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Must Do Problems Of Vector Algebra For NDA

Must Do Problems Of Vector Algebra

Must Do Problems
Q 2876101976

ABCD is a parallelogram. L is a point on BC which divides BC in the ratio `1 : 2`. AL intersects BD at P. M is a point on DC which divides DC in the ratio `1 : 2` and AM intersects BD in Q.
Point P divides AL in the ratio

(A)

`1 :2`

(B)

`1 :3`

(C)

`3:1`

(D)

`2:1`

Solution:

`BL = 1/3 b , AL = a + 1/3 b`

Let `AP = lamda AL` and `P` divides `DB` in the

ratio `mu : ( 1 - mu)`. Then,

`AP = lamda a + lamda/3 b` ... (i)

Also, `AP = mu a + (1 - mu ) b` ... (ii)

From Eqs. (i) and (ii),

`lamda a + lamda/3 b = mu a + (1 - mu) b`

`:. lamda = mu` and `lamda/3 = 1 - mu => A = 3/4`

Hence, P divides AL in the ratio `3:1` and

P divides DB in the ratio `3:1` similarly, Q

divides DB in the ratio `1:3`.

Thus, `DQ = 1/4 DB` and `PB = 1/4 DB`.

`:. PQ = 1/2 DB,` i.e. `PQ : DB = 1: 2`
Correct Answer is `=>` (C) `3:1`
Q 2612423330

Write a vector of magnitude `15` units in the direction of vector `hat i - 2 hat j + 2 hat k`
CBSE-12th 2010
Solution:

Unit vector along the direction of vector ` vec a , hat a = vec a/(|vec a|)`

Let `vec a = hat i − 2 hat j + 2 hat k`

` | vec a| = sqrt ((1)^2 + (-2)^2 + (2)^2) = pm 3`

i.e. ` hat a = 1/3 ( hat i − 2 hat j + 2 hat k )`

So, the vector whose magnitude is `15` and has direction along the vector `hat i − 2 hat j + 2 hat k` is

given by,

` 15 xx (1/3) ( hat i − 2 hat j + 2 hat k )`

` = 5 (hat i − 2 hat j + 2 hat k )`

` = ( 5 hat i − 10 hat j + 10 hat k )`

So the required vector is ` 5 hat i − 10 hat j + 10 hat k`
Q 2835891762

If `p, q, r` and `s` are respectively the magnitudes of the vectors `3 hat i - 2 hat j , 2 hat i + 2hat j + hat k , 4 hat i -hat j + hat k, 2 hat i + 2 hat j + 3 hat k`. Then, which one of the following is correct?

(A)

`r > s > q > p`

(B)

`s > r > p > q `

(C)

`r > s > p > q`

(D)

`s > r > q > p`

Solution:

Given p, q, r,and s are the magnitude

of vectors `3 hat i - 2 hat j , 2 hat i + 2 hat j + hat k`,

`4 hat i - hat j + hat k, 2 hat i + 2 hat j + 3 hat k`, respectively.

`=> p = | 3 hat i - 2 hat j | = sqrt(9 + 4) = sqrt(13) = 3.61`

`=> q = | 2 hat i + 2 hat j + hat k |`

`= sqrt( 4 + 4 + 1) = 3 = 3.00`

`=> r = | 4 hat i - hat j + hat j | = sqrt(16 + 1 + 1)`

`= 3 sqrt 2 = 4.24`

`=> s = | 2 hat i + 2 hat j + 3 hat k | = sqrt(4 + 4 + 9)`

`= sqrt(17) = 4.12`

`:. r > s > p > q`
Correct Answer is `=>` (C) `r > s > p > q`
Q 2846112073

Consider four points P, Q, R and S with their position vectors `- (hat j + hat k) , 4 hat i + 5 hat j + lamda hat k, 3 hat i + 9 hat j + 4 hat k` and `- 4 hat i + 4 hat j + 4 hat k`, respectively.
Find RS.

