Mathematics Tricks & Tips Of Vector Algebra For NDA
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Tricks & Tips Of Vector Algebra For NDA

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Different types of vectors

Q 2368723605

The vector `2j - k` lies
NDA Paper 1 2012
(A)

in the plane of XY

(B)

in the plane of YZ

(C)

in the plane of XZ

(D)

along the X·axis

Solution:

The vector `2j- k` lies in the plane of YZ because its
x-coordinate is zero.
Correct Answer is `=>` (B) in the plane of YZ
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 2318734600

If `c` is the unit vector perpendicular to both the
vectors `a` and `b`, then what is another unit vector
perpendicular to both the vectors `a` and `b`?
NDA Paper 1 2011
(A)

`cxxa`

(B)

`cxxb`

(C)

`-((axxb))/(|axxb|)`

(D)

`((axxb))/(|axxb|)`

Solution:

If `c` is perpendicular to both vectors `a` and `b`, then

`c=axxb`

But `c` is the unit vector

Then, `c = (axxb)/(|axxb|)` in the vertical upper direction and the other perpendicular unit vector `c` on both vectors `a` and `b` is

`c = - ((axxb))/(|axxb|)`


which is vertically below direction
Correct Answer is `=>` (C) `-((axxb))/(|axxb|)`
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`

Algebra of vectors and section formula

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 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 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 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 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 2318723609

`ABCD` is a parallelogram. If `AB =a, BC = b`, then
what is `BD` equal to ?
NDA Paper 1 2012
(A)

`a + b`

(B)

`a - b`

(C)

`-a - b`

(D)

`-a + b`

Solution:

Since, opposite sides of parallelogram are same

`therefore AB = a => CD = - a`
and `BC = b => DA = - b`
Applying addition formula in `ABCD`,

`BD = BC + CD`
`=b-a=-a+b`
Correct Answer is `=>` (D) `-a + b`
Q 2348023803

If `a = (2,1,-1), b = (1,- 1,0)` and `c = (5,-1,1)`, then
what is the unit vector parallel to `a+ b- c` in the
opposite direction?
NDA Paper 1 2012
(A)

`(i+ j-2k)/3`

(B)

`(i-2j+2k)/3`

(C)

`(21- j+ 2k)/3`

(D)

None of the above

Solution:

Given, `a = (2, '1,-1), b = (1, -1, 0)` and
`c = (5, -1, 1)`
Now, `a + b - c = (2 + I - 5, 1 - 1 + 1, - 1 + 0 - 1)`

`= (- 2, 1, -2) = d` (say)


`therefore` Unit vector(d) = `d/(|d|)`


` = (-2,1,-2)/sqrt((-2)^2+(1)^2+(-2)^2)`

` = ((-2 ,1 , -2))/sqrt(4+1+4) = 1/3(-2,1,-2)`

But in opposite direction. `d = - d`


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

`(2i-j+2k)/3`
Correct Answer is `=>` (C) `(21- j+ 2k)/3`
Q 2368123905

If `O` be the origin and `P, Q` and `R` be the points
such that `PO + OQ = QO +OR`, then which one
of the following is correct'?
NDA Paper 1 2012
(A)

P, Q and R are the vertices of an equilateral triangle

(B)

P, Q and R are the vertices of an isosceles triangle

(C)

P, Q and Rare collinear

(D)

None of the above

Solution:

Given condition is,


`PO +OQ = QO +OR`

`=> - OP + OQ = - OQ + OR`

`OQ + OQ = OP + OR`


`=> 2OQ =OP +OR`

`therefore OQ = (OP+OR)/2`

which represent Q it is the mid-point of P and Q. i.e. P, Q and R
are collinear
Correct Answer is `=>` (C) P, Q and Rare collinear

Scalar product of two vector and its application (Angel between two vector)

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 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 2328434301

If the vectors `i - x j- y k` and `i + x j + y k` are
orthogonal to each other, then what is the locus of
the point `(x, y)?`
NDA Paper 1 2012
(A)

A parabola

(B)

An ellipse

(C)

a circle

(D)

a straight line

Solution:

Since, the vectors `i - xj - yk` and `(i + x j + yk)` are
orthogonal to each other, then
`(i- xj- yk) *(i + xj + yk) = 0`

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

which represent a circle with centre at origin and having radius is 1.
Correct Answer is `=>` (C) a circle
Q 2358634504

If the vectors `- i - 2xj - 3yk` and `i - 3xj- 2yk` are
orthogonal to each other, then what is the locus of
the point `(x, y)?`
NDA Paper 1 2011
(A)

A straight line

(B)

An ellipse

(C)

A parabola

(D)

A circle

Solution:

Let `a = -i- 2xj- 3yk`
and `b = i- 3x j- 2yk`
If both the vectors are orthogonal to each other, then

`a *b = 0`

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


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

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

`=> x^2+y^2 = 1/6`
which is the equation of a circle
Correct Answer is `=>` (D) A circle
Q 2328434301

If the vectors `i - x j- y k` and `i + x j + y k` are
orthogonal to each other, then what is the locus of
the point `(x, y)?`
NDA Paper 1 2012
(A)

A parabola

(B)

An ellipse

(C)

a circle

(D)

a straight line

Solution:

Since, the vectors `i - xj - yk` and `(i + x j + yk)` are
orthogonal to each other, then
`(i- xj- yk) *(i + xj + yk) = 0`

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

which represent a circle with centre at origin and having radius is 1.
Correct Answer is `=>` (C) a circle
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 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 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 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 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 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 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 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 2348034803

Which one of the following is the unit vector
perpendicular to the vectors `4i + 2j` and `-3i + 2j'?`
NDA Paper 1 2011
(A)

`(i+j)/sqrt2`

(B)

`(i-j)/sqrt2`

(C)

