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previous year questions Of Vector Algebra For NDA

previous year questions Of Vector Algebra

Set - 1
Q 2723391241

If `veca = hati - hatj + hatk` , `vecb = 2 hati +3 hatj +2 hatk` and `vecc = hati +m hat j+n hatk` are three coplanar vectors and `|vecc| = sqrt6`, then which one of the following is correct?
NDA Paper 1 2017
(A)

`m=2` and `n=±1`

(B)

`m=±2` and `n=-1`

(C)

`m = 2` and `n = -1`

(D)

`m = ±2` and `n = 1`

Solution:

If `vec a , vec b , vec c` are coplanar

`vec a * (vec b xx vec c)=0`

`|(1,-1, 1),(2,3,2),(1,m,n)|=0`

`1(3n- 2m)+ 1 (2n - 2)+ 1( 2m -3)=0`

`5n-5=0`

`n=1`

Now `| vec c| = sqrt 6`

`sqrt(1^2 +m^2 +n^2) = sqrt 6`

`1+1+m^2 = 6`

`m= pm 2`
Correct Answer is `=>` (D) `m = ±2` and `n = 1`
Q 2753391244

Let ABCD be a parallelogram whose diagonals intersect at `P` and let `O` be the origin what is `vec(OA) + vec(OB)+vec(OC)+vec(OD)` equal to?
NDA Paper 1 2017
(A)

`2 vec(OP)`

(B)

`4 vec(OP)`

(C)

`6 vec(OP)`

(D)

`8 vec(OP)`

Solution:

`vec(OA)+vec(OB)+vec(OC)+vec(OD)=Q`

`=(vec (OP) + vec(PA) )+(vec(OP) + vec(PB))+(vec(OP) + vec (PC))+ (vec (OP)+ vec (PD))`

`4 vec(OP) + ( vec(PA)+ vec(PD)+ vec(PB) + vec(PC))`

`= 4 vec(OP) +(vec(PA) - vec(PA) + vec(PB)- vec(PB))`

(`∵` For a parallogram `P` bisects the two diagonal

`= 4 vec( OP)`
Correct Answer is `=>` (B) `4 vec(OP)`
Q 2773391246

`ABCD` is a quadrilateral whose diagonals are `AC` and `BD`. Which one of the following is correct?
NDA Paper 1 2017
(A)

`vec(BA)+vec(CD) = vec(AC)+vec(DB)`

(B)

`vec(BA)+vec(CD) = vec(BD)+vec(CA)`

(C)

`vec(BA)+vec(CD) = vec(AC)+vec(BD)`

(D)

`vec(BA)+vec(CD) = vec(BC)+vec(AD)`

Solution:

`vec (BA) = vec(BD)+ vec (DA)`

`vec (CD) = vec(CA) + vec (AD)`

`vec (BA) + vec(CD) =vec (BD) + vec(CA)`
Correct Answer is `=>` (B) `vec(BA)+vec(CD) = vec(BD)+vec(CA)`
Q 2713491340

If `veca xx vecb = vec c` and `vecb xx vecc = vec a`, then which one of the following is correct?
NDA Paper 1 2017
(A)

`veca , vecb , vecc` are orthogonal inpairs and `|veca| = | vec c|` and `|vecb| = 1`

(B)

`veca , vecb , vecc` are non-orthogonal to each other

(C)

`veca , vecb , vecc` are orthogonal in pairs but `|veca| ne | vecc||`

(D)

`veca , vecb , vecc` are orthogonal in pairs but `| vecb| ne 1`

Solution:

`vec a xx vec b = vec c`

`vec a * vec c = vec a * (vec a xx vec b)=0`

Hence `vec a bot vec c`

as `vec a xx vec b= c` & `vec b xx vec c = vec a`

`vec c` is `bot` to `vec a ` and `vec b`

`vec a` is `bot` to `vec b` and `vec c`

Hence, `vec a bot vec b bot vec c`

Now `vec a xx vec b=vec c`

`ab=c`...............(i)

& `vec b xx vec c=vec a`

`bc=a`..................(ii)

by (i) & (ii)

`|vec b|^2 =1`

`|vec b|=1` (`|vec b|` can't be `-1`)

and `|vec a|=|vec c|`
Correct Answer is `=>` (A) `veca , vecb , vecc` are orthogonal inpairs and `|veca| = | vec c|` and `|vecb| = 1`
Q 2801080828

If `veca xx vecb = vec c` and `vecb xx vecc = vec a`, then which one of the following is correct?
Paper 1 2017
(A)

`veca , vecb , vecc` are orthogonal inpairs and `|veca| = | vec c|` and `|vecb| = 1`

(B)

`veca , vecb , vecc` are non-orthogonal to each other

(C)

`veca , vecb , vecc` are orthogonal in pairs but `|veca| ne | vecc||`

(D)

`veca , vecb , vecc` are orthogonal in pairs but `| vecb| ne 1`

Solution:

`vec a xx vec b = vec c`

`vec a * vec c = vec a * (vec a xx vec b)=0`

Hence `vec a bot vec c`

as `vec a xx vec b= c` & `vec b xx vec c = vec a`

`vec c` is `bot` to `vec a ` and `vec b`

`vec a` is `bot` to `vec b` and `vec c`

Hence, `vec a bot vec b bot vec c`

Now `vec a xx vec b=vec c`

`ab=c`...............(i)

& `vec b xx vec c=vec a`

`bc=a`..................(ii)

by (i) & (ii)

`|vec b|^2 =1`

`|vec b|=1` (`|vec b|` can't be `-1`)

and `|vec a|=|vec c|`
Correct Answer is `=>` (A) `veca , vecb , vecc` are orthogonal inpairs and `|veca| = | vec c|` and `|vecb| = 1`
Q 2723491341

If `veca = 2hati+3hatj+4hatk` and `vecb= 3hati+2hatj-lamda hatk` are perpendicular, then what is the
value of `lamda`?
NDA Paper 1 2017
(A)

`2`

(B)

`3`

(C)

`4`

(D)

`5`

Solution:

`vec a bot vec b`

`vec a * vec b =0`

`6+6 - 4 lambda=0`

`lambda=3`
Correct Answer is `=>` (B) `3`
Q 2106434378

Consider a parallelogram whose
vertices are `A(1, 2), B(4, y) C(x, 6)` ard `D(3, 5)` taken in order.

What is the area of the parallelogram?
NDA Paper 1 2016
(A)

`7/2` sq units

(B)

`4` sq units

(C)

`(11)/2` sq units

(D)

`7` sq units

Solution:

Since, in a parallelogram diagonals bisect each other.

:. Mid-point of `AC` = Mid-point of `DB`

Area of parallelogram = Area of `Delta ABD +` Area of `Delta BCD`

` = 1/2 | (1, 2, 1),( 4, 3, 1),(3,5,1) | + 1/2 | (4, 3, 1),( 6, 6, 1),(3,5,1) |`

` = 1/2 [ | (1, 2, 1),( 4, 3, 1),(3,5,1) | + | (4, 3, 1),( 6, 6, 1),(3,5,1) | ]`

` = 1/2 [{1(3 - 5) - 2(4 - 3) + 1 (20 - 9)}`

`+ {4(6 - 5) - 3(6 - 3) + 1 (30 - 18)}]`

` = 1/2 [7 + 7] = (14)/2 = 7` sq units
Correct Answer is `=>` (D) `7` sq units
Q 2781680527

Let `veca = hati+hatj , vecb = 3hati+4hatk` and `vecb = vecc +vecd` where `vecc` is parallel to `veca` and `vecd` is perpendicular to `veca`
What is `vecc` equal to ?
NDA Paper 1 2016
(A)

`(3 ( hati+hatj))/2`

(B)

`(2 ( hati+hatj))/3`

(C)

`((hati+hatj))/2`

(D)

`((hati+hatj))/3`

Solution:

Given ,

`a = hat i + hat j ` ..................(i)
`b = 3 hat i + 4 hat k ` ................(ii)

and `b = c +d ` ..............(ii)

Since, cis parallel to a and d is perpendicular to a, then
`c = lamda a` .........................(iv)

and `d * a = 0` .................(v)

Now

`b = lamda a + ( x hat i + y hat j + z hat k )`

By using Eq (ii), (iii) and (iv) , we get

`( 3 hat i + 4 hat k ) = lamda ( hat i + hat j ) + ( x hat i + y hatj + 2 hat k )`

`=> (3 hat i + 4 hatk ) = ( lamda + x ) hat i + ( lamda + y ) hat j + z hat k`

Equating both sides , we get ` lamda + x = 3` ........(vi)

`lamda +y = 0` ............. (vii)

and `z = 4 ` viii

From Eq (vi) and (vii). we get

x -y =3 ..............(ix)

Now , `d * a = 0`

`=> (x hat i + y hat j + z hat k ) * (hat i + hat j ) = 0`

`=> x +y = 0` .......................(x)

On solving Eqs (ix) and (x). we get

`x = 3/2 ` and `y = -3/2`

`:. lamda = 3 -x = 3 -3/2 = 3/2`

`because c = lamda a `

`=> c = lamda ( hat i + hat j)`

`:. c = 3/2 ( hat i + hat j )`
Correct Answer is `=>` (B) `(2 ( hati+hatj))/3`
Q 2781680527

Let `veca = hati+hatj , vecb = 3hati+4hatk` and `becb = vecc +vecd` where `vecc` is parallel to `veca` and `vecd` is perpendicular to `veca`
If `vecd = xhati+y hatj+z hatk` then which of the following equations is / are correct ? .

1. `y - x = 4`

2. `2z -3 = 0`

Select the correct answer using the code given below .