(A)

`7 hat i + 5 hat k`

(B)

`7 hat i + 5 hat j`

(C)

`5 hat i + 7 hat k`

(D)

`- 7 hat i - 5 hat j`

Solution:

We have , `P (- hat j - hat k) , Q ( 4 hat i + 5 hat j + lamda hat k)`,

`R(3 hat i + 9 hat j + 4 hat k)` and `S( -4 hat i + 4 hat j + 4 hat k)`.

Then, `PQ = 4 hat i + 6 hat j + ( lamda + 1 ) hat k` ... (i)

`QR = - hat i + 4 hat j + (4 - lamda ) hat k` ... (ii)

`RS = -7 hat i - 5 hat j + 0 hat k` ... (iii)

From Eq. (iii), `RS = - 7 hat i - 5 hat j`
Correct Answer is `=>` (D) `- 7 hat i - 5 hat j`
Q 2816001879

Consider the regular hexagon ABCDEF with centre at O (origin).
`AD + EB + FC` is equal to

(A)

`2 AB`

(B)

`3 AB`

(C)

`4 AB`

(D)

None of these

Solution:

Consider the regular hexagon

ABCDEF with centre at O (origin).

`AD + EB + FC = 2AO + 2OB + 2OC`

`= 2(AO + OB) + 2OC`

`= 2AB + 2AB [ ∵ OC = AE ]`

`= 4 AB`
Correct Answer is `=>` (C) `4 AB`
Q 2835580462

If a is a non-zero vector of modulus a and `lamda` is a non-zero scalar and `lamda`, a is a unit vector, then

(A)

`lamda = pm 1`

(B)

` a = | lamda |`

(C)

`a = 1/(| lamda |)`

(D)

`a = 1/ lamda `

Solution:

Since, `lamda` a is a unit vector.

`:. | lamda a | = 1`

`=> | lamda | | a | = 1`

` => a = 1/(|lamda|)`
Correct Answer is `=>` (C) `a = 1/(| lamda |)`
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`
Q 2885191967

ABCD is a quadrilateral. Forces AB, CB, CD and DA act along its sides. What is their resultant?

(A)

`2 CD`

(B)

`2 DA`

(C)

`2 BC`

(D)

`2 CB`

Solution:

In DACD,

`CD + DA = CA` .......(i)

Now , in `Delta ABC`,

`CA+ AB = CB` ... (ii)

From Eqs. (i) and (ii),

`CD + DA + AB = CB`

`=> CB + CD + DA + AB = 2 CB`
Correct Answer is `=>` (D) `2 CB`
Q 2427001881

If `vec a, vec b, vec c` are the position vectors of the vertifes
of an equilateral triangle whose orthocentre is at
the origin, then
UPSEE 2010
(A)

`vec a+ vec b+vec c=vec 0`

(B)

`vec a- vec b=vec c`

(C)

`vec a+ vec b=vec c`

(D)

None of these

Solution:

The position vector of the centroid of thy triangle
is `(vec a+vec b+vec c)/3` . since the triangle is an equiteral
therefore the orthocentre coincides yvith the
centroid and hence

`(vec a+vec b+vec c)/3 = vec 0 => vec a+vec b+ vec c= vec 0`

`
Correct Answer is `=>` (A) `vec a+ vec b+vec c=vec 0`
Q 2855580464

If two concurrent forces be represented by n `OP` and m `OQ` respectively, then their resultant is given by `(m + n) OR`, where `R` is such that

(A)

`m : n = RQ : PR`

(B)

`m : n = PR : RQ`

(C)

`R` is mid-point of `PQ`

(D)

None of these

Solution:

We have, `OP = OR + RP` ... (i)

and `OQ = OR + RQ` ... (ii)

On multiply Eq. (i) by n and Eq. (ii) by

m and add, we get the result as given, i.e.

`(m + n)` OR.