`k`

(D)

`(i+j+k)/sqrt3`

Solution:

Let vector `x i+ yj + zk` be a perpendicular to vectors
`4 i + 2 j` and `-3i + 2j.`

`therefore 4x+2y = 0` ..........(i)

and `-3x+2y = 0` .................(ii)

Frcm Eqs. (i) and (ii),
`x = 0` and `y= 0`
Hence, the required vector is k.
Correct Answer is `=>` (C) `k`
Q 2308034808

Consider the following statements in respect of
the vectors `u_1 = (1,2,3), u_2 = (2,3,1), u_3 = (1,3,2)`
and `u_4 = (4, 6, 2)`.
I. `u_1` is parallel to `u_4`
II. `u_2` is parallel to `u_4` .
III. `u_2` is parallel to `u_3` .
Which of the statements given above is/are correct?
NDA Paper 1 2011
(A)

Only I

(B)

Only II

(C)

Only Ill

(D)

Both I and Ill

Solution:

Since, `u_4 = 2u_2 =) (4, 6, 2) = 2(2, 3, 1)`
So, `u_2` is parallel to `u_4 .`
Correct Answer is `=>` (B) Only II
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 2338123902

If the magnitudes of two vectors a and bare equal,
then which one of the following is correct?
NDA Paper 1 2012
(A)

(a +b) is parallel to (a -b)

(B)

(a + b)· (a -b) = 1

(C)

(a + b) is perpendicular to (a -b)

(D)

None of the above

Solution:

Given that, `|a| = |b|` ... (i)
(a) If `(a+ b)` is parallel to `(a- b)`, then
`(a + b) xx (a - b)` should be equal to zero.
`therefore (a + b) xx (a - b)= a )( a + b xx a - a xx b - b xx b`
`=0-axxb-axxb-0`

`= -2a xx b ne 0`

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

`=1+ a ·b-a·b-1`

`= 0 ne 1`
i.e., `(a + b)` is perpendicular to `(a-b)`
Correct Answer is `=>` (C) (a + b) is perpendicular to (a -b)
Q 2328345201

What is the sine of angle between the vectors
`i + 2j + 3k and - i + 2j + 3k?`
NDA Paper 1 2011
(A)

`sqrt(13/7)`

(B)

`sqrt(13)/7`

(C)

`13/ sqrt7`

(D)

None of these

Solution:

We know that the angle between the Vectors
`a_1 i + b_1 j + c_1 k` and `a_2i + b_2 j + c_2 k` is given by


`costheta = | (a_1a_2+b_1b_2+c_1c_2)/(sqrt(a_1^2+b_1^2+c_1^2)sqrt(a_2^2+b_2^2+c_2^2))|`



`therefore` Angle between the vectors `i -t 2 j + 3k` and `- i + 2 j + 3k` is
.given by


`costheta = |(1xx(-1)+2xx2+3xx3)/(sqrt(1+4+9)(sqrt(1+4+9)))|`



` = (-1+4+9)/(14) = 12/14 = 6/7`

Now `sintheta = sqrt(1-cos^2theta)`

` = sqrt(1-(36)/(49)) = sqrt((49-36)/(49)) = sqrt(13/49) = sqrt(13)/7`
Correct Answer is `=>` (B) `sqrt(13)/7`
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 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 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 2424701651

The values of `lambda` and `mu` for which the vectors
`a= 2 hat i+ lambda hat j- lambda hat k` is perpendicular to the vector
`b = 3 hat i+ hat j+ mu hat k` with `|a |= |b|` are
UPSEE 2011
(A)

`lambda=41/12, mu =31/12`

(B)

`lambda=41/12, mu -31/12`

(C)

`lambda=-41/12, mu =31/12`

(D)

None of these

Solution:

Since `a* b= 0`

`:. (2 hat i + lambda hat j- hat k)* (3 hat i + hat j + mu hat k) = 0`

`=> 6 + lambda-mu = 0` ......... (i)

and `| a|= | b|`

`:. sqrt (2^2 + lambda^2 + 1^2) = sqrt (3^2 +1^2 +mu^2)`

`=> sqrt(5+lambda^2) =sqrt(10+mu^ 2)`

`=> lambda^2- mu^2=5`

`=>(lambda+ mu) (lambda-mu)= 5`

`=> -6(lambda+ mu)=5` ............. (ii)

On solving Eqs. (i) and (ii), we get

`lambda=-41/12 , mu=31/12`
Correct Answer is `=>` (C) `lambda=-41/12, mu =31/12`

Vector product of two vector and its application

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 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 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 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 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 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 2318234100

If `|a | = 1 0, | b | = 2` and `a · b = 12`, then what is the value of `| a xx b | ?`
NDA Paper 1 2012
(A)

`12`

(B)

`16`

(C)

`20`

(D)

`24`

Solution:

Given that, `| a | = 1 0, | b | = 2`
and `a · b = 12`
`=> |a || b | cos theta = 12`
`=> 10 · 2. cos theta = 12`
`=> costheta = 3/5`............(i)

`sintheta = sqrt(1-cos^2theta) = sqrt(1-9/25) = 4/5`


Now `|axxb| = |a||b|sintheta|hatn|`

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

` = 10*2*1*|sintheta| = 10*2*1*|4/5|`


` = 20xx4/5 = 4xx4 = 16`
Correct Answer is `=>` (B) `16`
Q 2660067815

Find λ and μ if `(hat i + 3 hat j + 9 k) xx ( 3 hat i - lamda hat j + mu k) = 0`.
CBSE-12th 2016
Solution:

`(hat i + 3 hat j + 9 k) xx ( 3 hat i - lamda hat j + mu k) = 0`

` | ( hat i , hat j , hat k) ,( 1,3,9) ,( 3 , -lamda , mu) | = vec 0`

` hat i ( 3 mu + 9 lamda) - hat j ( mu - 27) + hat k ( - lamda - 9) = vec 0`

`3 mu + 9 lamda = 0` ..........(i)

`mu - 27 = 0` ...........(ii)

` - lamda - 9 = 0` ...........(iii)

by `eq^n` (2) & (3) `mu = 27` and ` lamda = -9`

` lamda , mu ` value satisfy the `eq^n (1)`

So `mu = 27 , lamda = -9`
Q 2836501472

If `a = 2 hat i - 3 hat j - hat k, b = hat i + 4 hat j - 2 hat k`, then what is `(a+ b) xx (a- b)` is equal to ?