NDA Paper 1 2016
(A)

1 only

(B)

2 only

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

Given ,

`a = hat i + hat j ` ..................(i)
`b = 3 hat i + 4 hat k ` ................(ii)

and `b = c +d ` ..............(ii)

Since, cis parallel to a and d is perpendicular to a, then
`c = lamda a` .........................(iv)

and `d * a = 0` .................(v)

Now

`b = lamda a + ( x hat i + y hat j + z hat k )`

By using Eq (ii), (iii) and (iv) , we get

`( 3 hat i + 4 hat k ) = lamda ( hat i + hat j ) + ( x hat i + y hatj + 2 hat k )`

`=> (3 hat i + 4 hatk ) = ( lamda + x ) hat i + ( lamda + y ) hat j + z hat k`

Equating both sides , we get ` lamda + x = 3` ........(vi)

`lamda +y = 0` ............. (vii)

and `z = 4 ` viii

From Eq (vi) and (vii). we get

x -y =3 ..............(ix)

Now , `d * a = 0`

`=> (x hat i + y hat j + z hat k ) * (hat i + hat j ) = 0`

`=> x +y = 0` .......................(x)

On solving Eqs (ix) and (x). we get

`x = 3/2 ` and `y = -3/2`

`:. lamda = 3 -x = 3 -3/2 = 3/2`

We have ,

`y -x = -3` [ from Eq (ix) ]

and `2 z - 3 =5` [ from Eq (x) ]

Hence , both statement are incorrect
Correct Answer is `=>` (C) Both 1 and 2
Q 2157112084

What is a vector of unit length orthogonal to both the
vectors `hat i + hat j + hat k` and `2hat i + 3hat j - hat k?`
NDA Paper 1 2016
(A)

`(-4hat i + 3hat j - hat k)/sqrt(26)`

(B)

`(-4hat i + 3hat j + hat k)/sqrt(26)`

(C)

`(-3hat i + 2hat j - hat k)/sqrt(26)`

(D)

`(-3hat i + 2hat j - hat k)/sqrt(14)`

Solution:

Let `a = hat i + hat j + hat k` and `b = 2hat i + 3hat j - hat k`

Clearly, the vector which is orthogonal to both the
vectors, is

`vec a xx vec b= | (hat i , hat j , hat k) , ( 1,1,1) ,(2,3,-1)| = hat i (-1- 3)- hat j (-1- 2) +hat k (3- 2)`

` = -4 hat i + 3hat j + hat k`

Now, required vector = Unit vector along ` a xx b`

` = (-4 hat i + 3hat j + hat k)/sqrt((-4)^(2) + (3)^(2) + 1^(2) )`

` = (-4hat i + 3hat j + hat k)/sqrt(26)`
Correct Answer is `=>` (B) `(-4hat i + 3hat j + hat k)/sqrt(26)`
Q 2177212186

What is the area of the parallelogram having diagonals
`3 hat i +hat j - 2 hat k` and ` hat i - 3 hat j + 4 hat k?`
NDA Paper 1 2016
(A)

`5 sqrt(5)` sq units

(B)

`4 sqrt(5)` sq units

(C)

`5 sqrt(3)` sq units

(D)

`15 sqrt(2)` sq units

Solution:

Let diagonals of a parallelogram are

` d_(1) = 3 hat i + hat j - 2 hat k`

` d_(2) =hat i - 3hat j + 4 hat k`

Area of parallelogram ` = 1/2 | d_(1) xx d_(2) |`

Then, ` d_(1) xx d_(2) = | ( hat i , hat j , hat k) ,(3,1, -2), (1,-3,4) |`

` = hat i (4- 6)- hat j (12 + 2) + hat k (-9- 1)`

` => d_(1) xx d_(2) = -2 hat i - 14 hat j - 10 hat k`

Now, ` | d_(1) xx d_(2) | = sqrt ((-2)^(2) + (-14)^(2) + (-10)^(2) )`

` = sqrt ((4 + 196 + 100)) = sqrt(300) = 10sqrt(3)`

Hence, area of parallelogram `= 1/2 xx 10sqrt(3) = 5sqrt(3)` sq units
Correct Answer is `=>` (C) `5 sqrt(3)` sq units
Q 2117178989

Let `hat a, hat b` be two unit vectors and ` theta` be the angle between them.

What is `cos(theta/2)` equal to?
NDA Paper 1 2016
(A)

` (| hata - hatb |)/2`

(B)

` (| hata + hatb |)/2`

(C)

` (| hata - hatb |)/4`

(D)

` (| hata + hatb |)/4`

Solution:

Given a and bare unit vectors.

Now, `| hata+hat b | ^(2) = (hat a+hat b). (hat a+hat b)`

`= | hat a |^(2) + | hat b |^(2) + |hat a || hat b| cos theta`

`=1+1+2cos theta`

` => | hata+hat b | ^(2) = 2 + 2 cos theta = 2 xx 2 (cos^(2)) theta/2`

` => | hata+hat b | ^(2) = 4 (cos^(2)) theta/2`

`=>| hata+hat b | = (2 cos) theta/2`

`:. cos theta/2 = | hata+hat b |/2`
Correct Answer is `=>` (B) ` (| hata + hatb |)/2`
Q 2117180080

Let `hat a, hat b` be two unit vectors and ` theta` be the angle between them.

What is `sin(theta/2)` equal to?
NDA Paper 1 2016
(A)

` (| hata - hatb |)/2`

(B)

` (| hata + hatb |)/2`

(C)

` (| hata - hatb |)/4`

(D)

` (| hata + hatb |)/4`

Solution:

`= | hat a |^(2) + | hat b |^(2) + (hat a - hat b) . (hat a - hat b)`

`=| hata| ^(2)+ | hatb| ^(2) -2 |hat a| |hat b| cos theta`

` = 1 + 1 - 2cos theta`

` = 2-2 cos theta = 2 ( 1 - cos theta)`

` => | hata-hat b | ^(2) = 2 xx 2 cos^(2) (theta/2) => sin (theta/2) = ( | hata-hat b |)/2`
Correct Answer is `=>` (A) ` (| hata - hatb |)/2`
Q 1638280102

The projections of a directed
line segment on the coordinate axes are `12, 4, 3,`
respectively.

What is the length of the line segment?
NDA Paper 1 2015
(A)

19 units

(B)

17 units

(C)

15 units

(D)

13 units

Solution:

Length of the line segment

`= sqrt(a^2 + b^2 +c^2) = sqrt(12^2 + 4^ 2 + 3^2) = 13` units
Correct Answer is `=>` (D) 13 units
Q 2201034828

The area of the square, one of whose diagonals is `3 hat i + 4 hat j`, is
NDA Paper 1 2015
(A)

`12` sq units

(B)

`12.5` sq units

(C)

`25` sq units

(D)

`156.25` sq units

Solution:

The length of diagonal is ` 3 hat i + 4 hat j`

`:. sqrt(9 + 16) = sqrt(25) = 5`

Let the length of square be `e`.

`:. e sqrt(2) =5`

` => e = 5/sqrt(2)`

Now, area `= e^2 = (5/sqrt(2))^2 = (25) /2 = 12.5` sq units
Correct Answer is `=>` (B) `12.5` sq units
Q 2251145024

If `b` and `c` are the position vectors of the points `B` and `C`
respectively, then the position vector of the point `D` such
that `BD = 4 BC` , is
NDA Paper 1 2015
(A)

`4(c- b)`

(B)

`- 4(c- b)`

(C)

`4c - 3b`

(D)

`4c + 3b`

Solution:

Given, `BD = 4 BC`

It means `D` divides the join of `BC` externally in the ratio
`4:3`.

`:.` Position vector of `D = (4c -3b)/(4-3) = 4c - 3b`
Correct Answer is `=>` (C) `4c - 3b`
Q 2281145027

If the position vector `a` of the point `(5, n)` is such that
` |a| = 13` , then the value `(s)` of `n` can be
NDA Paper 1 2015
(A)

`pm 8`

(B)

`pm 12`

(C)

Only `8`

(D)

Only `12`

Solution:

We have, `a = 5 hat i + n hat j`

`:. | a | = sqrt(25 + n^2) = 13`

`=> 25 + n^2 = 169 => n^2 = 169 -25 = 144`

`=> n = pm 12`
Correct Answer is `=>` (B) `pm 12`
Q 2251245124

If `|a | = 2` and `|b | = 3`, then `|a xx b |^2 + |a . b |^2` is equal to
NDA Paper 1 2015
(A)

`72`

(B)

`64`

(C)

`48`

(D)

`36`

Solution:

We have, `|a| = 2, | b| = 3`

`:. |a xx b|^2 + |a. b|^2 = |a |^2 |b|^2 sin^ 2 theta + |a|^ 2 |b|^2 cos^2 theta`

`= |a |^2 |b|^2 (sin^2 theta + cos^2 theta )`

`= |a |^2 |b|^2`

`= 4 xx 9 =36`
Correct Answer is `=>` (D) `36`
Q 2251178024

If the magnitude of difference of two unit vectors is `sqrt(3)`,
then the magnitude of sum of the two vectors is
NDA Paper 1 2015
(A)

`1/2` unit

(B)

`1` unit

(C)

`2` units

(D)

`3` units

Solution:

Given, `| hat(a) - hat(b) | = sqrt(3)`

Squaring on both sides, we get

`| hat a | ^2 + | hat b | ^2 - 2 hat a . hat b = 3`

`[ ∵ hat a` and `hat b` are unit vectors, sol `| hat a| = | hat b | = 1`

` => 2 hat a . hat b = -1`

Now ` | hat a + hat b |^2 = | hat a | ^2 + | hat b | ^2 + 2 hat a . hat b`

`=1+1-1= 1`

` => hat a + hat b = 1` unit
Correct Answer is `=>` (B) `1` unit
Q 2261178025

If the vectors `alpha hat i + alpha hat j + gamma hat k, hat i + hat k` and `gamma hat i + gamma hat j + beta hat k` lie on
a plane, where `alpha , beta` and `gamma` are distinct non-negative
numbers, then `gamma` is
NDA Paper 1 2015
(A)

arithmetic mean of `alpha` and `beta`

(B)

geometric mean of `alpha` and `beta`

(C)

harmonic mean of `alpha` and `beta`

(D)

None of the above

Solution:

Since, the vectors are coplanar.

`:. |(alpha, alpha, gamma),(1,0,1),(gamma ,gamma, beta )| = 0`

`=> alpha (0 - gamma ) - alpha (beta - gamma ) + gamma (gamma) = 0`

` => -alpha gamma - alpha beta + alpha gamma + gamma^2 = 0`

`=> gamma^2 =alpha beta`

`=> gamma = sqrt(alpha beta) `

Hence, `gamma` is `GM` of `alpha` and `beta` .
Correct Answer is `=>` (B) geometric mean of `alpha` and `beta`
Q 2201178028

The vectors `a, b, c` and `d` are such that `a xx b = c xx d` and
`a xx c= b xx d`. Which of the following is/are correct?
1. `(a- d)xx(b- c)= 0`
2.`(a xx b)xx(c xx d)= 0`

Select the correct answer using the code given below.
NDA Paper 1 2015
(A)

Only `1`

(B)

Only `2`

(C)

Both `1` and `2`

(D)

Neither `1` nor `2`

Solution:

We have, `axxb= cxxd` and `axxc = bxxd`

1. `(a- d) xx (b- c)`

`=a xx b -a xx c- d xx b + d xx c`

`=cxxd - bxxd + bxxd- cxxd = 0`

`:.` Statement `1` is correct.

2. `(a xx b) xx (C xx d)`

Take `axxb =X`

`:. X xx (cxxd)= (X.d) c - (X.c)d`

`=[(a xx b). d) c -[(a xx b) .c) d`

`= [(c xx d). d) c -[(c xx d) . c] d`

`= 0 . c -0 . d = 0`

`:.` Statement `2` is also correct.
Correct Answer is `=>` (C) Both `1` and `2`
Q 1629145011

The adjacent sides `AB` and `AC` of a triangle `ABC` are represented by the vectors `-2 hat i + 3 hat j + 2 hat k` and
`-4 hat i + 6 hat j + 2 hat k` respectively. The area of the triangle `ABC` is
NDA Paper 1 2015
(A)

`6` sq units

(B)

`5` sq units

(C)

`4` sq units

(D)

`3` sq units

Solution:

Given, In `Delta ABC, AB = 2 hat i + 3 hat j + 2 hat k`

and `AC = -4 hat i + 5 hat j + 2 hat k`

To find Area of ` Delta ABC`.