If `n RP + m RQ = 0`

or `- n PR + m PQ = 0`

or `m : n = PR : RQ`
Correct Answer is `=>` (B) `m : n = PR : RQ`
Q 2835191062

If `a = hat i + hat j + p hat k` and `b = hat i + hat j + hat k`, then `| a+ b | = | a | +| b |`, holds for

(A)

all real `p`

(B)

no real `p`

(C)

`p =- 1`

(D)

`p = 1`

Solution:

` sqrt ( { 4 + 4 + (p + 1)^2 }) = sqrt (p^2 + 2) + sqrt3`

On squaring both sides, we get

`p^2 + 2p + 9 = p^2 + 2`

`+ 3 + 2 sqrt 3 sqrt ( p^2 + 2)`

`:. 2 (p + 2) = 2 sqrt 3 sqrt ( p^2 + 2)`

Again, on squaring both sides, we get

`p^2 + 4p + 4 = 3p^2 + 6 => p = 1`
Correct Answer is `=>` (D) `p = 1`
Q 1856212174

For given vectors ` vec a = 2 hat i - hat j + 2 hat k` and
`b = - hat i + hat j - hat k`, find the unit vector in the
direction of the vector `vec a + vec b`.
Class 12 Exercise 10.2 Q.No. 9
Solution:

Now ` vec a = 2 hat i - hat j + 2 hat k` and ` vec b = - hat i + hat j - hat k`

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

Now ` | vec a + vec b | = sqrt (1^2 + 1^2) = sqrt(2)`

Hence uint vector along `vec a + vec b = (vec a + vec b)/( |vec a + vec b|)`

` = 1/sqrt(2) hat i + 1/sqrt(2) hat k`
Q 2426412371

Forces of magnitude `5` and `3` units acting in
the directions `6 hat i + 2 hat j + 3 hat k` and `3 hat i - 2 hat j + 6 hat k`
respectively act on a particle which is
displaced from the point `(2, 2, - 1)` to `(4, 3, 1)`.
The work done by the forces is
UPSEE 2014
(A)

`148 ` units

(B)

`(148)/7 ` units

(C)

`(78)/7 ` units

(D)

None of these

Solution:

Let `F` be the resultant force and `d` be the

displacement vector.

Then `F = 5 ((6 hat i + 2 hat j + 3 hat k))/sqrt( 36 + 4 + 9) + 3 ((3 hat i - 2 hat j + 6 hat k))/sqrt(9+ 4 + 36)`

`= 1/7 (39hat i + 4 hat j + 33 hat k)`

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

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

`:.` Total work done `= F · d`

`= 1/7 [(39 hat i + 4 hat j + 33 hat k)· (2hat i + hat j + 2 hat k )]`

`= 1/7 [78 + 4 + 66] = (148)/7` units
Correct Answer is `=>` (B) `(148)/7 ` units
Q 2815791660

If the position vector of a point `P` with respect to the origin `O` is `hat i + 3 hat j - 2 hat k` and that of a point Q is
`3 hat i + hat j - 2 hat k`, then what is the position vector of the bisector of the `angle POQ`?

(A)

`hat i - hat j - hat k`

(B)

`hat i + hat j - hat k`

(C)

`hat i + hat j + hat k`

(D)

None of these

Solution:

`∵ OP = hat i + 3 hat j - 2 hat k`

and `OQ = 3 hat i + hat j - 2 hat k`

Let `hat i + hat j - hat k` be required position vector

of the bisector of the `angle POQ`. It will

make equal angles with OP and OQ.