(A)

`2 (a xx b)`

(B)

`- 2 (a xx b)`

(C)

`(a xx b)`

(D)

`- (a xx b)`

Solution:

`:. a + b = (2 hat i - 3 hat j - hat k)`

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

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

` = hat i - 7 hat j + hat k`

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

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

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

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

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

Now, `a xx b = | ( hat i , hat j , hat k) , (2 , -3 , -1),(1 , 4 ,-2)|`

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

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

`= 10 hat i + 3 hat j + 11 hat k`

Hence, `( a + b ) xx ( a - b) = -2 (a xx b)`
Correct Answer is `=>` (B) `- 2 (a xx b)`
Q 2358623504

What is the value of `lamda` for which

`(lamdai+j-k)xx(3i-2j+4k) = (2i-11j-7k) ?`


NDA Paper 1 2012
(A)

`2`

(B)

`-2`

(C)

`1`

(D)

`7`

Solution:

Given, `(lamdai+j-k)xx(3i-2j+4k) = (2i-11j-7k) `

`=> |(i , j , k) , (lamda , 1 , -1) , (3 , -2 , 4)| = (2i-11j-7k)`


`=> 2i-(4lamda+3)j+(-2lamda-3)k = 2i-11j-7k`


On comparing the coefficient of `f`, we get

`(4lamda+3) = 11 => 4lamda = 8`

`=> lamda = 2`
Correct Answer is `=>` (A) `2`
Q 1749423313

If `|veca| = 10, |vecb| = 2` and `veca . vecb = 12`, then the value of `|veca xx vecb|` is
NCERT Exemplar
(A)

`5`

(B)

`10`

(C)

`14`

(D)

`16`

Solution:

Here, `|veca| = 10, |vecb| = 2` and `veca . vecb = 12 ` [given]

`therefore veca . vecb = |veca| |vecb| cos theta `

`12 = 10 xx 2cos theta`

=> `cos theta = 12/20 = 3/5`

=> `sin theta = sqrt(1 - cos^2 theta) = sqrt(1 - 9/25)`

`sin theta = pm4/5`

`therefore |veca xx vecb| = |veca| |vecb| |sin theta|`

= `10 xx 2 xx 4/5`

= `16`
Correct Answer is `=>` (D) `16`
Q 2815391269

If `u = a - b, v = a + b` and `|a| = |b | =2`, then `|u xx v|` is equal to

(A)

`2 sqrt(16 - (a.b)^2)`

(B)

`2 sqrt(4 - (a.b)^2)`

(C)

` sqrt(16 - (a.b)^2)`

(D)

`sqrt(4 - (a.b)^2)`

Solution:

`u xx v = 2(a xx b)`

`:. | u xx v |= 2 sqrt(a^2 b^2 sin^2 theta)`

`= 2 sqrt(a^2 b^2 (1 - cos^2 theta))`

`= 2 sqrt(a^2 b^2 - (a . b)^2)`

`= 2 sqrt(16 - (a . b)^2)`
Correct Answer is `=>` (A) `2 sqrt(16 - (a.b)^2)`
Q 1877534486

If `theta` is the angle between any two vectors `vec a` and
`vec b` then `| vec a . vec b| = | vec a xx vec b|` when `theta` is equal to
Class 12 Exercise ms Q.No. 19
(A)

`0`

(B)

`pi/4`

(C)

`pi/2`

(D)

`pi`

Solution:

`theta` is the angle between `vec a` and `vecb`

`:. | vec a . vec b| = | vec a| | vec b| | cos theta | ` and

`| vec a xx vec b| = | vec a | | vec b| | sin theta |`

we have `| vec a . vec b| = | vec a xx vec b| `

`:. | vec a | | vec b| | cos theta | = | vec a | | vec b | sin theta`

`=>| cos theta | = | sin theta|` or ` | tan theta| = 1`

or `tan theta = 1 => theta = pi/4`
Correct Answer is `=>` (B) `pi/4`
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 2378223106

lf `a * b = 0` and `a xx b = 0`, then which one of the
following is correct'?
NDA Paper 1 2012
(A)

a is parallel to b

(B)

a is perpendicular to b

(C)

a= 0 or b = 0

(D)

None of these

Solution:

Given. that, `a · b = 0`

`a` and `b` are perpendicular to each other
and `a xxb = 0`
i.e., `a` and bare parallel to each other.
So, both conditions are possible iff `a = 0` and `b = 0`
Correct Answer is `=>` (C) a= 0 or b = 0

Projection of one vector to another

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 2388445307

What is the projection of the vector` i - 2 j + k` on
the vector `4i - 4j + 7k ?`
NDA Paper 1 2011
(A)

`sqrt5/2`

(B)

`19/9`

(C)

`sqrt5/4`

(D)

`11/3`

Solution:

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


`therefore` Projection of a on b = `(a * b)/(|b|)`



` = (4xx1+(-2)(-4)+1xx7)/sqrt(16+16+49)`


` = (4+8+7)/sqrt(81) = 19/9`
Correct Answer is `=>` (B) `19/9`
Q 2456312274

The projection of the vector `a = hat i - 2 hat j + hat k` on
the vector `b = 4 hat i - 4 hat j + 7 hat k` is
UPSEE 2014
(A)