Clearly, area of ` Delta ABC = 1/2 | AB xx AC |` .........(1)

Let us first find `AB xx AC`, which is given by

`AB xx AC = | (hat i , hat j , hat k) ,( -2, 3, 2) ,(-4 ,5 ,2) |`

`=hat i (6- 10)- hat j (-4 + 8) +hat k (-10 + 12)`

`= -4 hat i - 4 hat j + 2 hat k`

`=> | AB xx AC | = sqrt((-4)^2 + (-4)^2 + (2)^2) = sqrt(32 + 4) = sqrt(36) = 6`

:. From Eq. (i),

Area of ` DeltaABC =1/2 xx = 3` sq units
Correct Answer is `=>` (D) `3` sq units
Q 1609145018

A force `F =3hat i + 4hat j -3 hat k` is applied at the point `P`,
whose position vector is `r = 2 hat i - 2 hat j -3 hat k`. What is
the magnitude of the moment of the force about the
origin?
NDA Paper 1 2015
(A)

`23` units

(B)

`19` units

(C)

`18` units

(D)

`21` units

Solution:

We have , `F =3hat i + 4hat j -3 hat k`

and `OP = r = 2 hat i - 2 hat j -3 hat k`

Clearly, the magnitude of moment of the force about

origin `= | r xx F |` ..........(1)

Let us first find `r xx F = | ( hat i ,hat j ,hat k) ,( 2 , -2 , -3) ,( 3,4, -3) |`

` = hat i (6 + 12)- hat j (- 6 + 9) + hat k (8 + 6) = 18 hat i- 3 hat j + 14 hat k`

`:. ` From Eq. (i),

` |hat r xx hat F | = sqrt((18)^2 + (-3)^2 + (14)^2)`

`= sqrt(324 + 9 + 196) = sqrt(529) = 23` units
Correct Answer is `=>` (A) `23` units
Q 1639245112

Given that the vector `alpha` and `beta ` are non-collinear. The
values of `x` and `y` for which `u - v = w` holds true if
`u = 2xalpha + ybeta, v = 2yalpha + 3xbeta` and `w = 2 alpha - 5beta`, are
NDA Paper 1 2015
(A)

`x = 2, y = 1`

(B)

`x = 1, y = 2`

(C)

`x = -2 , y = 1`

(D)

`x = -2 , y = - 1 `

Solution:

Given, `alpha` and `beta` are non- collinear

i.e., `alpha pm gamma beta` for any `gamma`.

`=> alpha` and `beta` are linear `y` independent vectors.

Also, we have `u = 2xalpha + y beta ; v = 2y alpha + 3x beta` and

`w =2alpha- 5beta` such that `u- v= w`.

Consider, `u - v = w`

`= (2xalpha + ybeta)- (2xalpha + 3xbeta) = 2alpha - 5beta`

`=(2x - 2y) alpha + (y- 3x )beta =2alpha -5beta =0`

`= (2x- 2y- 2)alpha + (y- 3x + 5) beta = 0`

Since, the vector `alpha` and `beta` are linearly independent

therefore, we have

`2x-2y-2=0 => x-y=1 ... (i)`

and `y - 3x + 5 = 0 => y - 3x = - 5 ... (ii)`

On solving Eqs. (i) and (ii), we get `x = 2` and `y = 1`
Correct Answer is `=>` (A) `x = 2, y = 1`
Q 1689345217

If `| a + b | = | a - b |` then which one of the following
is correct?
NDA Paper 1 2015
(A)

`a = lamda b` for some scalar `lamda`.

(B)

`a` is parallel to `b`

(C)

`a` is perpendicular to `b`

(D)

`a = b = 0`

Solution:

We have, `| a + b | = |a - b|`

`=> |a + b|^ 2 = |a -b|^2`

`= (a + b) . (a + b) = (a -b) (a -b)`

`=> |a|^ 2 + 2(a .b)+ |b|^2 = | a|^ 2 - 2(a . b)+ |b|^2`

(:.dot product is commutative)

`=> 2(a .b)= -2(a . b) => 4(a . b) = 0`

`=> a. b = 0`

`=> a` and `b` are perpendicular to each other.
Correct Answer is `=>` (C) `a` is perpendicular to `b`
Q 1711201120

What is the area of `Delta OAB`, where `O` is the origin,
`OA = 3 hat i - hat j + hat k` and `OB = 2 hat i + hat j- 3 hat k?`
NDA Paper 1 2014
(A)

`5 sqrt(6)` sq units

(B)

`(5 sqrt(6))/2` sq units

(C)

`sqrt(6)` sq units

(D)

`sqrt(30)` sq units

Solution:

Since, area of `Delta OAB = 1/2 | OA xx OB`

` :. OA xx OB = | (hat i , hat j , hat k) ,( 3, -1, 1) ,( 2,1,-3) |`

` = hat i [3 -1]- hat j [-9 - 2] + hat k [3 + 2]`

` = 2 hat i + 11 hat j + 5 hat k`

`:. | OA xx OB | = sqrt(2^2 + 11^2 + 5^2)`

` = sqrt(150) = 5 sqrt(6)`

`:.` Required area ` = 1/2 xx 5 sqrt(6) = (5 sqrt(6))/2` sq units
Correct Answer is `=>` (B) `(5 sqrt(6))/2` sq units
Q 1771201126

Which one of the following is the unit vector
perpendicular to both `a = - hat i + hat j + hat k` and `b = hat i - hat j + hat k`?
NDA Paper 1 2014
(A)

`( hat i + hat j)/sqrt(2)`

(B)

`hat k`

(C)

`(hat j + hat k)/sqrt(2)`

(D)

`( hat i - hat j)/sqrt(2)`

Solution:

Since, unit vector perpendicular to both `a` and `b`

` = pm ( a xx b)/(| a xx b |)`

`:. a xx b = | ( hat i , hat j , hat k) ,( -1,1,1) ,(1,-1,1) |`

` = hat i [1 + 1]- hat j [-1 -1] + hat k [1 -1]`

` = 2 hat i + 2hat j + 0 = 2 (hat i + hat j)`

and ` | a xx b | = sqrt ( 4 + 4) = 2 sqrt(2) `

`:.` Required unit vector ` = pm (2 ( hat i + hat j))/sqrt(2) = pm ( hat i + hat j)/sqrt(2)`
Correct Answer is `=>` (A) `( hat i + hat j)/sqrt(2)`
Q 1711301220

What is the interior acute angle of the parallelogram
whose sides are represented by the vectors

`1/sqrt(2) hat i + 1/sqrt(2) hat j + hat k` and `1/sqrt(2) hat i - 1/sqrt(2) hat j + hat k`?
NDA Paper 1 2014
(A)

`60^0`

(B)

`45^0`

(C)

`30^0`

(D)

`15^0`

Solution:

Let `a = 1/sqrt(2) hat i + 1/sqrt(2) hat j + hat k`

and ` b = 1/sqrt(2) hat i - 1/sqrt(2) hat j + hat k`

`:. cos theta = (a.b)/( | a| |b|)`

` = ((1/sqrt(2) hat i + 1/sqrt(2) hat j + hat k) . (1/sqrt(2) hat i - 1/sqrt(2) hat j + hat k))/(sqrt (1/2 + 1/2 + 1) sqrt (1/2 + 1/2 + 1))`

` = 1/2 [1/2 - 1/2 + 1] = 1/2 = cos 60^0`

` :. theta = 60^0`
Correct Answer is `=>` (A) `60^0`
Q 1721301221

For what value of `lamda` are the vectors
`lamda hat i + (1 + lamda ) hat j + (1 + 2 lamda ) hat k` and `(1- lamda ) hat i + lamda hat j + 2 hat k`
perpendicular?
NDA Paper 1 2014
(A)

` -1/3 `

(B)

` 1/3 `

(C)

`2/3`

(D)

`1`

Solution:

Let `a = lamda hat i + (1 + lamda ) hat j + (1 + 2 lamda ) hat k`

and ` b = (1- lamda ) hat i + lamda hat j + 2 hat k`

For `a` and `b` to be perpendicular we should have,

` a.b = | a | | b| cos (pi/2) =0`

` =>[ lamda hat i + (1 + lamda ) hat j + (1 + 2 lamda ) hat k] . [(1- lamda ) hat i + lamda hat j + 2 hat k] = 0`

` => lamda - lamda ^2 + lamda + lamda ^2 + 2 + 4 lamda = 0 => 6 lamda = -2 `

`:. lamda = -2/3 = -1/3`
Correct Answer is `=>` (A) ` -1/3 `
Q 1771301226

`veca + vecb + vecc = 0` such that `| veca | = 3, | vecb | = 5` and `| vecc | = 7`

What is cosine of the angle between `vecb` and `vecc?`
NDA Paper 1 2014
(A)

`(11)/(12)`

(B)

`(13)/(14)`

(C)

`- (11)/(12)`

(D)

`- (13)/(14)`

Solution:

We have, `veca + vecb + vecc = 0` ............(1)

On squaring both sides, we get

`veca.a + vecb.b + c.c + 2veca.vecb + 2vecb.vecc + 2vecc.veca = 0`

`(∵ veca. vecb = vecb.veca , vecb.vecc = vecc.vecb` and `vecc .veca =veca .vecc)`

`=> | veca | ^2 + | b|^2 + | c ^2 = - 2 [veca.vecb +vecb.vecc + vecc.veca]`

`=> (3)^2 + (5)^2 + (7)^2 = -2 [veca.vecb + vecvecb.vecc + vecc .a]`

` => veca.vecb +vecb.vecc +vecc.veca = (9+ 25+ 49)/(-2) = - (83)/2`

Also `veca + vecb + vecc = 0`

`=> veca + vecb = -vecc`

`=> veca^2 + vecb^2 + 2veca.vecb = c^2`

`=> 2veca.vecb = 49 - 9 - 25 = 15 => veca.vecb = (15)/2`

`=> | veca | | vecb | cos theta = (15)/2 => 3.5 . cos theta = (15)/2`

`=> cos theta = 1/2 =cos pi/3`

`:. theta = pi/3`

From Eq. (i),

` b + vecc = -a`

`=> b^2 + c^2 + 2vecb.vecc = a^2`

` => 2vecb.vecc = a^2 - b^2 - c^2 = 9 - 25 - 49 = - 65`

`=> b . c = - (65)/2 => | vecb | | vecc | cos theta = -(65)/2`

` => cos theta = -(65)/2 xx 1/5 xx 1/7 = - (13)/(14)`

Also, `| veca+ vecb | = | -c | = | vecc | = 7`
Correct Answer is `=>` (D) `- (13)/(14)`
Q 1781301227

`veca+ vecb + vecc = 0` such that `| veca | = 3, | vecb | = 5` and `| vecc | = 7`

What is `| veca+ vecb |` equal to?
NDA Paper 1 2014
(A)

`7`

(B)

`8`

(C)

`10`

(D)

`11`

Solution:

We have, `veca+ vecb + vecc = 0` ............(1)

Also, `| veca+ vecb | = | -vecc | = | vecc | = 7`
Correct Answer is `=>` (A) `7`
Q 1723167041

The vertices of a `Delta ABC` are `A(2, 3, 1)`,
`B( -2, 2, 0)` and `C(0, 1, -1)`.