` :.` Angle between `hat i + 3 hat j - 2 hat k`

and `hat i + hat j + hat k` ,

` theta = cos^(-1) ( (1 + 3 + 2)/( sqrt (1 + 9 + 4) sqrt (1 + 1 + 1) ))`

` = cos^(-1) ( 6/(sqrt(14) sqrt3))`

and angle between ` 3 hat i + hat j - 2 hat k`

and ` hat i + 3 hat j - hat k`

`phi = cos^(-1) ( (1 + 3 + 2)/( sqrt (1 + 9 + 4) sqrt (1 + 1 + 1) ))`

` = cos^(-1) ( 6/(sqrt(14) sqrt3))`

Hence, `theta = phi`
Correct Answer is `=>` (B) `hat i + hat j - hat k`
Q 2865780665

If `a, b` and `c` are three mutually perpendicular vectors of equal magnitude, then the angle `theta` which `a + b + c` makes with any one of three given vectors is given by

(A)

`cos^(-1) (1 /sqrt3)`

(B)

`cos^(-1) (1 /3)`

(C)

`cos^(-1) (2 /sqrt3)`

(D)

None of these

Solution:

Let ` |a| = |b| = |c| = lamda`

and ` a·b = b·c = c·a = 0`

`:. (a + b + c)^2 = 3 lamda^2`

`=> | a + b + c | = lamda sqrt3`

Now , `(a + b + c ) · a = lamda sqrt 3 · lamda cos theta`

or `lamda^2 + 0 + 0 = lamda^2 sqrt 3 cos theta_1`

`:. cos theta_1 = 1/sqrt3 = cos theta_2 = cos theta_3`

` => Q_1 = Q_2 = Q_3 = cos^(-1) (1 /sqrt3)`
Correct Answer is `=>` (A) `cos^(-1) (1 /sqrt3)`
Q 2856001874

For any two vectors a and b, consider the following statements
I. `|a + b| = | a - b | ` <=> `a, b` are orthogonal.
II. `| a + b | = | a | +| b |` <=> `a, b` are orthogonal
III. `| a+ b|^ 2 = |a|^ 2 + |b|^ 2` <=> ` a, b` are orthogonal.
Which of the above statement(s) is/are correct?

(A)

I and II

(B)

I and III

(C)

II and III

(D)

I, II and III

Solution:

I . `| a+b | = | a - b |`

On squaring both sides, we get

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

`=> |a|^2 + | b|^ 2 + 2a· b`

`= |a|^ 2 + |b|^2 - 2a·b`

`=> 4a · b = 0 => a · b = 0`

Hence, 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 + 2a·b`

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

`=> 2 | a | | b | cos theta = 2 | a | | b |`

`=> cos theta = 1 = cos theta => theta = 0`

Hence, a and bare parallel to each other.

III . `|a+ b|^ 2 = |a|^ 2 + |b|^ 2`

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

`=> a· b = 0`

Hence, a and b are orthogonal to

each other.

Hence, Statements I and III are

correct statements.
Correct Answer is `=>` (B) I and III
Q 2826301271

If `(a+ b)` is perpendicular to `b` and `a· (a + 2a) = 0` then

(A)

`2a = b`

(B)

`a = 2 sqrt b`

(C)

`a = 2b`

(D)

`a = b`

Solution:

Given,

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

`:. (a + b) . b = 0`

` => a· b + b^2 = 0`

`=> a · b = - b^2`

Also, `a (a + 2b) = 0`

`=> a^2 + 2 a · b = 0 => a · b = - a^2/2`

`:. b^2 = 1/2 a^2 => a^2 = 2b^2 => a = sqrt(2 b)`
Correct Answer is `=>` (B) `a = 2 sqrt b`
Q 2539680512

If `hat a` and `hat b` are two unit vectors such that `hat a + 2 hat b` and `5 hat a - 4 hat b` are perpendicular to each other, then the angle between `hat a` and `hat b` is
BCECE Mains 2015
(A)

`45°`

(B)

`60^0`

(C)

` cos^(-1) (1/3)`

(D)

` cos^(-1) (2/7)`

Solution:

Let ` u = hat a + 2 hat b` and `v = 5 hat a -4 hat b` and let `theta`

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

It is given that `u` and `v` are perpendicular to each other.