` 9/(19)`

(B)

`(19)/9`

(C)

`9`

(D)

`sqrt(19)`

Solution:

We know that, projection of a on b `= (a.b)/(|b|)`

` :. a. b = (hat i - 2 hat j + hat k) · (4 hat i - 4 hat j + 7 hat k)`

`= 4 + 8 + 7 =19`

and `|b| = sqrt( 4^2 + 4^2 + 7^2) = sqrt(81) = 9`

Hence, projection of a and b is `(19)/9`
Correct Answer is `=>` (B) `(19)/9`
Q 2805591468

What is the projection of the vector `hat i - 2 hat j + hat k` on the vector `4 hat i - 4 hat j + 7 hat k`?

(A)

`sqrt 5 //2`

(B)

` (19) /9`

(C)

`sqrt 5 //4`

(D)

`11//3`

Solution:

Let `a = 4 hat i - 4 hat j + 7 hat k`

and `b = hat i - 2 hat j + hat k`

Now, the projection of `b` on `a = (b· a)/(|a|)`

` = ( ( hat i - 2 hat j + hat k) . (4 hat i - 4 hat j + 7 hat k))/sqrt( 16 + 16+ 49)`

` = (4 + 8 + 7)/sqrt(81) = (19)/9`
Correct Answer is `=>` (B) ` (19) /9`
Q 1728756601

Find a vector of magnitude 6, which is perpendicular to both the vectors `2hati - hatj + 2hatk` and `4hati - hatj + 3hatk`.
NCERT Exemplar
Solution:

Let ` veca = 2hati - hatj + 2hatk` and `vecb = 4hati - hatj + 3hatk`.

So, any vector perpendicular to both the vectors `veca` and `vecb` is given by

`veca xx vecb = [(hati , hatj , hatk) , (2, -1 , 2) , (4 , -1 , 3)]`

= `hati(-3 + 2) -hatj(6 - 8) + hatk(-2 + 4)`

= `-hati + 2hatj + 2hatk = vecr` [say]

A vector of magnitude 6 in the direction of `vecr`

= `(vecr)/|vecr| . 6 = (-hati + 2hatj + 2hatk)/(sqrt(1^2 + 2^2 + 2^2)) . 6`

= `-6/3 hati + (12)/3 hatj + (12)/3 + hatk`

= `-2hati + 4hatj + 4hatk`
Q 2416112079

A girl walks `4` km towards West, then she
walks `3` km in a direction `30°` East of North
and stops. Then, the girl's displacement from
her initial point of departures is
UPSEE 2014
(A)

` - 5/2 hat i + (3sqrt3)/2 hat j`

(B)

` 1/2 hat i + (sqrt3)/2 hat i`

(C)

` - 1/2 hat i + (3sqrt3)/2 hat j`

(D)

None of these

Solution:

Let `O` and `B` be the initial and final positions of

the girl, respectively.

Then, the girl's position can be shown as in the

figure.

Now, we have `OA = 4 hat i`

`AB = hat i | AB | cos 60° + hat j |AB | sin 60°`

`(AB cos 60^0` is component of AB along X-axis

and `AB sin 60°` is component of AB along

Y-axis).

`= hat i 3 xx 1/2 + hat j 3 xx sqrt(3)/2 = 3/2 hat i + (3sqrt3)/2 hat j`

By the triangle law of vector addition, we have

`OB = AO + AB`

`= (- 4 hat i ) + (3/2 hat i + (3sqrt3)/2 hat j)`

` = ( - 4 + 3/2) hat i + (3sqrt3)/2 hat j`

` = ( ( - 8 + 3)/2) hat i + (3sqrt3)/2 hat j = (-5)/2 hat i + (3sqrt3)/2 hat j`

Hence, the girl's displacement from her initial

point of departure is `(- 5)/2 hat i + (3sqrt3)/2 hat j`.
Correct Answer is `=>` (A) ` - 5/2 hat i + (3sqrt3)/2 hat j`
Q 2461823725

Let `u, v` and w be such that `|u|= 1, | v|= 3` and
`| w| = 2`. If the projection of `v` along `u` is equal
to that of `w` along `u` and vectors `v` and w are
perpendicular to each other, then `|u- v + w|`
equals
UPSEE 2015
(A)

`2`

(B)

`sqrt 7`

(C)

`sqrt 14`

(D)

`14`

Solution:

Given, `v · u = w · u` and `v .bot w => v · w = 0`

Now, consider

`|u- v + w|^2 = |u|^2+ |v|^2 + |w|^z`
`- 2u · v - 2w · v + 2u · w`

`=1+9+4=14`

`=> |u- v +w| = sqrt 14`
Correct Answer is `=>` (C) `sqrt 14`

Area of parallelogram and triangle

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 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 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 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 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 2348534403

What is the area of the rectangle having vetikes
`A, B, C` and `D` with position vectors `i+1/2j+4k` ,`i+1/2j+4k , i-1/2j+4k` and `-i-1/2j+4k ?`


NDA Paper 1 2012
(A)

`1/2` sq unit

(B)

`1` sq unit

(C)

`2` sq units

(D)

`4` sq units

Solution:

Let the position vectors of the vertices of a rectangle A.
B, C and D are

`OA = -i+1/2j+4k => OB = i+1/2j+4k`

`OC = i-1/2j+4k => OD = -i+1/2j+4k`


Now `AB = OB -OA = (i+j/2+4k)-(-i+j/2+4k) = 2i`

`BC = OC -OB = (i-j/2+4k)-(i+j/2+4k) = -j`

`therefore` Area of rectangle = `|AB| * |BC|`

` = |(2i)| * |(-j)|`

` = 2*1*1 = 2` sq units
Correct Answer is `=>` (C) `2` sq units
Q 2322756631

Area of parallelogram whose diagonals
are `vec a` and `vec b` is :
BITSAT Mock
(A)

