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51.

Using vectors, find the area of the triangle with vertices `A(1,1,2), B(2,3,5)a n d C(1,5,5)`

Answer» The vertices of triangle ABC are given as A (1,1,2) , B (2,3,5) and C (1,5,5).
The adjacent sides `vec(AB) and vec(BC) of triangleABC` are given as
`vec(AB)=(2-1)hati+(3-1)hatj+(5-2)hatk=hati+2hatj + 3hatk`
`vec(BC)=(1-2)hati+(5+3)hatj+(5-5)hatk=-hati+2hatj`
`Area of triangleABC=1/2|vec(AB)xxvec(BC)|`
`vec(AB)xxvec(BC)=|{:(hati,hatj,hatk),(1,2,3),(-1,2,0):}|=hati(-6)-hatj(3)+hatk(2+2)=-6hati-3hatj+4hatk`
`|vec(AB)xxvec(BC)|=sqrt((-6)^(2)+(-3)^(2)+4^(2))=sqrt(36+9+16)=sqrt61`
Hence , the area of `triangle is sqrt61/2` square units.
Discussion
52.

If `veca` and `vecb` are vectors in space given by `veca=(hati-23hatj)/(sqrt(5))` `vecb=(2hati+hatj+3hatk)/(sqrt(14))` then the value of `(2veca+vecb).[(vecaxxvecb)xx(veca-2vecb)]`, is

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
53.

The volume of the tetrahedronwhose vertices are the points with position vectors `hati-5hatj+10hatk, -hati-3hatj+7hatk, 5hati-hatj+lamdahatk` and `7hati-4hatj+7hatk ` is 11 cubic units then the value of `lamda` is (A) 7 (B) 1 (C) -7 (D) -1

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
54.

`A , B , Ca n dD`are any four points in thespace, then prove that `| vec A Bxx vec C D+ vec B Cxx vec A D+ vec C Axx vec B D|=4`(area of ` A B C`.)

Answer» Let P.V of A,B, C d and D be `veca, vecb,vecc and vec0` , respectively. Then `vec(AB)xxvec(CD)=(vecb-veca)xx(-vecc), vec(BC)xxvec(AD)=(vecc-vecb)xx(-veca)`
`vec(CA)xxvec(BD)=(veca-vecc)xx(-vecb)`
`vec(AB)xxvec(CD)+vec(BC)xxvec(AD)+vec(CA)xxvec(BD)`
`veccxxvecb+vecaxxvecc+vecaxxvecc+vecbxxveca-vecaxxvecb+veccxxvecb`
`2(veccxxvecb+ vecbxxvecbxxveca+vecaxxvecc)`
`2(veccxx(vecb-veca)-vecaxx(vecb-veca))`
`= 1(vec(AC)xxvec(AB))`
`|vec(AB)xxvec(CD)+vec(BC)xxvec(AD)+vec(CA)xxvec(BD)|=4|1/2(vec(AC)xxvec(AB))|=4triangleABC`
Discussion
55.

If the scalar projection of vector `xhati-hatj+hatk1` on vector `2hati-hatj+5hatk is 1/sqrt30`. The find the value of x.

Answer» Projection of `xhati-hatj+hatk "on" 2hati-hatj+5hatk= ((xhati-hatj+hatk).(2hati-hatj+5hatk))/(sqrt(4+1+25))`
`(2x+1+5)/(sqrt10)`
But, given `(2x+6)/(sqrt30)=1/sqrt30 Rightarrow 2x+6=1 or x=(-5)/2`
Discussion
56.

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
57.

If `theta` is the angle between the unit vectors `veca and vecb`, then prove that i. `cos((theta)/2)=1/2|veca+vecb|`` ii. sin (theta/2)=1/2|veca-vecb|`

Answer» i. `(veca+vecb).(veca+vecb)= |veca|^(2)+|vecb|^(2)+2veca.vecb`
`=1 +1+2(1)(1) cos theta`
`cos(theta/2)=1/2|veca+vecb|`
ii. `(veca-vecb).(veca-vecb)=|veca|^(2)=|vecb|^(2)-2veca.vecb`
`= 1+1-2(1)(1)cos theta`
`=2-2costheta`
`or sin"theta/2=1/2|veca+vecb|`
Discussion
58.