`:. u·v = 0`

`=> ( hat a + 2 hat b) . (5 hat a - 4 hat b) = 0`

`=> 5 | hat a |^2 - 8 | hat b |^2 + 10 ( hat a . hat b) - 4 ( hat a . hat b) = 0`

`=> - 3 + 6 ( hat a . hat b) = 0`

`[ ∵ | hat a | = | hat b | = 1]`

` => - 3 + 6 cos theta = 0`

`[∵ hat a· hat b = | hat a | | hat b | cos theta =cos theta ]`

`=> cos theta = 1//2`

`:. theta = 60^0`
Correct Answer is `=>` (B) `60^0`
Q 2427001881

If `vec a, vec b, vec c` are the position vectors of the vertifes
of an equilateral triangle whose orthocentre is at
the origin, then
UPSEE 2010
(A)

`vec a+ vec b+vec c=vec 0`

(B)

`vec a- vec b=vec c`

(C)

`vec a+ vec b=vec c`

(D)

None of these

Solution:

The position vector of the centroid of thy triangle
is `(vec a+vec b+vec c)/3` . since the triangle is an equiteral
therefore the orthocentre coincides yvith the
centroid and hence

`(vec a+vec b+vec c)/3 = vec 0 => vec a+vec b+ vec c= vec 0`

`
Correct Answer is `=>` (A) `vec a+ vec b+vec c=vec 0`
Q 2835191062

If `a = hat i + hat j + p hat k` and `b = hat i + hat j + hat k`, then `| a+ b | = | a | +| b |`, holds for

(A)

all real `p`

(B)

no real `p`

(C)

`p =- 1`

(D)

`p = 1`

Solution:

` sqrt ( { 4 + 4 + (p + 1)^2 }) = sqrt (p^2 + 2) + sqrt3`

On squaring both sides, we get

`p^2 + 2p + 9 = p^2 + 2`

`+ 3 + 2 sqrt 3 sqrt ( p^2 + 2)`

`:. 2 (p + 2) = 2 sqrt 3 sqrt ( p^2 + 2)`

Again, on squaring both sides, we get

`p^2 + 4p + 4 = 3p^2 + 6 => p = 1`
Correct Answer is `=>` (D) `p = 1`
Q 2865580465

In a `Delta ABC, D, E, F` are the mid-points of the sides `BC, CA` and `AB`, respectively, the vector `AD` is equal to

(A)

`BE + CF`

(B)

`BE - CF`

(C)

`CF - BE`

(D)

`- BE - CF`

Solution:

`AD = 3 GD = 3 · 1//2 (GB + GC)`

` = 2/3 ( 2/3 EB + 2/3 FC) = - BE - CF`
Correct Answer is `=>` (D) `- BE - CF`
Q 2584634557

Magnitudes of vectors `vec(a) , vec(b), vec (c)` are `3,4,5` respectively. If

`vec(a)` and `vec(b) + vec(c) , vec(b)` and `vec(c) + vec(a) , vec(c)` and

`vec(a) + vec(b)` are mutually perpendicular, then magnitude of

`vec(a) + vec(b) + vec(c)` is
UPSEE 2008
(A)

`4 sqrt (2)`

(B)

`3 sqrt (2)`

(C)

`5 sqrt (2)`

(D)

`3 sqrt (3)`

Solution:

Key Idea If `vec(a)` and `vec(b)` are mutually perpendicular, then

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

Since, `vec(a)` and `vec(b) + vec(c)` are mutually perpendicular.