`vec a + vec b`

(B)

`vec a . vec b`

(C)

`1/2 |vec a xx vec b|`

(D)

` |vec a xx vec b|`

Solution:

Area of parallelogram

`= 1/2 |d_1 × d_2 |`

(where `d_1` and `d_2` are diagonals)

`1/2 |vec a xx vec b|`
Correct Answer is `=>` (C) `1/2 |vec a xx vec b|`
Q 2348534403

What is the area of the rectangle having vetikes
`A, B, C` and `D` with position vectors `i+1/2j+4k` ,`i+1/2j+4k , i-1/2j+4k` and `-i-1/2j+4k ?`


NDA Paper 1 2012
(A)

`1/2` sq unit

(B)

`1` sq unit

(C)

`2` sq units

(D)

`4` sq units

Solution:

Let the position vectors of the vertices of a rectangle A.
B, C and D are

`OA = -i+1/2j+4k => OB = i+1/2j+4k`

`OC = i-1/2j+4k => OD = -i+1/2j+4k`


Now `AB = OB -OA = (i+j/2+4k)-(-i+j/2+4k) = 2i`

`BC = OC -OB = (i-j/2+4k)-(i+j/2+4k) = -j`

`therefore` Area of rectangle = `|AB| * |BC|`

` = |(2i)| * |(-j)|`

` = 2*1*1 = 2` sq units
Correct Answer is `=>` (C) `2` sq units
Q 1826891771

Find the area of the triangle with vertices
`A(1, 1,2), B (2,3, 5)` and `C (1, 5,5)`.
Class 12 Exercise 10.4 Q.No. 9
Solution:

As `A( 1, 1,2), B(2,3,4), C(1,5,5)`

` vec (OA) = hat i + hat j + 2 hat k , vec ( OB) = 2hat i + 3 hat j + 5 hat k`

and `vec (OC) = hat i + 5 hat j +5 hat k`

` vec (AB) = vec (OB) - vec( OA) = hat i + 2 hat j + 2 hat k`

` vec (AC) = vec (OC) - vec (OA) = 4 hat j +3 hat k`

Now `vec (AB) xx vec (AC) = | ( hat i , hat j , hat k ) ,(1, 2, 2),(0 , 4 ,3) | = -6 hat i - 3 hat j + 4 hat k`

Area of `Delta ABC = 1/2 | vec (AB) xx vec (AC) | = 1/2 sqrt (61)` sq. units.
Q 2339367212

What is the area of the triangle with vertices
`(0, 2, 2), (2, 0, -1)` and `(3, 4, 0)`?
NDA Paper 1 2010
(A)

`15/2` sq units

(B)

`15` sq units

(C)

`7/2` sq units

(D)

`7` sq units

Solution:

Let `A (0, 2, 2), 8 = (2, 0, -1)` and `C = (3, 4, 0)`

Now, `AB = (2. -2,- 3)` and `AC = (3, 2,-2)`

`:.` Area of triangle `= 1/2 | | AB xx AC | |`

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

`= 1/2 | [ i (4+6) + j (-4 + 9) + k (4+6) ] |`

`= 1/2 | 10 i + 5j + 10 k |`

`= 1/2 sqrt ( (10)^2 + (5)^2 + (10)^2) =1/2 sqrt (225)`

`= 15/2` sq units
Correct Answer is `=>` (A) `15/2` sq units

Scalar triple product And application

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 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 2318845700

If `a = i - k, b = x i + j + (1 - x)k` and
`c = yi + x i + (1 + x- y)k`, then `a · (b xx c)` depends
upon
NDA Paper 1 2010
(A)

Only x

(B)

Only y

(C)

Both x and y

(D)

Neither x nor y

Solution:

`because a= i- k,b =x i+ j + (1- x)k`
and `c = yi + xj + (1 + x - y) k`


`therefore a * (b xx c)`= `| ( 1 , 0 , -1 ) , ( x , 1 , 1-x ) , ( y , x , 1+x-y ) | `


Expanding along `R_1 = 1 (1 + x -y - x + x^2) - 1 (x^2 - y)`

`= 1- y + x - x^2 + y = 1`.

which shows that `a · (b xx c)` does not depend on `x` and `y`.
Correct Answer is `=>` (D) Neither x nor y
Q 2239580412

Which of the following expressions are meaningful?
BITSAT Mock
(A)

` vec(u) * (vec(v) xx vec(w))`

(B)

`(vec(u) xx vec(v))* vec(w)`

(C)

` (vec(u) * vec(v) ) xx vec(w)`

(D)

`vec(u) xx (vec(v) * vec(w))`

Solution:

`bar(u) * (bar(v) xx bar(w))` is meaningful.
Correct Answer is `=>` (A) ` vec(u) * (vec(v) xx vec(w))`
Q 2283012847

Let `vec(a) = 3 hat(i) + 2 hat(k)` and `vec(b) = 2 hat(j) + hat(k)`. If `vec(c)` is a unit vector , then

the maximum value of the scalar triple product `[ vec(a)vec(b)vec(c)]` is `....................`

(A)

`-2sqrt(16)`

(B)

`2sqrt(16)`

(C)

`sqrt(-16)`

(D)

`sqrt(16)`

Solution:

`vec(a) xx vec(b) = -4 hat(i) - 3 hat(j) + 6 hat(k)`

`[ vec(a)vec(b)vec(c) ] = ( vec(a) xx vec(b) ) * vec(c)`

`= (-4 hat(i) -3 hat (j) + 6 hat (k) ) * vec(c)`

`=> ` The maximum value of

`[ vec(a)vec(b)vec(c) ] = | -4 hat(i) - 3 hat (j) + 6 hat (k) | |vec c | = sqrt(61)`.
Correct Answer is `=>` (D) `sqrt(16)`
Q 2123491341

Value of `[veca xx vecb \ \ \ \ veca xx vecc \ \ \ \vecd]` is always equal to
JEE Mock Mains
(A)

`(veca vecd) [veca vecb vecc]`

(B)

`(veca vecc) [veca vecb vecc]`

(C)

`(veca vecb) [veca vecb vecc]`

(D)

None of these

Solution:

`(veca xx vecb)* ((veca xx vecc) xx vecd)`

`= (veca xx vecb)*((veca.vecd) vecc - (vecc.vecd)veca)`

`(veca.vecd) [vecavecbvecc- (vecc.vecd)[vecavecbvecc]]`

`= (veca.vecd)[vecavecbvecc]`
Correct Answer is `=>` (A) `(veca vecd) [veca vecb vecc]`
Q 2108834708

Let `vec(v) = 2 hat(i) + hat(j) - hat(k) ` and `vec(w) = hat(i) +3 hat(k)`. If `vec(u)` is a unit vector,

then maximum value of the scalar triple product `[ vec(u) vec(v) vec(w) ]` is


(A)

`-1`

(B)

`sqrt(10) + sqrt (6)`

(C)

`sqrt(59)`

(D)

`sqrt(60)`

Solution:

`[ vec(u) vec(v) vec(w) ] = |vec(u) * ( vec(v) xx vec(w) ) | `

`= | vec(u) * ( 3 hat(i) -7 hat(j) - hat(k)) |`

`= | vec(u) | sqrt(59) cos theta`

`:. ` Maximum `[ vec(u) vec(v) vec(w) ] = sqrt (59)`

` ( :. | vec(u) | =1 , cos theta le 1 )`
Correct Answer is `=>` (C) `sqrt(59)`
Q 2365580465

Let `vec(v) = 2 hat(i) + hat(j) - hat(k)` and `vec(w) = hat(i) + 3 hat(k)`

If `vec(u)` is a unit vector, then the maximum
value of the scalar triple product

`[vec(u) vec(v) vec(w) ]` is
BITSAT Mock
(A)

`-1`

(B)

`sqrt (10) + sqrt (6)`

(C)

`sqrt (58)`

(D)

`sqrt (59)`

Solution:

As `[vec(u) vec(v) vec(w)]` represents the volume
of parallelopiped, it would be maximum if

`vec(u)` is perpendicular to

the plane of `vec(v)` and `vec(w)`.

So height is `1` and the area is base `x` height.

i.e., `|vec(v) xx vec(w) | xx 1 =` volume of

parallelopiped `= sqrt (59)`
Correct Answer is `=>` (D) `sqrt (59)`

VECTOR TRIPLE PRODUCT AND Application

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 1611191920

A vector `vecR` is given by` vec R =vec A xx(vecB xxvec C)`
Which of the following is true?
UPSEE 2016
(A)

`vecR `is parallel to` vecA`

(B)

`vecR `must be parallel to` vecB`

(C)

`vecR `must be perpendicular to ` vecB`

(D)

none of the options

Solution:

`vecR` must be perpendicular to `vecA` as well as perpendicular to `(vecBxxvecC)`
let `vecA = veci, vecB =(veci+vecj), vecC = veck`
`vecR =vecAxx(vecBxxvecC) =vecixx[(veci+vecj)xxveck] =vecixx[(-vecj+veci)] =-veck`

hence `vecR` is neither parallel nor perpendicular to` vecB`
Correct Answer is `=>` (D) none of the options
Q 2449080813

If `a = i + 2j + 3k` and `b = i xx (a xx i) + j xx (a xx j) + k xx (a xx k)` then length of `b` is equal to
BCECE Stage 1 2012
(A)

`sqrt(12)`

(B)

`2 sqrt(120`

(C)

`3 sqrt(14)`

(D)

`2 sqrt(14)`

Solution:

We have `a = i + 2j + 3k`


`therefore b = i xx (a xx i)+ j xx (a xx j) +kxx (a xx k)` ........(i)

Now `ixx (axxi) = (i * i) a -(i * a) i`


` = 1(i+2j+k) -(1)i`


` = 2j+3k`

Simi.larly, `j xx (axx j) = i + 3k`
and ` kxx(axxk)= i+2j`

`therefore` From Eq. (i),

`b = 2j + 3k + i + 3k + i + 2j`
`= 2i + 4j + 6k`

`=> |b| = sqrt(4+16+36) = 2 sqrt(14)`
Correct Answer is `=>` (D) `2 sqrt(14)`
Q 2418167900

If `vec a` is any vector, then
`hat i xx ( vec a xx hat i) + hat j xx (vec a xx hat j) + hat k xx (vec a xx hat k)` is equal
to
UPSEE 2009
(A)

`vec a`

(B)

`2 vec a`

(C)

`3 vec a`

(D)

` vec 0`

Solution:

Let ` vec a = a_1 hat i + a_2 hat j + a_3 hat k`

Now, ` hat i xx (vec a xx hat i ) = (hat i· hat i) vec a -( hat i · vec a) hat i`

` = vec a - a_1 hat i`

Similarly, ` hat j xx (vec a xx hat j) = vec a - a_2 hat j`

and `hat k xx (vec a xx hat k) = vec a - a_3 hat k`

` :. hat i xx ( vec a xx hat i) + hat j xx (vec a xx hat j) + hat k xx (vec a xx hat k)`