What is the area of the triangle?
NDA Paper 1 2014
(A)

`6sqrt(2)` sq units

(B)

`3sqrt(2)` sq units

(C)

`10sqrt(3)` sq units

(D)

None of these

Solution:

Given that vertices of a triangle are,

let `(x_1 , y_1 , z_1) = A (2, 3, 1)`,

`(x_2 ,y_2 , z_2 ) = B (- 2, 2, 0)`,

and `(x_3, y_3, z_3) = C (0, 1, -1)`,

Now, we find

`Delta x = 1/2 |(y_1 ,z_1 ,1),(y_2 ,z_2 ,1) ,( y_3 ,z_3 ,1)| = 1/2 | (3,1,1),(2,0,1),(1,-1,1)|`

` = 1/2 {3(0+ 1)- 1(2 - 1)+ 1(-2-0)}`

` = 1/2 (3- 1 - 2) = 1/2 (3 - 3) = 1/2 xx 0 = 0`

` Deltay = 1/2 |(z_1 ,x_1 ,1),(z_2 ,x_2 ,1) ,( z_3 ,x_3 ,1)| = 1/2 | (1,2,1),(0,-2,1),(-1,0,1)|`

` 1/2 {1 (-2 -0)-2 (0 + 1)+ 1 (0 -2)}`

` 1/2 (-2 -2 -2) = 1/2 xx - (6) = -3`

and ` Delta z = 1/2 |(x_1 ,y_1 ,1),(x_2 ,y_2 ,1) ,( x_3 ,y_3 ,1)| = 1/2 | (2,3,1),(-2,2,1),(0,1,-1)|`

` = 1/2 { 2 (-2 -1)- 3 (2- 0) + 1 (- 2 - 0 )}`

` 1/2 (-6 -6- 2) = 1/2 xx - 14 = -7`

`:.` Required area of `Delta ABC = sqrt(Delta_x^2 + Delta_y^2 + Delta_z^2)`

` = sqrt( (0)^2 + (- 3)^2 + (- 7)^2)`

` = sqrt (0 + 9 + 49)`

` = sqrt(58)`
Correct Answer is `=>` (D) None of these
Q 1713267140

Consider the vectors `a = i - 2 j + k` and `b = 4 i- 4j + 7k`.

What is the scalar projection of `a` on `b`?

NDA Paper 1 2014
(A)

`1`

(B)

`(19)/9`

(C)

`(17)/9`

(D)

`(23)/9`

Solution:

Given vectors are

`a = i - 2j + k` and `b = 4i - 4j + 7k`

Scalar projection of `a` on `b = (a.b)/(|b|)`

` = ( ( i - 2j + k). ( 4i - 4j + 7k))/(| 4i - 4j + 7k |)`

` = (4+8+7)/sqrt((4)^2 + (-4)^2 + (7)^2)`

` = (19)/sqrt( 16 + 16 + 49) = (19)/sqrt(81) = (19)/9`

which is the required scalar projection of `a` on `b`.
Correct Answer is `=>` (B) `(19)/9`
Q 1753267144

Consider the vectors `a = i - 2 j + k` and `b = 4 i- 4j + 7k`.

What is the vector perpendicular to both the vectors?
NDA Paper 1 2014
(A)

`-10 i - 3 j + 4 k`

(B)

`-10 i + 3 j + 4 k`

(C)

`10 i - 3 j + 4 k`

(D)

None of the above

Solution:

Given vectors are

`a = i - 2j + k` and `b = 4i - 4j + 7k`

The vector perpendicular to both the vectors `a` and `b`

`= a xx b`

` = |(i ,j,k),(1,-2,1),(4,-4,7)|`

` = i (-14 + 4)- j (7- 4) + k (- 4 + 8)`

` = - 10i - 3j + 4k`
Correct Answer is `=>` (A) `-10 i - 3 j + 4 k`
Q 1713067849

Let `| veca | = 7, | vecb | = 11`,
`| veca +vec b | = 10 sqrt(3)`

What is `| veca- vecb |` equal to?
NDA Paper 1 2014
(A)

`2sqrt(2)`

(B)

`2sqrt(10)`

(C)

`5`

(D)

`10`

Solution:

Given that, `| veca | = 7, | vecb | = 11 , | veca +vec b | = 10 sqrt(3)`

We have,

`| veca +vec b|^2 = | veca|^ 2 + |b|^2 - 2veca.vecb` ...........(1)

`:. | veca +vec b | = 10sqrt(3)`

`=> | veca +vec b|^2 = 100 xx 3`

`=> |a|^2 + |b|^2 + 2veca.vecb = 300`

` => (7)^2 + (11)^2 + 2a. b = 300`

` => 49 + 121 + 2veca.vecb =300`

` => 2veca .vec b = 300 - 170`

` => 2veca.vecb = 130`

Now, put the value of `|a|, |b|` and `2a .b` in Eq. (1), we get

` | veca - vecb|^2 = (7)^2 + (11)^2 - 130`

` = 49 + 121 - 130 = 170 - 130 = 40`

` :. | veca - vecb| = sqrt(40) = 2sqrt(10)`
Correct Answer is `=>` (B) `2sqrt(10)`
Q 1733167942

Let `| a | = 7, | b | = 11`,
`| a + b | = 10 sqrt(3)`

What is the angle between `(a+ b)` and `(a- b)`?
NDA Paper 1 2014
(A)

`pi/2 `

(B)

`pi/3`

(C)

`pi/6 `

(D)

None of these

Solution:

Given that, `| a | = 7, | b | = 11 , | a + b | = 10 sqrt(3)`

We have, `(a+ b). (a- b) = | a|^ 2- | b|^2`

Let `theta ` be the angle between `(a + b)` and `a -b`.

Then,` cos theta = ( (a + b). (a -b))/(|a + b | | a -b|)`

` = (|a|^2 - |b|^ 2)/(|a + b | | a -b|)`

` = ( (7)^2 - (11)^2)/(10sqrt(3) xx 2 sqrt(10) ) = ( (7 + 11) (7 - 11) )/(20sqrt(3) xx sqrt(10) )`

` = (18 xx (-4))/(20sqrt(30)) = ( -18)/(5sqrt(30))`

` = (-6 xx 3)/(5sqrt(30)) xx sqrt(30)/sqrt(30) - ( -6 xx 3sqrt(30))/(5 xx 30)`

` = -(3sqrt(30))/(25)`

`:. theta = cos^(-1) ( (-3)/5 sqrt(6/5))`

which is the required angle.
Correct Answer is `=>` (D) None of these
Q 2308112008

If the angle between the vectors `i - m j` and `j + k` is
`pi/3` , then what is the value of `m?`
NDA Paper 1 2013
(A)

`0`

(B)

`2`

(C)

`-2`

(D)

None of these

Solution:

Let `a = i - mj` and `b = j + k`

Given that `pi/3` is the angle between `a+b`


`therefore cos pi/3 = |(a*b)/(|a||b|)|`

` => 1/2 = ((i-mj)*(j+k))/(sqrt(1+m^2) sqrt(1+1))`

` => 1/2 = (-m)/(sqrt2 sqrt(1+m^2))`

`=> 1/sqrt2 = (-m)/sqrt(1+m^2)`

On squaring both sides, we get


`1+m^2 = 2m^2`

`=> m^2 = 1 => m = pm1`
Correct Answer is `=>` (D) None of these
Q 2318212100

What is the vector perpendicular to both the
vectors `i - j` and `i`?
NDA Paper 1 2013
(A)

`i`

(B)

`-j`

(C)

`j`

(D)

`k`

Solution:

The vector perpendicular to both the vectors `(i- j)`

and `i = (i-j)xxi = ixxi-jxxi = 0+ixxj = k`
Correct Answer is `=>` (D) `k`
Q 2338212102

If the position vectors of the points A and B are
respectively, `3i -5 j + 2 k` and `i + j- k`, then what
is the length of `AB?`
NDA Paper 1 2013
(A)

`11`

(B)

`9`

(C)

`7`

(D)

`6`

Solution:

Given that,
Position vector of A, `OA = 3i - 5j + 2 k`
and position vector of B `OB = i+j-k`


`therefore AB = OB-OA`

` = (i+j-k)-(3i-5j+2k)`

` = -2i+6j-3k`

`therefore` Length of `AB = |AB|`

` = sqrt(4+36+9) = sqrt(49) = 7`
Correct Answer is `=>` (C) `7`
Q 2358212104

If the vectors` i- 2xj- 3y k` and `i + 3xj + 2y k` are
orthogonal to each other, then the locus of the
point `(x, y)` is
NDA Paper 1 2013
(A)

hyperbola

(B)

ellipse

(C)

parabola

(D)

Circle

Solution:

Let `a = i -2xj - 3yk` and `b = i + 3xj + 2 yk`
Since, `a + b` are orthogonal to each other

`therefore a*b = 0`

`=> (i-2xj-3yk) * (i+3xj+2yk) = 0`

`=> 1-6x^2-6y^2 = 0`

` => x^2+y^2 = 1/6`

Hence, the locus of point (x,y) is a circle.
Correct Answer is `=>` (D) Circle
Q 2378212106

What is the value of `p` for which the vector
`p (2 i- j + 2 k)` is of `3` units length?
NDA Paper 1 2013
(A)

`1`

(B)

`2`

(C)

`3`

(D)

`6`

Solution:

Let `a = p (2 i - j + 2 k)`
Given that, length of `a` is `3` units

`|a| = 3 => |p(2i-j+2k)| = 3`

`=> sqrt(4p^2+p^2+4p^2) = 3`

` => 9p^2 = 9 => p^2 = 1`

`p = pm1`
Correct Answer is `=>` (A) `1`
Q 2389123017

If the projections of a straight line segment on the
coordinate axes are `2, 3` and `6`, then the length of
the segment is
NDA Paper 1 2013
(A)

5 units

(B)

7 units

(C)

11 units

(D)

49 units

Solution:

The vector projection equation of a straight line is

`r= a i + bj+ck`, ... (i)

where, `a = 2, b = 3` and `c = 6`

`:. r = 2i + 3j + 6k`

`:. ` Length of the line segment is `| r | = sqrt ((2)^2 + (3)^2 +(6)^2)`

`= sqrt (4+9 +36)`

`= sqrt (49) = 7` units
Correct Answer is `=>` (B) 7 units
Q 2308212108

If `a = 2 i + 2 j + 3 k, b = - i + 2 j + 3 k` and
`c = 3 i + j` are three vectors such that `a + t b` is
perpendicular to `c`,then what is `t` equal to?
NDA Paper 1 2013
(A)

`8`

(B)

`6`

(C)

`4`

(D)

`2`

Solution:

Given that,
`a=21+2J+3k`,
`b=-i+2J+k`
and `c = 3i+ j`
Now, `a +tb = (2i+ 2j+ 3k)+ t(-i+ 2j+ k)`
`= (2 -t) i + (2 + 2t) j + (3 + t) k`

Since the vectors `(a + t b)` and `c` are perpendicular to each other.
`therefore (a+ t b) * c = 0`
`=> { (2 - t )i + (2 + 2 t) j + (3 + t )k} * (3i + j) = 0`
`=> 3(2-t)+(2+2t)=0`
`6 - 3t + 2 + 2 t = 0`
`t = 8`
Correct Answer is `=>` (A) `8`
Q 2348312203

If `vecbeta` is perpendicular to both `vec alpha` and `gamma` , where `alpha = k`
and `vec gamma = 2i + 3j + 4k`, then what is `beta` equal to
NDA Paper 1 2013
(A)

`3i +2j`

(B)

`- 3i +2j`

(C)

`2i -3j`

(D)

`-2i +3j`

Solution:

Given that `vec alpha = k` and `vec gamma = 2i+3j+4k`

Since `vec beta` is perpendicular to both `vec alpha` and `vec gamma`


`vec beta = pm (vecalphaxxvecgamma) = pm |(i , j , k) (0 , 0 1) (2 ,3 4)|`


` = pmi(0-3)-j(0-2)+k(0-0) = pm (-3i+2j)`
Correct Answer is `=>` (B) `- 3i +2j`
Q 2378412306

For any vector `vecalpha`, what is `(vecalpha * i) i + (vecalpha * j) j + (vecalpha * k) k`
equal to?
NDA Paper 1 2013
(A)

`vecalpha`

(B)

`3vecalpha`

(C)

`-vecalpha`

(D)

`0`

Solution:

Let `vecalpha = x i+yj+zk`

Then `(vec alpha * i)i = +(vec alpha *j)j+(vec alpha *k)k`

` = {(x i+yj+zk)i }*i+{(x i+yj+zk)j} *j+{(x i+yj+zk)k} *k`

` = (x)i+(y)j+(z)k = vecalpha`
Correct Answer is `=>` (A) `vecalpha`
Q 2378512406

If the magnitude of `a xx b` equals to `a * b`, then
which one of the following is correct?
NDA Paper 1 2013
(A)

`a= b`

(B)

The angle between `a` and `b` is `45^0`

(C)

`a` is parallel to `b`

(D)

`a` is perpendicular to `b`

Solution:

Given that, Magnitude of `(axx b) `= Magnitude of `(a * b)`

`=> |a xx b| = |a * b|`

`|a||b| sintheta||hatn| = |a||b||costheta|`

`{because axxb = |a||b||sintheta|hatn| text(and) a *b = |a||b||costheta|}`

`=> |sintheta| * 1 = | costheta| (because |hatn| = 1)`

`=> |tantheta| = 1`

`=> tantheta = 1 = tanpi/4`

`therefore theta = pi/4`
So, the angle between a and b is `pi/4`
Correct Answer is `=>` (B) The angle between `a` and `b` is `45^0`
Q 2318712600

If `|a| = sqrt2 , |b| = sqrt3` and `|a+b| = sqrt6` then what is `|a-b|` equal to ?
NDA Paper 1 2013
(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

Solution:

Given that `|a| = sqrt2 , |b| = sqrt3`

and `|a+b| = sqrt6`


We know that

`|a+b|^2+|a-b|^2 = |a|^2+|b|^2+2a * b +|a|^2+|b|^2-2a *b`

` = 2(|a|^2+|b|^2)`

`therefore |a|^2-|b|^2 = 2(|a|^2+|b|^2)-|a+b|^2`


` = 2[(sqrt2)+(sqrt3)^2] -(sqrt6)^2`

` = 2(2+3)-6 = 10-6 = 4`

`=> |a-b| = 2`
Correct Answer is `=>` (B) `2`
Q 2308812708

Which one of the following vectors is normal to
the vector `i + j + k?`
NDA Paper 1 2013
(A)

`i + j- k`

(B)

`i- j + k`

(C)

`i- j- k`

(D)

None of these

Solution:

Let `a = i + j + k`

If any vector normal to a, then dot product of both vector should
be zero

(a) `(i+ j + k)·(i + j -k)= 1+ 1-1= 1ne0`

(b) `(i + j + k). (i-j+ k)::: 1- 1 + 1 = 1 ne 0`

(c) `(i+ j+ k)-,(i-j-k)=1-1-1= -1ne0`
Correct Answer is `=>` (A) `i + j- k`
Q 2318112900

If `theta` is the angle between the vectors `4 (i - k)` and
`i + j + k`, then what is `(sintheta+ costheta)` equal to?
NDA Paper 1 2013
(A)

`0`

(B)

`1/2`

(C)

`1`

(D)

`2`

Solution:

Let `a= 4i-4k` and `b= i+ j + k`

Let `theta` be the angle between `a` and `b`.

`therefore costheta = ( a * b)/(|a||b|) = ((4i+4k) *(i+j+k))/(|a|*|b|)`

` = (4+0-4)/(|a||b|) = 0`

` = cos90^0 =>theta = 90^0`

`therefore sintheta+costheta = sin90^0+cos90^0 = 1+0 = 1`
Correct Answer is `=>` (C) `1`
Q 2318123009

`EFGH` is a rhombus, such that the `angle EFG` is `60^0`. The magnitude of vectors `FH` and `{mEG}` are equal, where `m` is a scalar. What is the value of `m?`
NDA Paper 1 2012
(A)

`3`

(B)

`1.5`

(C)

`sqrt2`

(D)

`sqrt3`

Solution:

Let !he side of rhombus be `a`.
Then in `Delta GEF , sin60^0 = (OF)/a => OF = axx sqrt3/2`


We know that, the diagonal of rhombus bisect each other
perpendicularly

`because FH = 20F = 2 a sqrt3/2 = sqrt3` .............(i)

Again in `DeltaO E F sin30 = (OE)/a => OE = axx1/2`

`therefore EG = 2EO = 2 * a/2 = a`

Given, magnitude of FH =Magnitude of {mEG}

`therefore asqrt3 = ma`

On comparing, we get

`m = sqrt3`
Correct Answer is `=>` (D) `sqrt3`

Set - 2

Q 2318156909

Let `a` and `b` be two unit vectors and `alpha` be the angle
between them. If `(a+ b)` is also the unit vectors,
then what is the value of `a.?`
NDA Paper 1 2010
(A)

`pi/4`

(B)

`pi/3`

(C)

`(2pi)/3`

(D)

`pi/2`

Solution:

`because a * b = |a||b| cosalpha`

`=> cosalpha = a * b` ...................(i)

`(because |a| = |b| = 1` where, a and b are unit vectors)
[since, (a + b) is unit vector]


Now `|a+b| = 1`

`=> |a|^2+|b|^2+2a * b = 1`

`=> 1+1+2cosalpha = 1`

`=> 2cosalpha = -1`

`cosalpha = -1/2 = cos ((2pi)/3)`

`=> alpha = (2pi)/3`
Correct Answer is `=>` (C) `(2pi)/3`
Q 2358267104

What is the value of `lamda` for which the vectors
`i- j + k,2 i + j- k` and `lamda i- j + lamdak` are coplanar'?
NDA Paper 1 2010
(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

Solution:

Given vectors are coplanar, if


`|(1 , -1 , 1 ) , (2 , 1 , -1) , (lamda , -1 , lamda )| = 0`



Expanding along `R_1`

`1(lamda-1) +1(2lamda+lamda)+1(-2-lamda) = 0`


`=> lamda -1+3lamda-2-lamda = 0`

`=> 3lamda = 3 => lamda = 1`
Correct Answer is `=>` (A) `1`
Q 2328467301

What is the geometric interpretation of the
identity `(a - b) xx (a + b) = 2(a xx b)?`
I. If the diagonals of a given parallelogram are
used as sides of a second parallelogram, then
the area of the second parallelogram is twice
that of the given parallelogram.
II. If the semi-diagonals of a given parallelogram
are used as sides of a second parallelogram,
then the area of the second parallelogram is
half that of the given parallelogram.
Select the correct answer using the codes given below
NDA Paper 1 2010
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

If a and b are the adjacent sides of any parallelogram,
then its area `= | a xx b |`
`therefore` Area of `ABCE = | a xx b |`


Now `AC =a + b`
Then, `CS = a - b`
So, area of `ACSR = | (a +b) xx (a -b) |`
`=2|axxb|`
`because (a+ b) xx (a -b)= 2(a xx b)`


Hence, Statement I is true.
Now, side

`AD = ((a+b)/2)`

`AQ = ((a-b)/2)`

So area of `AQPd = | ((a+b)/2) xx ((a-b)/2)|`

` = 1/4| (a+b) xx (a-b)|`

` = 1/4xx2|a xx b| = 1/2|axxb|`
Hence, Statement II is also true
Correct Answer is `=>` (C) Both I and II
Q 2388467307

A vector b is collinear with the vector a and
satisfies the condition `a · b = 3`. ·what is b equal to?
NDA Paper 1 2010
(A)

`(1, 1/2,-1/2)`

(B)

`(2/3, 1/3, -1/3)`

(C)

`(1 / 2, 1 / 4, - 1 / 4)`

(D)

`(1. 1, 0)`

Solution:

Let vector `b = x i + y j + zk`
and `a = 2i + j - k`
Given that, `a . b = 3` ... (i)
`=> (x i+ yj + zk) · (2i + j- k) = 3`
`=> 2x + y- z = 3`
Since, vectors a and bare collinear, i.e, angle between both the
vectors should be `0^0`.
Then, `a · b = | a || b | cos 0`

`=> a * b = sqrt(4+1+1) sqrt(x^2+y^2+z^2xx1)`

`=> a * b = sqrt6 sqrt(x^2+y^2+z^2)` ........(ii)

From Eqs. (i) and (ii),


`3 = sqrt6 sqrt(x^2+y^2+z^2)`

`=> x^2+y^2+z^2 = 3/2` .............(iii)

Here `b = (1 , 1/2 , -1/2)` will satisfy Eq (iii)
Correct Answer is `=>` (A) `(1, 1/2,-1/2)`
Q 2388567407

The vectors `a = x i + yj + zk, b = k` and `c` are such
that they form a right handed system. What is `c`
equal to?
NDA Paper 1 2010
(A)

`j`

(B)

`yj-xk`

(C)

`yi-xj`

(D)

`x i-yj`

Solution:

If `a= x i+ yj + zk`
and `b = k, c` form a right handed system.


`therefore c = (axxb) = | (i , j , k) , (x , y , k) , (0 , 0 , 1) |`


`=> c = i(y-0)-j(x-0)+k(0-0)`


` c = yc - xj`
Correct Answer is `=>` (C) `yi-xj`
Q 2348767603

If a and b are the unit vectors along a and b
respectively, what is the projection of b on a?
NDA Paper 1 2009
(A)

`a * b`

(B)

`a · b`

(C)

`a · b`

(D)

`| axx b|`

Solution:

The projection of b on a = `(a * b)/|a| = a * b` `(because |a| = 1)`
Correct Answer is `=>` (C) `a · b`
Q 2388867707

What are the unit vectors parallel to `xy` plane and
perpendicular to the vector `4i- 3j + k?`
NDA Paper 1 2009
(A)

`± (3 i + 4j)/5`

(B)

`± (4i +3j)/5`

(C)

`± (3i- 4j)/5`

(D)

`± (41- 3j)/5`

Solution:

By taking option (a),
Condition of perpendicularity `a . b = 0`

`pm ((3i+4j))/5 * (4i-3j+k) = 1/5(12-12) = 0`
Correct Answer is `=>` (A) `± (3 i + 4j)/5`
Q 2348067803

What is the vector in the `xy`-plane through origin
and perpendicular to the vector `r = ai + bj` and of
the same length?
NDA Paper 1 2009
(A)

`-ai- bj`

(B)

`ai- bj`

(C)

`-ai + bj`

(D)

`bi- aj`

Solution:

Let `r_1 = bi - aj` [by option (d)]

Condition of perpendicularity, `a · b = 0`
Now, `r_1 · r = (bi-aj) · (ai+ bj)`
`= ab-ab = 0`
Correct Answer is `=>` (D) `bi- aj`
Q 2338167902

