-veca|`
` |vecb + vecc|^(2) =|veca|^(2)`
` |vecb|^(2) +|vecc|^(2) + 2(vecb .vecc) = |veca|^(2) `
` 48 + |vecc|^(2) + 48 = 144 `
` |vecc| = 4sqrt3`
` 1/2 |vecc|^(2) - |veca| = 24 -12 =12 and 1/2 |vecc|^(2) + | veca| = 24 + 12 =36`
So, option (a) is true and option (b) is not true. Again.
` veca + vecb + vecc= vec0`
` Rightarrow |veca + vecb| = | -vecc|`
` Rightarrow |veca + vecb|^(2) +2 (veca.vecb) = |vecc|^(2)`
` Rightarrow 144 + 48+2 (veca .vecb) = 48`
` Rightarrow veca.vecb = -72`
So, option (d) is true.
Again
` veca + vecb + vecc= vec0`
` Rightarrow veca xx ( veca + vecb + vecc) = veca xx vec0`
` Rightarrow veca xx veca + veca xx vecb + veca xx vecc= vec0`
` Rightarrow veca xx vecb = vecc xx veca `
` therefore |veca xx vecb + vecc xx veca| = |2 (veca xx vecb)| = 2|veca xx vecb|`
` 2 sqrt(|veca|^(2) |vecb|^(2) -(veca -vecb)^(2) )`
` 2 sqrt(144xx 48 -(-72) ^(2)) = 48sqrt3`
So, option (c) is correct
Hence, option (a),(c) and (d) are ture.
Discussion
31.

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. ` a lt x lt 1//2`B. ` 1//2 lt x lt 15`C. ` x gt 1//2 or x lt 0`D. none of these

Answer» Correct Answer - D
The angle between `veca and vecb` is obtuse.
`veca.vecb lt 0`
` Rightarrow 14x^(2) - 8x +x gt 0 Rightarrow 7x(2x -1) gt 0 Rightarrow 0 lt x lt 1//2 …(1)`
The angle between `vecb` and z-axis is acute and less thann `pi//6` .
` ( vecb.veck)/(|vecb||veck|) gt cos "" pi/6 " " [ therefore, theta , pi/6 Rightarrow cos theta gt cos "" pi/6]`
` x/ (sqrt(x^(2) +53)) gt sqrt3/2`
` Rightarrow 4x^(2) gt 3x^(2) + 159`
` Rightarrow x^(2) gt 159 Rightarrow gt sqrt(159) or x gt - sqrt159 `
Clearly, (i) and (ii) cannot hold togerther.
Discussion
32.

`(veca.hati)(vecaxxhati)+(veca.hatj)+(veca.hatk)(vecaxxhatk)` is equal toA. ` 3veca`B. `veca`C. `vec0`D. `2veca`

Answer» Correct Answer - C
Let `veca =a_(1)hati+a_(2)hati +a_(2)hatj +a_(3)hatk`, Then,
` veca.hati=a_(1),veca.hatj=a_(2), veca.hatk =a_(3)`
` veca xx hati =- a_(2) hatk + a_(2) hatj ,veca xx hatj = a_(1)hatk -a_(3)hati , veca xx hatk = - a_(1)hatk + a_(2)hati `
` (veca .hati) (veca xx hati) + (veca .hatj) (veca xx hatj) + ( veca .hatk) (veca xx hatk)`
` = a_(1) (-a_(2)hatk + a_(3)hatj) + a_(2) (a_(1)hatk -a_(3) hati) + a_(3) (-a_(1)hatj + a_(2)hati) = vec0`
Discussion