If three unit vectors `veca, vecb and vecc " satisfy" veca+vecb+vecc= vec0`. Then find the angle between `veca and vecb`.

Answer» `veca+vecb=-vecc`
`|veca+vecb|^(2)+2veca.vecb=1`
` veca.vecb=-1/2`
`|veca||vecb|cos theta =-1/2`
or `cos theta = 1/2`
`theta= (2pi)/3`
Discussion
59.

Let `veca` and `vecb` be two non-collinear unit vectors. If `vecu=veca-(veca.vecb)vecb` and `vec=vecaxxvecb`, then `|vecv|` isA. `|vecu|`B. `|vecu|+ |vecu.vecb|`C. `|vecu| + |vecu .veca|`D. none of these

Answer» Correct Answer - b,d
`vecu = vecas - ( veca . Vecb) vecb`
` vecb xx (veca xx vecb)`
` Rightarrow |vecu| = |vecb xx (veca xx vecb)|`
` |vecb| |veca xx vecb| sin 90 ^(@)`
` |vecb||veca xx vecb|`
` |vecv|`
Also ` vecu . Vecb = vecb.vecb xx (veca xx vecb)`
` [vecb vecb veca xx vecb]`
=0
` Rightarrow |vecv| = |vecu| + |vecu. vecb|`
Discussion
60.

Vectors perpendicular to`hati-hatj-hatk` and in the plane of `hati+hatj+hatk and -hati+hatj+hatk` are (A) `hati+hatk` (B) `2hati+hatj+hatk` (C) `3hati+2hatj+hatk` (D) `-4hati-2hatj-2hatk`A. `hati + hatk`B. `2hati + hatj + hatk`C. `3hati+ 2 hatj + hatk`D. `-4 hati - 2hatj - 2hatk`

Answer» Correct Answer - b,d
Let `alpha =hati - hatj -hatk , vecbeta = hati + hatj +hatk`
` and vecgmma = -hati +hatj +hatk`
Let requried vector, `veca=xhati + yhatj +zhatj`
then `vecalpha, vecbeta, vecgamma` are coplanar, i.e.
`|{:(x,y,z),(1,1,1),(-1,1,1):}|=0or y=z`
Also `veca and vecalpha` are perpendicular,
` Rightarrow x-y -z =0`
`Rightarrow x = zy`
Discussion
61.

If `veca + 2 vecb + 3 vecc = vec0 " then " veca xx vecb + vecb xx vecc + vecc xx veca= `A. `2 (veca xx vecb)`B. ` 6( vecb xx vecc)`C. `3 ( vecc xx veca)`D. `vec0`

Answer» Correct Answer - c,d
we know that `veca + vecb + vecc = vec0` then
`veca xx vecb =vecb xx vecc = vecc xx veca`
Given `veca + 2vecb + 3 vecc = vec0`
`2 veca xx vecb = 6 vecb xx vecc = 3 vecc xx veca`
Hence `veca xx vecb + vecb xx vecc + vecc xx veca `
` 2 (veca xx vecb) or 6 (vecb xx vecc) or 3 (vecc xx veca)`
Discussion
62.

`veca ,vecb and vecc` are unimdular and coplanar. A unit vector `vecd` is perpendicualt to them , `(veca xx vecb) xx (vecc xx vecd) = 1/6 hati - 1/3 hatj + 1/3 hatk` , and the angle between `veca and vecb is 30^(@)` then `vecc` isA. `(hati-2 hatj + 2 hatk)//3`B. `(-hati +2hatj - 2 hatk)//3`C. `(-hati +2hatj - hatk)//3`D. `(-2hati -2hatj + hatk)//3`

Answer» Correct Answer - a,b
Given `1/6hati -1/3 hatj + 1/3hatk = (veca xx vecb) xx (veccxxvecd)`
`[ veca vecb vecd]vecc - [veca vecb vecc] vecd`
`[veca vecb vecd]vecc`
`[ veca, vecb and vecc` are coplanar]
`[veca vecb vecd] = (veca xx vecb).vecd`
` |veca xx vecb| |vecd| cos theta`
`(therefore vecd bot veca,vecd,bot vecb, therefore vecd||vecaxxvecb)`
`ab sin 30^(@) .1. (+-1) ( theta = 0 or pi) `
`1.1. 1/2 .1(+-1)= +- 1/2`
form (i) m we have
`vecc= +- (1/3 hati -2/3hatj +2/3hatk) =+- (hati -2hatj + 2hatk)/3`
Discussion
63.