`:. vec(a) * (vec(b) + vec(c)) =0`

`=> vec(a) * vec(b) + vec(c) * vec(a) =0`..........(i)

Similarly, `vec(b )* vec(c) + vec(a) * vec(b) = 0`..........(ii)

and `vec(c) * vec(a) + vec(b) * vec(c) = 0`...........(iii)

On adding Eqs. (i), (ii) and (iii) , we get

`2 (vec(a) * vec(b) + vec(b) * vec(c) + vec(c) * vec(a)) =0`

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

`+ 2 (vec(a) * vec(b) + vec(b) * vec(c) + vec(c) * vec(a) )`

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

`=9 +16 +25`

`=50`
`=> | vec(a) + vec(b) + vec(c) | = 5 sqrt (2)`
Correct Answer is `=>` (C) `5 sqrt (2)`
Q 2856001874

For any two vectors a and b, consider the following statements
I. `|a + b| = | a - b | ` <=> `a, b` are orthogonal.
II. `| a + b | = | a | +| b |` <=> `a, b` are orthogonal
III. `| a+ b|^ 2 = |a|^ 2 + |b|^ 2` <=> ` a, b` are orthogonal.
Which of the above statement(s) is/are correct?

(A)

I and II

(B)

I and III

(C)

II and III

(D)

I, II and III

Solution:

I . `| a+b | = | a - b |`

On squaring both sides, we get

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

`=> |a|^2 + | b|^ 2 + 2a· b`

`= |a|^ 2 + |b|^2 - 2a·b`

`=> 4a · b = 0 => a · b = 0`

Hence, 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 + 2a·b`

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

`=> 2 | a | | b | cos theta = 2 | a | | b |`

`=> cos theta = 1 = cos theta => theta = 0`

Hence, a and bare parallel to each other.

III . `|a+ b|^ 2 = |a|^ 2 + |b|^ 2`

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

`=> a· b = 0`

Hence, a and b are orthogonal to

each other.

Hence, Statements I and III are

correct statements.
Correct Answer is `=>` (B) I and III
Q 2582180037

The vectors `A` and `B` are such that `| A + B | = | A - B |`· The angle between the two vectors will be
WBJEE 2016
(A)

`0^0`

(B)

`60^0`

(C)

`90^0`

(D)

`45^0`

Solution:

Given, `| A + B | = | A - B |`

`:. (A + B)^2 = (A - B)^2`

` A^2 + B^2 + 2AB cos theta = A^2 + B^2 - 2AB cos theta`

` 2AB cos theta + 2 AB cos theta = 0`

`4 AB cos theta = 0`

Here `A != 0 , B != 0`

`:. cos theta = 0`

or `theta = 90^0`
Correct Answer is `=>` (C) `90^0`
Q 2539680512

If `hat a` and `hat b` are two unit vectors such that `hat a + 2 hat b` and `5 hat a - 4 hat b` are perpendicular to each other, then the angle between `hat a` and `hat b` is
BCECE Mains 2015
(A)

`45°`

(B)

`60^0`

(C)

` cos^(-1) (1/3)`

(D)

` cos^(-1) (2/7)`

Solution:

Let ` u = hat a + 2 hat b` and `v = 5 hat a -4 hat b` and let `theta`

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

It is given that `u` and `v` are perpendicular to each other.

`:. u·v = 0`

`=> ( hat a + 2 hat b) . (5 hat a - 4 hat b) = 0`

`=> 5 | hat a |^2 - 8 | hat b |^2 + 10 ( hat a . hat b) - 4 ( hat a . hat b) = 0`

`=> - 3 + 6 ( hat a . hat b) = 0`

`[ ∵ | hat a | = | hat b | = 1]`

` => - 3 + 6 cos theta = 0`

`[∵ hat a· hat b = | hat a | | hat b | cos theta =cos theta ]`

`=> cos theta = 1//2`

`:. theta = 60^0`
Correct Answer is `=>` (B) `60^0`
Q 1123067841

If three unit vectors `veca,vecb,vecc` satisfy the condition `veca+vecb+vecc = vec0,`then the angle between `veca and vecb` is :
EAMCET 2010
(A)

`(2\pi)/3`

(B)

`(5\pi)/6`

(C)

` (\pi)/3`

(D)

`\pi/6`

Solution:

The three unit vectors `veca,vecb,vecc` are given.