Let `a= 2i- 3j + 4k` and `b` is a unit vector
codirectional with a. If `m` is a scalar such that
`b = ma`, then what is the value of `m?`
NDA Paper 1 2009
(A)

`1/5`

(B)

`1/sqrt5`

(C)

`1/29`

(D)

`1/sqrt(29)`

Solution:

Given, `a = 2i- 3j + 4k`
Also, `b = ma`
`= m(2i- 3j + 4k)`
As `b` is a unit vector.
Now, `| 2i - 3j + 4k | = sqrt(4+9+16) = sqrt(29)`

Therefore, `m` should be `1/sqrt(29)` `(because |b| = 1)`
Correct Answer is `=>` (D) `1/sqrt(29)`
Q 2348178003

The magnitude of the vectors `a` and `b` are equal
and the angle between them is `60^0`. If the vectors
`lamda a + b` and `a - lamda.b` are perpendicular of each other,
what is the value of `lamda.?`
NDA Paper 1 2009
(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

Solution:

Since, `(lamda.a+ b)·(a- lamda..b) = 0`
`=> lamda. | a |^2 + (1 - lamda^2)a · b - lamda. | b |^2 = 0`
`=> (1 - lamda..)^2 |a || b | cos 60^0 = 0` `(because |a| = |b|)`
`=> lamda. = ± 1 or lamda. = 1` (given, `theta = 60^0`)
Correct Answer is `=>` (A) `1`
Q 2328278101

If `|a | = 3, |b | = 4` and `| a- b | = 7`, what is the value of `|a - b| ?`
NDA Paper 1 2009
(A)

`3`

(B)

`2`

(C)

`1`

(D)

`0`

Solution:

Since, `| a + b |^2 + | a - b |^2 = 2 [| a |^2 + | b |^2]` (by parallelogram law)

`=> |a+b|^2+7^2 = 2(3^2+4^2)`

`=> |a+b|^2 = 1`


`therefore |a+b| = 1`
Correct Answer is `=>` (C) `1`
Q 2348278103

Consider the diagonals of a quadrilateral formed
by the vectors `3i + 6j- 2k` and `4 i - j + 3k`. The
quadrilateral must be a
NDA Paper 1 2009
(A)

square

(B)

rhombus

(C)

rectangle

(D)

None of these

Solution:

Let `d_1 = 3i+6j-2k`

`d_2 = 4i-j+3k`

`d_1 * d_2 = 3(4)+6(-1) -2(3) = 0`

Hence `|d_1| = sqrt(3^2+6^2+2^2 ) = 7`


`|d_2| = sqrt(4^2+1^2+3^2) = sqrt(26)`


`|d_1| ne |d_2|`


Hence, given quadrilateral is a rhombus
Correct Answer is `=>` (B) rhombus
Q 2328478301

vVhich one of the following vectors of magnitude
`sqrt(51)` makes equal angles with three vectors

`a = (i-2j+2k)/2 , b = (-4i-3k)/5` and `c = j ?`
NDA Paper 1 2009
(A)

`5i- j- 5`

(B)

`5i + j + 5k`

(C)

`-5i - j + 5k`

(D)

`5i + 5j- k`

Solution:

From option (a).

Let `d = 5i - j - 5k => |d| = sqrt(51)`

Then `costheta_1 = (a * d)/(|a||d|) = |( (i-2j+2k)/3 * (5i-j-5k))/(3/(1 * sqrt(51)))|`


` = |(-4+3)/sqrt(51)| = 1/sqrt(51)`


Similarly, `costheta_2 = (b* d)/(|b||d|) = |((-4i-3k)/5 * (5i-j-5k))/(1* sqrt(51))|`

` = |(-4+3)/sqrt(51)| = 1/sqrt(51)`



`cos theta_3 = (c * d)/(|c||d|) = |(j * (5i-j-5k))/(1* sqrt(51))|`

` = |-1/sqrt(51)| = 1/sqrt(51)`

Here `theta_1 = theta_2 = theta_3 = cos^(-1)(1/sqrt(51))`

So, the vector `5i- j- 5k` makes an equal angles with three
vectors a, b and c.
Correct Answer is `=>` (A) `5i- j- 5`
Q 2378478306

If `|a| = 2 , |b| = 5 ` and `|axxb| = 8` then what is the value of `a*b ?`
NDA Paper 1 2009
(A)

`4`

(B)

`6`

(C)

`8`

(D)

`10`

Solution:

We know that

`|axxb|^2+|a*b|^2 = (|a|^2xx|b|^2)`


`therefore 64+|a*b|^2 = (4xx25)`

`=> |a* b|^2 = 36`

`=> a * b = 6`
Correct Answer is `=>` (B) `6`
Q 2328578401

If `|a+b| = |a-b|` then which one of the
following is correct?
NDA Paper 1 2009
(A)

a is parallel to b

(B)

;a is perpendicular to b

(C)

a is equal to b

(D)

Both a and b are unit vectors

Solution:

`because |a+b| = |a-b|`

`=> |a+b|^2 = |a-b|^2`


`=> |a|^2+|b|^2+2|a| * |b| = |a|^2+|b|^2-2|a| *|b|`


`=> 4|a| * |b| = 0`

`=> a bot B`

Hence, a is perpendicular to b.
Correct Answer is `=>` (B) ;a is perpendicular to b
Q 2348578403

If `a = i - 2j + 5k` and `b = 2i + j - 3k`, then what is
`(b- a) · (3a + b)` equal to?
NDA Paper 1 2009
(A)

`106`

(B)

`-106`

(C)

`53`

(D)

`-53`

Solution:

`because a = i- 2j + 5k`
and `b = 2i + j- 3k`
`therefore b-a=2i+j--3k-i+2j-5k`
`= i + 3j- 8k`
and `(3a +b)= (3i- 6j + 15k) + (2i + j- 3k)`
`=5i-5j+12k`
Hence, `(b- a)· (3a +b)= (i + 3j- 8k) · (5i- 5j + 12k)`
`= 5- 15- 96= -106`
Correct Answer is `=>` (B) `-106`
Q 2368578405

Let a, b and c be the position vectors of points A,
B and C, respectively. Under which one of the
following conditions are the points A, B and C
collinear?
NDA Paper 1 2009
(A)

a x b is equal to 0

(B)

b x c is parallel to a x b

(C)

a x b is perpendicular to b x c

(D)

(a x b)+ (b x c)+ (c x a) is equal to 0

Solution:

The points A, Band Care collinear, if
`(a xx b)+ (b xx c)+ (c xx c)= 0` (by property)
Correct Answer is `=>` (D) (a x b)+ (b x c)+ (c x a) is equal to 0
Q 2318578409

If `a = i + j + k, b = i - j + k` and `c = i + j - k`, then
what is `a xx (b +c)+ b xx (c +a) + c xx (a+ b)` equal
to?
NDA Paper 1 2009
(A)

`2i + 3j- k`

(B)

`2i- 3j- k`

(C)

`3i + j + k`

(D)

`0`

Solution:

Since, `a = i + j + k, b = i- j + k` and `c = i+j-k`

`therefore axx(b+c)+bxx(c+a)+cxx(a+b)( (because axxb = -bxxa) , (bxxc = -cxxb) , (cxxa = -axxc))`

`=axxb+axxc+bxxc+bxxa+cxxa+cxxb`
`=axxb-cxxa+bxxc-axxb+cxxa-bxxc=0`
Correct Answer is `=>` (D) `0`
Q 2328678501


NDA Paper 1 2009

Assertion : (A) The work done when the force and displacement are perpendicular to each other is zero.

Reason : (R) The dot product `A· B` vanishes, is the vectors A and B are perpendicular.

(A) Both A and R individually true and R is the correct explanation of A
(B) Both A and R are individually true but R is not the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Solution:

A. We know that,
Work done `= F . d = | F | . | d | cos theta`
Since, `theta = 90^0`
`= F · d = | F | ·| d | cos 90^0 = 0`
R. `A·B = 0`
Since, A and Bare perpendicular.
Hence, both A and R are true and R is the correct explanation
of A.
Correct Answer is `=>` (B)
Q 2318734609

A rectangular box with a cover is to have a square
base. The volume is to be `10` cu cm. The surface
area of the box in terms of the side `x` is given by
which one of the following functions?
NDA Paper 1 2008
(A)

`f(x)=(40//x)+2x^ 2`

(B)

`f(x)=(40//x)+x ^2`

(C)

`f(x)=(40//x)+ x`

(D)

`f(x) =(60//x)+ 2x`

Solution:

Let the height bf rectangular box be `y` cm

`:.` Volume of rectangular box `= x xx x xx y = 10` (given)

`=> y=10/x^2` (volume `=l xx b xx h`)............(i)

Now, surface area of box

`=2(x^2+xy+yx)`

[surface area `= 2 (lb + bh + hl)`]

`=2(x^2+2xy)=2(x^2+20/x)` [from Eq. (i)]

`=2x^2+40/x`

`:.` Required fanction is `f(x) =(2x^2+40/x)`
Correct Answer is `=>` (A) `f(x)=(40//x)+2x^ 2`
Q 2308778608

If `a = (1- 2, 3), b = (3, 1, 2)` be two vectors and `c` be
a vector of length ` l` and parallel to `(a + b)`, then
what is `c` equal to'?
NDA Paper 1 2008
(A)

`1/sqrt(14) (-2 , -3 ,1)`

(B)

`1/sqrt2 (1 , 0 , 1)K`

(C)

`1/sqrt(42) (-5 ,-4 , -1)`

(D)

None of these

Solution:

`a = i - 2 j + 3k` and `b = 3i + j + 2 k`
`therefore a + b = 4i - j + 5k`

Now, `C=lamda(a+b)` `[because c || (a+b)]`


` = lamda(4i-j+5k)`

`|c| = |4lamdai-lamdaj+5lamdak|`

`l = sqrt(4lamda^2+lamda^2+25lamda^2) ` `[because |c| = l`(given)]


`=> l = sqrt(42lamda)`

`=> lamda = l/sqrt(42)`


`therefore c = l/sqrt(42)(4i-j+5k)`

` = l/sqrt(42)(4 ,-1 ,5)`
Correct Answer is `=>` (D) None of these
Q 2338878702

If `r_1 = lamda.i + 2j + k` and `r_2 = i + (2 - lamda.)j + 2k` are
such that `| r_1 | > | r_2 | ` then `lamda`. satisfies which one of
the following '?'
NDA Paper 1 2008
(A)

`lamda = 0`

(B)

`lamda = 1`

(C)

`lamda < 1`

(D)

`lamda > 1`

Solution:

Given, `r_1 = lamdai + 2j + k`
and `r_1 = i + (2 - lamda) j + 2 k`
`| r_1| > | r_2|` given


`=> sqrt(lamda^2+(2)^2+(1)^2) > sqrt((1)^2+(2-lamda)^2+(2)^2)`


` => lamda^2+4+1 > 1+4+lamda^2-4lamda+4`

`=> 5 > 9-4lamda`

`=> 4lamda > 4 => lamda > 1`
Correct Answer is `=>` (D) `lamda > 1`
Q 2359301214

If `P , Q` and `R` are the mid-points of the sides `AB`,
`BC` and `CA` respectively of `a` `DeltaABC` and if `a, p` and
`q` are the position vectors of `A, P` and `Q`
respectively, what is the position vector of `R?`
NDA Paper 1 2008
(A)

`2a - (p- q)`

(B)

`{p- q)- 2a`

(C)

`a - (p - q)`

(D)

`a/2 -(p-q)/2`

Solution:

Let the position vectors of `B, C` and `R` are `b, c` and `r`,
respectively.