The volume of heparallelepiped whose sides are given by ` vec O A=2i-2, j , vec O B=i+j-ka n d vec O C=3i-k`is`4//13`b. `4`c. `2//7`d. `2`A. `4//13`B. 4C. `2//7`D. 2

Answer» Correct Answer - d
Volume of parallelepiped = `[veca vecb vecc]`
`|{:(2,-2,0),(1,1,-1),(3,0,-1):}|=2(-1)+_2(-1+3)=2`
Discussion
64.

If `vecA=(1,1,1) and vecC=(0,1,-1)` are given vectors then find a vector `vecB` satisfying equations `vecAxxvecB=vecC and vecA.vecB=3`

Answer» Correct Answer - `5/3hati+2/3hatj + 2/3 hatk`
Given `vecA = hati + hatj - hatk and vecC = hatj -hatk`
Let `vecB =x hati + yhatj +zhatk`
Given that `vecAxxvecB=vecC Rightarrow |{:(hati,hatj,hatk),(1,1,1),(x,y,z):}|=hatj=hatk`
`or (z-y)i +(x-z)hatj +(y-x)hatk=hatj-hatk`
z-y=0 , x-z=1 and y-x =-1
Also , `vecA. vecB=3`
`Rightarrow x+y +z=3`
Using (i) and(ii) , we get
`y=2//3,xx=5//3,z=2//3`
`vecB = 5/3 hati + 2/3hatj + 2/3 hatk`
Discussion
65.

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` equals (A) `+-(hati+hatj-2hatk)/sqrt(6)` (B) `+-(hati+hatj-hatk)/sqrt(3)` (C) `+-(hati+hatj+hatk)/sqrt(3)` (D) `+-hatk`A. `+- (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
66.

If `veca, vecb and vecc` are three non-coplanar vectors, then find the value of `(veca.(vecbxxvecc))/(vecb.(veccxxveca))+(vecb.(veccxxveca))/(vecc.(vecaxxvecb))+(vecc.(vecbxxveca))/(veca.(vecbxxvecc))`

Answer» Since ` (veca vecb vecc] ne 0`, we have
`(veca.(vecbxxvecc))/(vecb.(veccxxveca))+(vecb.(veccxxveca))/(vecc.(vecaxxvecb))+(vecc.(vecbxxveca))/(veca.(vecbxxvecc))=([veca vecb vecc])/([vecb vecc veca])+([vecbveccveca])/([vecc veca vecb])+([vecc vecb veca])/([veca vecb vecc])`
`= ([vecavecb vecc])/([veca vecbvecc])+([vecavecbvecc])/([vecavecbvecc])-([vecavecbvecc])/([veca vecb vecc])`
= 1+ 1 -1=1
Discussion
67.

For non zero vectors `veca,vecb, vecc` `|(vecaxxvecb).vec|=|veca||vecb||vecc|` holds iffA. `veca.vecb=0 , vecb .vecc=0`B. `vecb.vecc = 0, vecc, veca =0`C. `vecc.veca =0, veca,vecb =0`D. `veca.vecb= vecb.vecc= vecc.veca =0`

Answer» Correct Answer - d
`|(veca xx vecb).vecc|=|veca||vecb||vecc|`
`or||veca||vecb|sin theta cos alpha|=|veca||vecb||vecc|`
` or |sinstheta||cos alpha|=1`
`Rightarrow theta=pi//2 and alpha=0`
` Rightarrow veca bot vecband vecc||hatn`
or perpendicular to both `veca and vecb`
`Rightarrow veca.vecb=vecb.vecc=vecc.veca=0`
Discussion
68.