`= 3 vec a - ( a_1 hat i + a_2 hat j + a_3 hat k)`

` = 3 vec a - vec a = 2 vec a`
Correct Answer is `=>` (B) `2 vec a`
Q 2418801700

Let `vec A = 2 hat i + hat j + 2 hat k` and `vec b = hat i + hat j` If `vec c` is a
vector such that `vec a . vec c = | vec c | , | vec c - vec a | = 2sqrt2` and
the angle between `vec a xx vec b` and `vec c` is `30°`, then
`| (vec a xx vec b) xx vec c |` is equal to
UPSEE 2009
(A)

`2/3`

(B)

`3/2`

(C)

`2`

(D)

`3`

Solution:

Now, `vec a xx vec b = | ( hat i , hat j , hat k),(2,1,-2),(1,1,0)|`

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

` => |vec a xx vec b| = sqrt(4 + 4 +1) = 3`

Now, ` | ( vec a xx vec b) xx vec c | = | ( vec a xx vec b)||vec c| . sin 30^0`

` = 3 | vec c | 1/2 ` ....(i)

Also, ` | vec c - vec a| = 2sqrt2`

` => | vec c|^2 + | vec a|^2 - 2 .vec c . vec a = 8`

` => | vec c|^2 + 9 - 2 | vec c| = 8`

`=> | vec c|^2 - 2 | vec c| + 1 = 0`

` => ( | vec c| - 1 )^2 = 0 => | vec c| = 1`

`:.` From Eq. (i)

` | ( vec a xx vec c) xx vec c | = 3 xx 1 xx 1/2 = 3/2`
Correct Answer is `=>` (B) `3/2`

Condition for coplanarity and colinearity of vectors

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 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 2358323204

The vector `a xx (b xx a)` is coplanar with
NDA Paper 1 2012
(A)

Only a

(B)

Only b

(C)

Both a and b

(D)

Neither a nor b

Solution:

Given that, `a xx (b xx a)`
which is the vector triple product `= (a · a )b - (a · b) a`
`= lamdab -mua`
where, `lamda` and `mu` are scalar quantity.
`=> a xx (b xx a)` is coplanar with both `a` and `b`.
Correct Answer is `=>` (C) Both a and b
Q 2328134001

What is the value of `m`, if the vectors
`2i- j + k, i + 2j- 3k` and `3i + mj +5k` are coplanar ?
NDA Paper 1 2012
(A)

`-2`

(B)

`2`

(C)

`-4`

(D)

`4`

Solution:

We know that, if three vectors `a, b` and `c` are coplanar,
then `[abc]= 0`


Let `a = 2i-j+k, b = i+2j-3k` and `c = 3i+j+5k`

Then `[abc] = |(2,-1 ,1) , (1 , 2 , -3) , (3 , m , 5)| = 0`


`=> 2(10+3m)+1(5+9)+1(m-6) = 0`


`=> 20+3m+14+m -6 = 0`

`=> 7m +28 = 0`

`therefore m = -4`
Correct Answer is `=>` (C) `-4`
Q 2368734605

For what value of `m` are the points with position
vectors `10i + 3j, l2i - Sj` and `mi + 11j` collinear'`?`
NDA Paper 1 2011
(A)

`-8`

(B)

`4`

(C)

`8`

(D)

`12`

Solution:

Let `OA = 10i + 3j`
`OB = 12 i - 5j` and `OC = mi + 11 j`
Since, A, B and Care collinear.
We have, `AB = lamdaBC`
`=> (OB - OA) = A(OC - OB)`
`=> (2i- 8j) =lamda {(m- 12)i + 16j}`
On comparing the coefficients of `i, j` and `k`, we get
`lamda(m - 12) = 2` ... (i)

`16lamda = -8 => lamda = -1/2`


from eq(i)

`-1/2(m-12) = 2 => m-12 = -4 => m = 8`

Alternate Method
If the given position vectors are collinear, then the area of triangle
should be zero.

`|(10, 3 , 1) , (12 , -5 , 1) , (m , 11 , 1)| = 0`

`=> |(10 , 3 , 1 ) , (2 , -8 , 1) , (m-10 , 8 , 1 ) | = 0`

` => 16+8(m-10) = 0`

`=> 8m = 64 => m = 8`
Correct Answer is `=>` (C) `8`
Q 2348145003

The points with position vectors `10i + 3j, 12i- 5j`
and `ai + 11j` are collinear, if the value of `a` is
NDA Paper 1 2011
(A)

`-8`

(B)

`4`

(C)

`8`

(D)

`12`

Solution:

Since, the points with position vectors `10i + 3j, 12 i- 5j`
and `ai + 11 j` are collinear i.e., area of triangle formed by these
position vectors should be zero.


therefore `1/2| (10 , 3 , 1) , (12 , -5 , 1) , (a , 11 , 1) | = 0`


`=> a (3+ 6)-1·1 (10-12)+ 1 (-50- 36)= 0`
`=> 8a + 22- 86 = 0`
` => 8a= 64`
`a=8`
Correct Answer is `=>` (C) `8`
Q 2308345208

It the vector a lies in the plane of vectors b and c,
then which one of the following is correct?
NDA Paper 1 2011
(A)

`a· (b xx c)= 0`

(B)

`a· (b xx c)= 1`

(C)

`a·(bxxc)=-1`

(D)

`a·(bxxc)=3`

Solution:

Since, vector a lies in the plane of vectors band c.
So, a, band care coplanar

`=> a · (bxxc) = 0` `( because [a b c] = 0)`
Correct Answer is `=>` (A) `a· (b xx c)= 0`
Q 2388734607

For what value of `m` are the vectors `2i - 3j + 4 k`,
`i + 2j- k` and `mi - j + 2k` coplanar'?
NDA Paper 1 2011
(A)

`0`

(B)

`5/3`

(C)