In `DeltaAOB , p = (a+b)/2` (by section formula)





`=> b = 2p -a` ... (i)

Now, in `Delta ABOC, q = (b+C)/2` (by section formula)


`=> c = 2q- b`

`=> c = 2q- (2p -a)` [from Eq. (i)]
`c = 2q-2p+a` ... (ii)

and In `DeltaAOC , r = (a+c)/2 ` (by section formula)

` = (2q-2p+a)+a)/2` [from Eq. (ii)]


`= q - p + a = a - (p- q)`
Correct Answer is `=>` (C) `a - (p - q)`
Q 2309301218

If `(3a - b) xx (a + 3b) = k a xx b`, what is the value of
`k?`
NDA Paper 1 2008
(A)

`10`

(B)

`5`

(C)

`8`

(D)

`-8`

Solution:

`(3a- b) xx (a+ 3b)`

`= (3a- b) xx a+ (3a- b) xx 3b`
`= 3a xx a - b xx a + 3a xx 3b - b xx 3b` `{(because axxa = 0 , bxxb = 0) , (axxb = -bxxa)}`


` = 0-(-axxb)+9axxb -0`


` = 10axxb = kaxxb`

`k = 10` (on comparing
Correct Answer is `=>` (A) `10`
Q 2369401315

What is the value of `lamda` if the triangle whose
vertices are `i, j ` and `i + j + 'lamdak`, will be right angled?
NDA Paper 1 2008
(A)

`2`

(B)

`0`

(C)

`-1`

(D)

`1`

Solution:

Given that the vertices of `Delta ABC` are` i, j` and `i + j + lamda.k`.

Then, `AB = sqrt2 , BC = sqrt(1+lamda^2) ` and `CA = sqrt(1+lamda^2)`


Since `DeltaABC` is right angled traingle


`therefore AB^2 = BC^2+CA^2`

`=> 2 = 1 +lamda^2+1+lamda^2`

`therefore lamda = 0`
Correct Answer is `=>` (B) `0`
Q 2319401319

The scalar triple product `(A xx B) · C` of three
vectors `A, B` and `C` determines
NDA Paper 1 2008
(A)

volume of a parallelopiped

(B)

volume of a tetrahedron

(C)

volume of an ellipsoid

(D)

None of the above

Solution:

Volume of parallelopiped = `(AxxB) * C = [ABC]`
Correct Answer is `=>` (A) volume of a parallelopiped
Q 2319501419

If `hata` and `hatb` are unit vectors, then what is the value
of `| a xx b |^2 + (a · b)^2 ?`
NDA Paper 1 2008
(A)

`0`

(B)

`2`

(C)

`1`

(D)

`1/2`

Solution:

`|axxb|^2+(a * b)^2`


` = (|a| * |b| * sintheta)^2+(|a| * |b| * costheta)^2` [`because a` and `b` are unit vectore `=> |a| = |b| = 1]`


`= (1*1* sintheta)^2+(1*1* costheta)^2`


`sin^2theta+cos^2theta = 1`
Correct Answer is `=>` (C) `1`
Q 2369601515

If two forces are equal to `2 OA` and `3 BO`, their
resultant being `'lamda OG`, where` G` is the point on` AB`
such that` (BG)/(AG) = - 2/3` then what is the value of `'lamda?`
NDA Paper 1 2008
(A)

`1`

(B)

`-1`

(C)

`2`

(D)

None of these

Solution:

Given `(BG)/(AG) = -2/3` ...........(i)


In `DeltaAOG , OG = (2xxBG-(-3)xxAG)/(AG-BG)` (by section formula)


` = (2xx(BG)/(AG)+3)/(1-(BG)/(AG)) = (2(-2/3)+3)/(1-(-2/3))` {fromEq.(i)]


` = ((-4/3)+3)/(1+2/3) = ((5/3)/(5/3)) = 1`


`because 1* OG = lamdaOG => lamda= 1`
Correct Answer is `=>` (A) `1`
Q 2319601519

What is the length of vector` (1, 1) '?`
NDA Paper 1 2008
(A)

`0`

(B)

`1`

(C)

`sqrt2`

(D)

`1/2`

Solution:

The position vector of given vector is

`OA = i+j`


`|OA| = sqrt(1^2+1^2) = sqrt2`


`therefore` Length of vector = `sqrt2 .`
Correct Answer is `=>` (C) `sqrt2`
Q 2379701616

If a and b are two unit vectors inclined at an angle
`60^0` to each other, which one of the following is
correct?
NDA Paper 1 2008
(A)

`|a+b| < 1`

(B)

`|a+b| > 1`

(C)

`|a-b| < 1`

(D)

`|a - b| > 1`

Solution:

`|a +b| = sqrt(|a+b|^2)`


` = sqrt(|a|^2+|b|^2+2|a||b|cos60^0)` `(because theta = 60^0)`

[`because a` and `b` are unit vectors `=>| a | =| b | = 1`]


`sqrt((1)^2+(1)^2+2*1*1/2) = sqrt3`


`therefore |a+b| > 1`
Correct Answer is `=>` (B) `|a+b| > 1`
Q 2339801712

If a is a position vector of a point `(1,- 3)` and `A` is
another point `(-1, 5),` then what are the
coordinates of the point `B` such that `AB = a?`
NDA Paper 1 2008
(A)

`(2,0)`

(B)

`(0, 2)`

(C)

`(-2, 0)`

(D)

`(0, - 2)`

Solution:

Given that,`a = i - 3j` and `OA = - i + 5 j`
Let the coordinates of `B` are `(x, y)`, then `OB =x i+ yj`
`therefore AB = a = OB - OA`

`=> (x + 1) i + (y - 5) j = i - 3 j`
`=> x + 1 = 1` and `y - 5 = - 3` (after comparing)
`=> x = 0` and `y = 2`
So, the coordinates of `B` are `(0, 2)`.
Correct Answer is `=>` (B) `(0, 2)`
Q 2319801719

If `a = 2 i - 3j - k` and `b = i + 4j - 2k`, what is
`(a+ b) xx (a- b)` equal to?
NDA Paper 1 2008
(A)

`2 (a xx b)`

(B)

`-2 (a xx b)`

(C)

`(a xx b)`

(D)

`-(axx b)`

Solution:

`because a = 2i-3j-k` and `b = i+ 4j-2k` (given)
`therefore a+b = (2i-3j-k)+ (i +4j-2k)`
`= 3i + j- 3k`

and `a - b = (2i- 3j- k)- (i + 4j- 2k)= i- 7 j + k`


`therefore (a+b)xx(a-b) = |(i , j , k) , (3 , 1 , -3) , (1 , -7 , 1) |`


` = i| (1 , -3) , (-7 , 1) | -j | (3 , -3 ) , (1 , 1) |+k| (3 , 1) ,(1 , -7 )|`

` = i(1-21)-j(3+3)+k(-21-1)`

` = -20i-6j-22k`

` = -2(10i+3j+11k)` ........(i)

Now `axxb = | (i , j , k) , (2 , -3 , -1 ) , (1 , 4 , -2) |`


` = i| (-3 , -1) , (4 , -2)| -j | (2 ,-1) , (1 ,-2) | +k|(2 ,-3 ) , (1 ,4)|`

` = i(6+4)-j(-4+1)+k(8+3)`

`= 10i+3j+11k`

` = -2(axxb) = -2(10i+3j+11k)` .........(ii)

Hence, `(a+ b) xx (a- b)= -2(a xx b)`
[after equating Eqs. (i) and (ii)]
Correct Answer is `=>` (B) `-2 (a xx b)`
Q 2319001819

If `a` is a non-zero vector of modulus `alpha`, where `lamda` is a
non-zero scalar and `lamda a` is `a` unit vector, then
NDA Paper 1 2008
(A)

`lamda = pm 1`

(B)

`a = |lamda|`

(C)

`a = 1/(|lamda|)`

(D)

` a = 1/lamda`

Solution:

Since, `lamda a` is a unit vector


`therefore |lamda a | = 1 => |lamda||a| = 1` (`because |a| = a` (given))


` => a = 1/(|a|)`
Correct Answer is `=>` (C) `a = 1/(|lamda|)`
Q 2329112011

Consider the following statements
If a and b are the vectors forming consecutive
sides of a regular hexagon `ABCDEF`; then

I. `CE = b - 2a`

II. `AE = 2b - a`
III. `FA = a- b`

Which of the above statements are correct?
NDA Paper 1 2008
(A)

I anci II

(B)

II and Ill

(C)

I and Ill

(D)

I, II and Ill

Solution:

In `DeltaABC`, `AB + BC = AC`
`=> AC = a + b` .......(i)
and `AD = 2'BC` (by property)
`therefore AD = 2b`

In `DeltaACD`

`AC +CD= AD`


`CD = 2 b - (a + b) = b - a`

In `DeltaCDE` , `CE = CD + DE = b - a - a`
`= b -2a`
`=> FA - CD = - (b - a ) = a- b`

In `DeltaAEF, AE = EF + FA = - BC + FA`
`=> AE ,= -b -(b-a)`
`=a -2b`
Correct Answer is `=>` (C) I and Ill
Q 2319112019

If `a, b` and `c` are unit vectors such that a is
perpendicular to the plane of `b, c` and the angle
between `b` and `c` is `pi/3` then what is the value of

`|a+b+c | ?`
NDA Paper 1 2008
(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

Solution:

Since, a is perpendicualr to `b` and `c`.