Let `veca=xhati+12hatj-hatk,=2hati+2xhatj+hatkand vecc=hati+hatk`. If the ordered set `[vecb vecc veca]` is left handed, then find the value of x.

Answer» for a left-handed system .
`[vecbveccveca]lt0`
`|{:(2,2x,1),(1,0,1),(x,12,-1):}|lt0`
`or 2(0-12)-2x(-1-x)+1(12)lt0`
` or x^(2)+x-6lt0`
`x in (-3,2)`
Discussion
69.

If `veca, vecb and vecc`1 are three non-coplanar vectors, then `(veca + vecb + vecc). [(veca + vecb) xx (veca + vecc)]` equals

Answer» Correct Answer - d
`(veca+ vecb+vecc).[(veca+vecb)xx(veca+vecc)]`
`= (veca+vecb+vecc).[veca xx veca +vecaxx vecc+vecb xx veca +vecbxxvecc]`
`= (veca +vecb +vecc).[veca xx vecc +vecb xx veca+vecb xxvecc]`
`[veca vecb vecc] -[veca vecb vecc]-[veca vecb vecc]`
`-[veca vecb vecc]`
Discussion
70.

The scalar `vecA.(vecB+vecC)xx(vecA+vecB+vecC)` equals (A) 0 (B) `[vecA vecB vecC]+[vecB vecC vecA]` (C) `[vecA vecB vecC]` (D) none of these

Answer» Correct Answer - a
`vecA.(vecB + vecC) xx (vecA + vecB +vecC)`
` vecA . ([vecB xx vecA +vecB xx vecB + vecC + vecC xx vecA + vecC xx vecB +vecC xx vecC]`
`= vecA.vecBxxvecA + vecA.vecBxxvecC+vecA.vecCxx vecA + vecA+ vecC xx vecB " " ( "using" veca xx veca =0)`
`0+[vecA vecB vecC]+0+ [vecA vecC vecB]`
`[vecA vecB vecC]-[vecA vecB vecC]`
=0
Discussion
71.

In a `/_ABC `points D,E,F are taken on the sides BC,CA and AB respectively such that `(BD)/(DC)=(CE)/(EA)=(AF)/(FB)=n` prove that `/_DEF= (n^2-n+1)/((n+1)^2)` /_ABC`

Answer» Take A as the origin and last the postion vectors of point B and C be `vecb and vecc`, Respectively.
Therefore, the position vectors, of D, E and F are, respectively, `(nvecc+vecb)/(n+1),vecc/(n+1)and (n vecb)/(n +1)`. Therefore,
`vec(ED)=vec(AD)-vec(AE)=((n-1)vecc+vecb)/(n+1)and vec(EF)=(nvecb-vecc)/(n+1)`
vector area of `DeltaABC=1/2(vecbxxvecc)`
vector area of `DeltaDEF=1/2(vec(EF)xxvec(ED))`
`1/(2(n+1)^(2))[(nvecb-vecc)xx{(n-1)vecc+vecb}]`
`= 1/(2(n+1)^(2))[(n^(2)-n)vecbxxvecc+vecbxxvecc]`
`1/(2(n+1)^(2))[(n^(2)-n+1)(vecbxxvecc)]=(n^(2)-n+1)/((n+1)^(2))Delta_(ABC)`
Discussion
72.

for any three vectors, `veca, vecb and vecc , (veca-vecb) . (vecb -vecc) xx (vecc -veca) = 2 veca.vecb xx vecc`.

Answer» Clearly, vectors `veca - vecb, vecb-vecc, vecc-veca` are coplanar,
`Rightarrow [ veca -vecb vecb- vecc vecc-veca] =0`
Therefore, the given statement is false.
Discussion
73.

If `vecb` is not perpendicular to `vecc` . Then find the vector `vecr` satisfying the equation `vecr xx vecb = veca xx vecb and vecr. vecc=0`

Answer» Given `vecrxxvecb=vecaxxvecbRightarrow (vecr-veca)xxvecb=0`
Hence, (`vecr - veca) and vecb` are parallel.
`Rightarrow vecr-veca=tvecb`
`vecr.vecc=0`
Therefore, taking dot product of (i) by `vecc`, we get `vecr.vecc -veca.vecc=t(vecb .vecc)`
`or 0-veca.vecc=r(vecb.vecc)ort=-((veca.vecc)/(vecb.vecc))`
from (i) and (ii) solution of `vecr" is " vecr =veca-((veca.vecc)/(vecb.vecc))vecb`
Discussion
74.