`1`

(D)

`8/5`

Solution:

Let the vectors `a = 2 i - 3j + 4k, b = i + 2 j - k` and
`c = mi - j + 2 k` be coplanar, then `[a b c] = 0`

` => | (2 , -3 , 4) , (1 , 2 , -1 ) , ( m , -1 , 2 )| = 0`

Expanding along `R_1`,

`2(4-1)+ 3(2 + m)+ 4(-1-2m)= 0`

`=> 6+6+3m-4-8m = 0 => -5m+8 = 0`


`therefore m = 8/5`
Correct Answer is `=>` (D) `8/5`
Q 2388734607

For what value of `m` are the vectors `2i - 3j + 4 k`,
`i + 2j- k` and `mi - j + 2k` coplanar'?
NDA Paper 1 2011
(A)

`0`

(B)

`5/3`

(C)

`1`

(D)

`8/5`

Solution:

Let the vectors `a = 2 i - 3j + 4k, b = i + 2 j - k` and
`c = mi - j + 2 k` be coplanar, then `[a b c] = 0`

` => | (2 , -3 , 4) , (1 , 2 , -1 ) , ( m , -1 , 2 )| = 0`

Expanding along `R_1`,

`2(4-1)+ 3(2 + m)+ 4(-1-2m)= 0`

`=> 6+6+3m-4-8m = 0 => -5m+8 = 0`


`therefore m = 8/5`
Correct Answer is `=>` (D) `8/5`
Q 2368734605

For what value of `m` are the points with position
vectors `10i + 3j, l2i - Sj` and `mi + 11j` collinear'`?`
NDA Paper 1 2011
(A)

`-8`

(B)

`4`

(C)

`8`

(D)

`12`

Solution:

Let `OA = 10i + 3j`
`OB = 12 i - 5j` and `OC = mi + 11 j`
Since, A, B and Care collinear.
We have, `AB = lamdaBC`
`=> (OB - OA) = A(OC - OB)`
`=> (2i- 8j) =lamda {(m- 12)i + 16j}`
On comparing the coefficients of `i, j` and `k`, we get
`lamda(m - 12) = 2` ... (i)

`16lamda = -8 => lamda = -1/2`


from eq(i)

`-1/2(m-12) = 2 => m-12 = -4 => m = 8`

Alternate Method
If the given position vectors are collinear, then the area of triangle
should be zero.

`|(10, 3 , 1) , (12 , -5 , 1) , (m , 11 , 1)| = 0`

`=> |(10 , 3 , 1 ) , (2 , -8 , 1) , (m-10 , 8 , 1 ) | = 0`

` => 16+8(m-10) = 0`

`=> 8m = 64 => m = 8`
Correct Answer is `=>` (C) `8`
Q 2805491368

For what value of `m` are the vector `2 hat i - 3 hat j + 4 hat k , hat i + 2 hat j - hat k` and `m hat i - hat j + 2 hat k` are coplanar ?

(A)

`0`

(B)

`5//3`

(C)

`1`

(D)

`8//5`

Solution:

Let the vectors `a = 2 hat j - 3 hat i + 4 hat k`,

`b = hat i + 2 hat j - hat k` and `c = m hat i - hat j + 2 hat k` are

coplanar, then `[a b c] = 0`

`|(2 , -3 , 4),(1,2,-1),(m , -1 , 2) | = 0`

`=> 2(4 - 1) + 3 (2 + m) + 4 (-1-2m) = 0`

`=> - 5m + 8 = 0 => m = 8/5`
Correct Answer is `=>` (D) `8//5`
Q 2815480369

If `a + b + c = pd, b + c + d = qa` and `a,b,c` are non-coplanar, then `a+ b + c + d` is equal to

(A)

`0`

(B)

`pa`

(C)

`qb`

(D)

`(p+ q) c`

Solution:

On putting the value of d from the

given relations, we have

`a+ b+ c+ d = a+ b+ c+ (qa - b - c)` ... (i)

`= (a+ b+ c) + 1/q (a+ b+ c) ` ... (ii)

`:. (1 + q) a + 0 b + 0 c`

` = (1 + 1/p) (a + b + c)`

Since, a, b and c arc non-coplanar, we

have on comparing the coefficients

`1 + q = 1 + 1/p` and `1 + 1/p = 0`

`:. p - 1` and `q = -1`

On putting for p in Eq. (ii) or `q` in

Eq. (i), we get `a + b + c + d = 0`
Correct Answer is `=>` (A) `0`
Q 2328134001

What is the value of `m`, if the vectors
`2i- j + k, i + 2j- 3k` and `3i + mj +5k` are coplanar ?
NDA Paper 1 2012
(A)

`-2`

(B)

`2`

(C)

`-4`

(D)

`4`

Solution:

We know that, if three vectors `a, b` and `c` are coplanar,
then `[abc]= 0`


Let `a = 2i-j+k, b = i+2j-3k` and `c = 3i+j+5k`

Then `[abc] = |(2,-1 ,1) , (1 , 2 , -3) , (3 , m , 5)| = 0`


`=> 2(10+3m)+1(5+9)+1(m-6) = 0`


`=> 20+3m+14+m -6 = 0`

`=> 7m +28 = 0`

`therefore m = -4`
Correct Answer is `=>` (C) `-4`
Q 2358423304

Consider the following statement
`I. 4i xx 3i = 0`

`II. (4i)/(3i) = 4/3`

Which of the above is/are correct ?
NDA Paper 1 2012
(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

Both statements are true.

`I. 4i xx 3i = 12 (i xx i) = 12 xx 0 = 0` `(because ixxi = 0)`


`II. (4i)/(3i) = 4/3`



Hence, divisibility in vectors are not possible.
Correct Answer is `=>` (C) Both I and II
 
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