`therefore a * b = |a||b| cos 90^0 = 0`

and `a * c = |a||c| cos90^0 = 0`

and anqle between `b` and `c = pi/3`


`therefore b * c = |b||c|cospi/3`

` = 1/2` (since, b and c are unit vectors)

`(because |b| = |c| = 1)`

Now `|a+b+c|^2 = |a|^2+|b|^2+|c|^2+2( a*b+b*c+c*a)`


` = 1+1+1+2 *(0+1/2+0)`

`[because a , b ` and `c` are unit vectors `=> |a| = |b| = |c| = 1]`


` = 1+1+1+1 = 4`

`=> |a+b+c| = 2`
Correct Answer is `=>` (B) `2`
Q 2359212114

What is the locus of the point `(x, y)` for which the
Vectors `(i - x j - 2k)` and `(2i + j + yk)` are
orthogonal?
NDA Paper 1 2008
(A)

A circle

(B)

An ellipse

(C)

A parabola

(D)

a straight line

Solution:

Since, `(i- xj- 2k)` and `(2i + j + yk)` are orthogonal.
`therefore (i-xj-2k) · (2i+j+yk) = 0`
`=> 2-x-2y=0 => x+2y=2`
which is an equation of straight line.
Thus, the locus of the point `(x, y)` is a straight line.
Correct Answer is `=>` (D) a straight line
Q 2309212118

What is the number of vectors of length `1` unit
perpendicular to the vectors `a = (1, 1, 0)` and
`b = (0, 1, 1)?`
NDA Paper 1 2008
(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

Solution:

`because a= i + j` and `b = j + k`


`axxb = |(i , j , k) , (1 , 1 , 0) , (0 , 1 , 1)| = i-j+k`

and `|axxb| = sqrt(1+1+1) = sqrt3`


Unit vector perpendicular to a and b


` = pm ((axxb))/(|axxb|) = pm (i-j+k)/sqrt3`


Thus, the number of vectors of length 1 unit perpendicular to the
vectors a and b is 2.
Correct Answer is `=>` (B) `2`
Q 2319312210

What is the area of the rectangle of which
`r =a i + bj` is a semi-diagonal?
NDA Paper 1 2008
(A)

`a^2+b^2`

(B)

`2(a^2+b^2)`

(C)

`4(a+b^2)`

(D)

`4ab`

Solution:

`because` Semi-diagonal, `r = ai + b j`
So, the sides of rectangle are `2a` and `2b`.
Hence, area of rectangle `= 2a xx 2b = 4ab`
Correct Answer is `=>` (D) `4ab`
Q 2359312214

Which one of the following is correct'? If the
vector c is normal to the vectors a and b, then c is
NDA Paper 1 2007
(A)

parallel to both a + b and a - b

(B)

normal to a -band parallel to a + b

(C)

normal to a +band parallel to a - b

(D)

normal of both a +band a - b

Solution:

Since, `c` is normal to the vectors `a` and `b`, then
`c · a = 0` and `c · b = 0` (given)
Now, `c · (a+b) = c·a+c * b = 0+0=0`
and `c · (a - b) = c ·a - c · b = 0 - 0 = 0`
Hence, `c` is normal to `(a + b)` and parallel to `(a - b)`.
Correct Answer is `=>` (D) normal of both a +band a - b
Q 2319412310

Which one of the following statements is not
correct'?
NDA Paper 1 2007
(A)

Vector product is commutative

(B)

Vector product is not associative

(C)

Vector product is distributive over addition

(D)

Scalar product is commutative

Solution:

We know that vector product is not commutative.
(because` axx b = -b xx a)`
Correct Answer is `=>` (A) Vector product is commutative
Q 2369412315

If `ai+j+k, i+bj+k` and `i+j+ck` are coplanar
vectors, what is the value of `a + b + c - abc?`
NDA Paper 1 2007
(A)

`0`

(B)

`1`

(C)

`2`

(D)

`-2`

Solution:

Since, the vectors `a i + j + k, i + b j + k` and
`i + j + ck` are coplanar vectors. Therefore,

`|(a , 1 , 1) , (1 , b , 1) , (1 , 1 , 1)| = 0`

Expand along `R_1`

`a(bc - 1)- l(c- 1) + 1(1- b)= 0`

`=> abc-a-c+1+1-b=0`

`=> a+b+c-abc=2`
Correct Answer is `=>` (C) `2`
Q 2309512418

If `a, b` and `c` are non-zero vectors and
`| (a xx b) · c | = | a || b || c |`, then which one of the
following is correct?
NDA Paper 1 2007
(A)

`a · b = b · c = c · a ne 0`

(B)

`a · b = 0`

(C)

`b · c= 0`

(D)

`a · b = b · c = c · a = 0`

Solution:

Since, `| (a xx b) · c |` is the volume `(V)` of the
parallelopiped whose adjacent edges are `a, b` and `c`.
i.e., `V =| a xx b | OL`
`=> V =| a || b | (sin theta)(cos phi) | c |`




where `0 le theta le pi` is the angle between `a` and `b`.
Now, `| (a xx b) · c | =| c || b || c |`
`=> | sin theta cos phi | = 1 => sin theta = 1. cos phi = 1`
`=> theta = pi/2 , phi = 0 => a · b = b · c = c · a = 0`
Correct Answer is `=>` (D) `a · b = b · c = c · a = 0`
Q 2369612515

If `a = i + 2j- 3k` and `b = 3i - j + lamdak` and `(a + b)` is
perpendicular to `a - b`, then what is the value of
`lamda.?`
NDA Paper 1 2007
(A)

`-2`

(B)

`pm2`

(C)

`3`

(D)

`pm3`

Solution:

`because a= i + 2j- 3k` and `b = 3i- j + lamdak`
`therefore a + b = i + 2 j- 3k + 3i- j + lamda k`
`= 4i + j + (lamda - 3)k`
and `a - b = i + 2 j - 3k - 3i + j- lamdak`
`= -2i + 3j- (3 + lamda)k`
Since, `(a + b)` is perpendicular to `(a -b).`

`therefore (a+b) · (a-b) = 0`
`=> { 4i + j +(lamda.- 3) k} { -2i + 3j- (3 + lamda.) k} = 0`

`=> -8 + 3 + (3^2- lamda^2}) = 0`

`=> 4- lamda^2 = 0 => lamda = ± 2`
Correct Answer is `=>` (B) `pm2`
Q 2379712616

If the vectors `AB = c, BC = a` and `CA = b` are the
sides of `a DeltaABC`, then which of the following
vectors represent(s) the median `AD?`
`I. 1/2a+c`

`II. -1/2b+1/2c`

`III. 1/2 a+b`

Select the correct answer using the codes given below
NDA Paper 1 2007
(A)

I and II

(B)

I and Ill

(C)

Only I

(D)

Only II

Solution:

In `DeltaABD`



`AB =AD+ DB`

`c = Ad - 1/2a` (since,D is the mid-point of BC)

`AD = 1/2 a+c`


Also In `DeltaACD , AD+DC=CA`


`=> AD +1/2a = b => AD = b -1/2a`


Hence, only Statement I represents median AD
Correct Answer is `=>` (C) Only I
Q 2309712618

`O(0, 0), A(0, 3)` and `B (4, 0)` are the vertices of
`Delta OAB`. If a force `1 0i` acts at `B`, then what is the
magnitude of moment of force about the vertex `A?`
NDA Paper 1 2007
(A)

0 unit

(B)

30 units

(C)

40 units

(D)

50 units

Solution:

Since, `O(0, 0), A(0, 3)` and `B(4, 0)` are the vertices of
`DeltaOAB`.
Moment of force about the vertex `A = r xx F`
`= 3j xx 10i` `(because OA = 3j)`
`= - 30k` `(because jxxi = -k)`
`therefore` Magnitude of moment `= | - 30k | = 30` units
Correct Answer is `=>` (B) 30 units
Q 2369012815

For any two vectors a and b, consider the following
statements
I. `| a + b | = | a - b |` <=> `a` and `b` are orthogonal.
II. `| a + b | = | a | + | b |` <=> `a` and `b`are orthogonal.
III. `| a + b |^2 = | a |^2 + | b|^2`
<=> `a` and `b` are orthogonal.
Which of the above statements are correct?
NDA Paper 1 2007
(A)

I and II

(B)

I and Ill

(C)

II and Ill

(D)

I, II and Ill

Solution:

I. `| a+ b| =| a- b|`
On squaring both sides, we get
`|a+b|^2 =|a-b|^2`
`=> | a |^2 + | b |^2 + 2a · b = | b |^2 + | b |^2- 2a · b`
`=> 4a · b = 0 => a · b = 0`
So, `a` and `b` are orthogonal to each other.
II. `|a+b|=|a|+|b|`

On squaring both sides, we get
`|a+b|^2 =(|a|+|b|)^2`

`=> |a|^2+|b|^2 +2 a · b = |a|^2 +|b|^2 +2|a||b|`
`=> 2| a || b | cos theta = 2 | a|| b |`
`=> cos theta = 1 =, cos 0 => theta = 0`
So, `a` and `b` are parallel to each other.
Ill. `| a + b |^2 =| a |^2 + | b |^2`
`=> | a |^2 + | b |^2 + 2 a · b = | a |^2 + | b|^2`
`=> a · b = 0`
Hence, `a` and `b` are orthogonal.
Correct Answer is `=>` (B) I and Ill
Q 2359112914

Two vectors `2i + mj - 3nk` and `5i + 3mj + nk` are
such that their magnitudes are respectively `sqrt(14)`
and `sqrt(35)`, where `m` and `n` are integers. Which one of
the following is correct?
NDA Paper 1 2007
(A)

m takes 1 value and n takes 1 value

(B)

m takes ·, value and n takes 2 values

(C)

m takes 2 values and n takes 1 value

(D)

m takes 2 values and n takes 2 values

Solution:

Given `|2i+mj-3nk|+sqrt(14)` ..............(i)


and `|5i+3mj+nk| = sqrt(35) ` ..........(ii)

from eq(i)

`|2i + mj- 3nk |^2 = 14`
`=> 4 + m^2 + 9n^2 = 14`
`=> m^2 + 9n^2 = 10` ...........(iii)

and from Eq(ii)

`25+9m^2+n^2 = 35`


`=> 9m^2+n^2 = 10` ............(iv)

From Eqs (iii) and (iv)

`m^2+9n^2 = 9m^2+n^2`

`=> 8n^2 = 8m^2`

`=> n = pm m` ..........(v)

So, m takes2· values and n takes 2 values form Eq. (iii) and (iv).
Correct Answer is `=>` (D) m takes 2 values and n takes 2 values
Q 2309112918

If two vectors `a` and `b` are non-zero and
non-collinear, then what is the value of `x` for which
the vectors `p = (x - 2) a + b` and `q = (x + 1) a - b`
are collinear?
NDA Paper 1 2007
(A)

`1`

(B)

`1/2`

(C)

`2/3`

(D)

`2`

Solution:

Since, `p` and `q` are collinear, then
`p= lamdaq`
`=> (x - 2)a + b = lamda(x + 1) a - lamda..b`
On equating the coefficients, we get

`x-2 = lamda(x+1)` and `-lamda = 1`

`=> x- 2 = - (x + 1)`
`=> 2x = 1`


`therefore x = 1/2`
Correct Answer is `=>` (B) `1/2`
Q 2359123014

If a and bare position vectors of the points `A` and
`B` respectively, then what is the position vector of
a point `C` on `AB` produced such that `AC = 2AB ?`
NDA Paper 1 2007
(A)

`2a - b`

(B)

`2b-a`

(C)

`a - 2b`

(D)

`a - b`

Solution:

Let `c` be the position vector of `C`.
`AB =OB -OA`
`=b-a`
and `AC = 2 AB = 2b- 2a` (given)


Now In `DeltaAOC , AC = OC-OA`


`=> 2b-2a=c-a`

`c = 2b-2a +a =2b-a`
Correct Answer is `=>` (B) `2b-a`
Q 2369223115

If `| a | = 3` and `| b | = 4,` then for what value of `lamda` is
`(a + lamda.b)` perpendicular to `(a - lamdab)?`
NDA Paper 1 2007
(A)

`3/4`

(B)

`4/3`

(C)

`9/16`

(D)

`3/5`

Solution:

Since, `(a + lamdab)` is perpendicular to `(a - lamdab)`

`therefore (a+lamdab) * (a- lamdab) = 0`

`=> |a|^2-lamda^2|b|^2+lamda a * b - lamda b * a = 0`


`=> |a|^2-lamda^2|b|^2 = 0` `(because a *b = b * a)`

`=> 9-16lamda^2 = 0`

`therefore lamda = 3/4`.
Correct Answer is `=>` (A) `3/4`
 
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