If `veca and vecb` are two given vectors and k is any scalar,then find the vector `vecr` satisfying `vecr xx veca +k vecr=vecb`.

Answer» `vecrxxveca+kvecr=vecb`
`or (vecrxxveca)xxveca+kvecrxxveca= vecbxxveca`
`or (vecr. veca)veca-(veca.veca)vecr+k(vecb-kvecr)=vecbxxveca`
`or (vecr. veca)veca+kvecb-vecbxxveca=(|veca|^(2)+k^(2))vecr`
`or vecr((vecr.veca)veca+kvecb-vecbxxveca)/(|veca|^(2)+k^(2))`
Also, in Eq, (i) taking dot product with `veca`., we have
`(vecrxxveca).veca+kvecr.veca=vecb.veca`
`or vecr. veca=(vecb.veca)/k`
`Rightarrow vecr=1/(k^(2)+|veca|^(2))[((veca.vecb)veca)/k+kvecb+(vecaxxvecb)]`
Discussion
75.

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

Answer» Correct Answer - 3
As `|veca-vecb|^(2)+|vecb-vecc|^(2)+|vecc-veca|^(2)`
`=3(|veca|^(2)+|vecb|^(2)+|vecc|^(2))-|veca+vecb+vecc|^(2)`
`Rightarrow 3xx 3-|veca+vecb+vecc|=9`
`or |veca +vecb+vecc|=0`
`veca +vecb +vecc=0`
`or vecb +vecc=-veca`
`Rightarrow |2veca+5(vecb +vecc)|=|-3veca|=3|veca|=3`
Discussion
76.

`hatu and hat v` are two non-collinear unit vectors such that `|(hatu+hatv)/2+hatuxxvecv|=1`. Prove that `|hatuxxhatv|=|(hatu-hatv)/2|`

Answer» Given that `|(hatu+hatv)/2+hatuxxhatv|=1`
` |(hatu+hatv)/2+hatuxxhatv|^(2)=1`
`(2+2costheta)/4+sin^(2)theta=1`
`cos^(2)(theta /2)=cos^(2)theta`
`theta=npi+-theta/2,n inZ`
`(2pi)/3`
`|hatuxxhatv|=sin((2pi)/3)=sin (pi/3)=|(hatu-hatv)/2|`
Discussion
77.

If `vecx.veca=0vecx.vecb=0 and vecx.vecc=0` for some non zero vector `vecx` then show that `[veca vecb vecc]=0`

Answer» Correct Answer - 1
` vecX. vecA = 0 Rightarrow " either " vecA = 0 or vecX bot vecA`
` vecX.vecC =0 Rightarrow " either" vecC=0 vecX bot vecC`
In any of the three cases,
`vecA, vecB, vecC =0 Rightarrow [vecA vecB vecC] =0`
Otherwise if `vecX bot vecA, vecX bot vecB and vecX bot vecC` , then
`vecA, vecB and vecC` are coplanar. then
` [ vecA vecB vecC] =0`
Therefore, the statement is true.
Discussion
78.

`[vecaxx vecb " " vecc xx vecd " " vecexx vecf]` is equal toA. `[veca vecb vecd] [vecc vece vecf]-[veca vecb vecc] [vecd vece vecf]`B. `[veca vecb vece ] [vecf vecc vecd] - [veca vecb vecf] [vece vecc vecd] `C. `[vecc vecd veca] [vecb vece vecf] - [veca vecd vecb] [vecavece vecf]`D. `[veca vecc vece] [vecb vecd vecf]`

Answer» Correct Answer - a,b,c
Let `vecA=vecaxxvecb, vecB=veccxxvecd and vecC=vecexxvecf`
we know that
`vecA.(vecB xx vecC)=vecB. (vecCxx vecA) = vecC.(vecAxx vec B) `
`(vecaxx vecb).[(veccxxvecd) xx (vecexxvecf)]`
`(vecaxxvecb) .[{(veccxx vecd).vecf} vece. {(vecc xx vecd) vece} vecf}`
`[vecc vecd vecf] [vecavecb vece] - [vecc vecd vece] [veca vecb vecf]`
similarly, other parts can be obtained .
Discussion
79.

If `vecr.veca=0, vecr.vecb=1and [vecr vecavecb]=1,veca.vecbne0,(veca.vecb)^(2)-|veca|^(2)|vecb|^(2)=1,` then find `vecr` in terms of `veca and vecb`.

Answer» Writing `vecr` as a linear combination of `veca , vecb and veca xx vecb` . We have
`vecr=xveca+yvecb+z(vecaxxvecb)`
for scalars x,y and z
`0=vecr.veca=x|veca|^(2)+yveca.vecb " " ("taking dot product with "veca)1`
`1- vecr.vecb=xveca.vecb+y|vecb|^(2)" " ("taking dot product with"vecb)`
Solving , we get y `(|veca|^(2))/(|veca|^(2)|vecb|^(2)-(veca.vecb)^(2))=|veca|^(2)`
`and x=(veca.vecb)/((veca.vecb)^(2)-|veca|^(2)|vecb|^(2))=veca.vecb`
Also `1=[vecr vecavecb]=z|vecaxxvecb|^(2) " " ("taking dot product with "vecaxxvecb)`
`z=1/(|vecaxxvecb|^(2))`
`vecr=((veca.vecb)veca-|veca|^(2)vecb)+(vecaxxvecb)/(|vecaxxvecb|^(2))`
`=vecaxx(vecaxxvecb)+(vecaxxvecb)/(|vecaxxvecb|^(2))`
Discussion
80.

Let `veca` and `vecb` be unit vectors such that `|veca+vecb|=sqrt(3)`, then the value of `(2veca+5vecb)`. `(3veca+vecb+vecaxxvecb)=`

Answer» `(2veca+5vecb).(3veca+vecb+vecaxxvecb)=6veca.veca+17veca.vecb+5vecb.vecb=11+17veca.vecb`
`(veca.(vecaxxvecb)=vecb.(vecaxxvecb)=0,as veca and vecb` are perpendicular to `veca xx vecb`)
`|veca+vecb|=sqrt3`
`|veca+vecb|^(2)=3`
`|veca|^(2)+|vecb|^(2)+2veca.vecb=3`
`veca.vecb=1/2`
`Rightarrow (2veca+5vecb).(3veca+vecb+veca xx vecb)11+17/2=39/2`
Discussion
81.

`veca , vecb and vecc` are three non-coplanar vectors and `vecr`. Is any arbitrary vector. Prove that `[vecbvecc vecr]veca+[vecc veca vecr]vecb+[vecavecbvecr]vecc=[veca vecb vecc]vecr`.

Answer» `Let" " vecr=x_(1)veca+x_(2)vecb+x_(3)veccRightarrow vecr.(vecbxxvecc)=x_(1)veca.(vecbxxvecc)or x_(1)=([vecr vecb vecc])/([veca vecb vecc])`.
`Also, " " vecr.(veccxxveca)=x_(2)vecb.(veccxxveca)orx_(2)=([vecrveccveca])/([vecavecbvecc])`
`and vecr.(vecaxxvecb)=x_(3)vecc. (vecaxxvecb)or x_(3)=([vecr vecavecb])/([veca vecbvecc])`
`Rightarrow vecr=([vecrvecb vecc])/([vecavecbvecc])veca+([vecrveccveca])/([vecavecbvecc])vecb+([vecrvecavecb])/([vecavecb vecc])vecc`
`or [vecbveccvecr]veca+[veccvecavecr]vecb+[vecavecbvecr]vecr=[vecavecbvecc]vecr`
Discussion
82.

if vector `vecx` satisfying `vecx xx veca+ (vecx.vecb)vecc =vecd` is given by `vecx = lambda veca + veca xx (vecaxx(vecdxxvecc))/((veca.vecc)|veca|^(2))`

Answer» `vecx xxveca=(vecx.vecb)vecc=vecd`
Discussion
83.

If `[ veca vecbvecc]=2` , then find the value of `[(veca+2vecb-vecc) (veca - vecb) (veca - vecb-vecc)]`

Answer» `[(veca+2vecb-vecc)(veca-vecb)(veca-vecb-vecc)]`
`=|{:(1,2,-1),(1,-1,0),(1,-1,-1):}|[veca vecbvecc]`
`= 2[veca vecb vecc] = 6`
Discussion
84.

If `veca,vecb and vecc` are three mutually perpendicular unit vectors and `vecd` is a unit vector which makes equal angal with `veca,vecb and vecc`, then find the value of `|veca+vecb+vecc+vecd|^(2)`.

Answer» `|veca+vecb+vecc+vecd|^(2)=sum|veca|^(2)+2sumveca.vecb=4+2vecd.(veca+vecb+vecc) " " (veca,vecb,vecc "are mutually perpendicular")`
`Let vecd=lambdaveca+muvecb+v vecc. Then vecd.veca=vecd.vecb=vecd.vecc=costheta`Therefore,
`lambda=mu=v=costheta`
` lambda^(2)+mu^(2)+v^(2)=1Rightarrow3cos^(2)theta=1costheta=1/sqrt3`
`|veca+vecb+vecc+vecd|^(2)=4+-(2.3)/sqrt3=4+-2sqrt3`
Discussion
85.

if `veca,vecb and vecc` are mutally perpendicular vectors of equal magnitudes, then find the angle between vectors and` veca+ vecb=vecc`.

Answer» since `veca,vecb and vecc` are mutually perpendicular , `veca.vecb=vecb.vecc=vecc.veca=0`
Angle between `veca and veca+vecb+vecc` is
`cos theta=(veca.(veca+vecb+vecc))/(|veca||veca+vecb=vecc|)`
`now |veca|=|vecb|=|vecc|=a`
`|veca+vecb=vecc|^(2)= |veca|^(2)|vecb|^(2)+|vecc|^(2)+2veca.vecb+2vecb.vecc+vecc.veca`
`=a^(2)=a^(2)+a^(2)+0+0+0`
`3a^(2)`
`|veca+vecb+vecc|=sqrt3a`
Putting this value in (i) , we get `theta = cos^(-1) 1/sqrt3`
Discussion
86.

Let `veca,vecb and vecc` be three vectors such that `vecane0, |veca|=|vecc|=1,|vecb|=4and |vecbxxvecc|=sqrt15`. If `vecb-2vecc=lambdaveca` then find the value of `lambda` .

Answer» Let the angle between `vecb and vecc be alpha`. Then
`|vecbxxvecc|=sqrt15`
`|vecb||vecc| sinalpha=sqrt15`
`sin alpha=sqrt15/4`
` cos alpha = 1/4`
`Rightarrowvecb-2vecc=lambdaveca`
`or|vecb-2vecc|^(2)=lambda^(2)|veca|^(2)`
`|vecb|^(2)+4||vecc|^(2)-4.vecb.vecc=lambda^(2)|veca|^(2)`
`or 16+ 4 -4 xx 4xx 1 xx 1/4=lambda^(2)`
`or lambda^(2)=16 or lambda = +-4`
Discussion
87.

Let `vecA , vecB and vecC` be vectors of legth , 3,4and 5 respectively. Let `vecA` be perpendicular to `vecB + vecC, vecB " to " vecC + vecA and vecC " to" vecA + vecB` then the length of vector `vecA + vecB+ vecC` is __________.

Answer» Correct Answer - `5sqrt2`
Given that `|vecA|=3, |vecB|=4, |vecC|=5`
`vecAbot (vecB + vecC) Rightarrow vecA. (vecB +vecC) =0`

`Rightarrow vecA.vecB + vecA.vecC=0`
` vecB bot (vecC +vecA)RightarrowvecB.(vecC+vecA_=0`
`Rightarrow vecB.vecC+vecB.vecA=0`
`vecCbot (vecA+vecB) RightarrowvecC. (vecA+vecB)=0`
` Rightarrow vecC.vecA+vecC.vecdB=0`
Adding (i), (ii) and (iii) we get
`2(vecA.vecBr+vecB.vecC+vecC.vecA)=0`
Now , `|vecA + vecB + vecC|^(2)`
`(vecA + vecB+vecC).(vecA + vecB+vecC)`
`|vecA|^(2)+|vecB|^(2)+|vecC|^(2)`
`+2(vecA.vecB + vecB.vecC+vecC.vecA)`
9+16+25+0
= 50
`|vecA + vecB +vecC|= 5sqrt2`
Discussion
88.

Let `veca,vecb and vecc`are vectors such that `|veca|=3,|vecb|=4and |vecc|=5, and (veca+vecb)` is perpendicular to `vecc,(vecb+vecc)` is perpendiculatr to `veca` and `(vecc+veca)` is perpendicular to `vecb`. Then find the value of `|veca+vecb+vecc|`.

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
89.

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

Answer» Correct Answer - b
`|(veca xxvecb)xxvecc|=|vecaxxvecb||vecc|sin30^(@)`
`=1/2 (vecp +vecq+vecr)a^(2)`
`or vecx=1/2 (vecp +vecq +vecr)`
we have `veca = 2hati+hatj -2hatk and vecb = hati+hatj`
`Rightarrow vecaxx vecb = 2hati -2hatj+hatk`
`or |veca xx vecb|=sqrt9=3`
Also given `|vecc-veca|^(2)=8`
`or |vecc|^(2)=|veca|^(2)-2veca.vecc=8`
Given `|aveca|=3 and veca. vecc =|vecc|` , using these we get
`|vecc|^(2) -2|vecc|+1=0`
`or (|vecc|-1)^(2)=0`
`or |vecc|=1`
Substituting values of `|veca xx vecb|and |vecc|` in (i), we get
`|(veca xx vecb)xxvecc|=1/2xx 3xx 1= 3/2`
Discussion
90.

Let `veca = 2i + j+k, vecb = i+ 2j -k and a` unit vector `vecc` be coplanar. If `vecc` is pependicular to `veca`. Then `vecc` isA. `1/sqrt2(-j+k)`B. `1/sqrt3(i-j-k)`C. `1/sqrt5(i-2j)`D. `1/sqrt3(i-j-k)`

Answer» Correct Answer - a
As `vecc` is coplanar with `veca and vecb` we take
`vecc = alpha veca + betavecb`
where `alpha and beta` are scalars.
As `vecc` is perpendicular to `veca` , using (i), we get
`0 = alphaveca.veca alpha +betavecb.veca`
`or 0 =alpha (6) +beta(2+2-1) =3 (2alpha+beta) `
`or beta = -2alpha`
Thus `vecc=alpha(veca -2vecb)=alpha(-3j+3k)=3alpha(-j+k)`
`or |vecc|^(2)=18alpha^(2)`
`or 1=18alpha^(2)`
`or alpha= +- 1/(3sqrt2)`
`vecc =+- 1/sqrt2 (-j+k)`
Discussion
91.

Let `vec(alpha)=ahati+bhatj+chatk, vec(beta)=bhati+chatj+ahatk` and `vec(gamma)=chati+ahatj+bhatk` be three coplnar vectors with `a!=b`, and `vecv=hati+hatj+hatk`. Then `vecv` is perpendicular toA. `vecalpha`B. `vecbeta`C. `vecgamma`D. none of these

Answer» Correct Answer - a,b,c
It is given that `vecalpha, vecb and vecgamma` are coplanar vectors, therefore
` [vecalpha vecbeta vecgamma] =0`
`Rightarrow |{:(a,b,c),(b,c,a),(c,a,b):}|=0`
` or 3abc -a^(3) -b^(3) -c^(3) =0`
` or a^(3) +b^(3) +c^(3) -3abc =0`
` or (a + b +C0 (a^(2) +b^(2) + c^(2) -ab -bc-ca)=0`
or a+b+c =0
` [ a^(2) +b^(2) +c^(2)-ab -bc-ca ne 0)`
` Rightarrow vecv. vecalpha = vecv. vecbeta = vecv. vecgamma = 0`
Hence, `vecv` is perpendicular to `vecalpha ,vecbeta and vecgamma`
Discussion
92.

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

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