The condition that is met by the unit vectors is `veca+vecb+vecc = vec0`.

Let `theta` be the angle between the vectors `veca & vecb.`

`veca +vecb = - vecc`
`|veca + vecb |² = | -vecc |²`
`|veca |² + | vecb |² + 2 a b cos(theta)= | vecc |²`
The magnitude of unit vectors is` |veca|=|vecb|=|vecc|=1`
Thus,`1+1+2(1)(1)cos( theta) = 1`
`=> cos (theta) = (-1/2)`

Therefore , `theta = (2pi)/3`
Correct Answer is `=>` (A) `(2\pi)/3`
Q 2427801781

The vector `vec a` is equal to
UPSEE 2010
(A)

` ( vec a * hat i) hat i + ( vec a * hat k) hat j +( vec a * hat i) hat k`

(B)

`( vec a * hat j) hat i-(vec a * hat k) hat j-(vec a * hat i) hat k`

(C)

`( vec a * hat j) hat i+(vec a * hat k) hat j+(vec a * hat i) hat k`

(D)

`( vec a * vec a) +(hat i+hat j+ hat j)`

Solution:

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

`vec a * hat i=a_1, vec a * hat j =a_2, vec a * hat k=a_3`

`:. vec a =( vec a * hat i) hat i + (vec a * hat j) hat j+(vec a * hat k) hat k`
Correct Answer is `=>` (A) ` ( vec a * hat i) hat i + ( vec a * hat k) hat j +( vec a * hat i) hat k`
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 1759223114

For any vector `veca`, the value of `(veca xx hati)^2 + (veca xx hatj)^2 + (veca xx hatk)^2` is
NCERT Exemplar
(A)

`veca^2`

(B)

`3veca^2`

(C)

`4veca^2`

(D)

`2veca^2`

Solution:

Let `veca = xhati + yhatj + zhatk`

`therefore veca^2 = x^2 + y^2 + z^2`

`therefore veca xx hati = |[(hati, hatj, hatk) , (x, y, z) , (1, 0, 0)]|`

= `hati[0] - hatj[-z] + hatk[-y]`

= `zhatj - yhatk`

`therefore (veca xx hati)^2 = (zhatj - yhatk) (zhatj - yhatk)`

= `y^2 + z^2`

Similarly, `(veca xx hatj)^2 = x^2 + z^2`

and `(veca xx hatk)^2 = x^2 + y^2`

`therefore (veca xx hati)^2 + (veca xx hatj)^2 + (veca + hatk)^2 = y^2 + z^2 + x^2 + z^2 + x^2 + y^2`

= `2(x^2 + y^2 + z^2) = 2veca^2`
Correct Answer is `=>` (D) `2veca^2`
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`
Q 2680012817

The vector `c`, directed along the internal bisector of the angle between the vectors `a = 7 hat i + 4 hat j - 4 hat k` and `b = - 2 hat j - hat j + 2 hat k` with `| c | = 5 sqrt6`, is
BCECE Mains 2015
(A)

` 5/3 ( hat i - 7 hat j + 2 hat k)`

(B)

` 5/3 (5 hat i + 5 hat j + 2 hat k)`

(C)

` 5/3 ( hat i + 7 hat j + 2 hat k)`

(D)

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

Solution:

The required vector `c` is given by

`c = lamda (a + b)`

` => c = lamda ( (a)/(|a|) + b/(|b|) )`

` => c = lamda { 1/9 (7 hat i - 4 hat j - 4 hat k) + 1/3 ( - 2 hat i - hat j + 2 hat k) }`

`=> c = lamda/9 ( hat i - 7 hat j + 2 hat k)`

` => |c| = pm lamda/9 sqrt( 1 + 49 + 4) = pm lamda/9 sqrt(54)`

But ` |c| = 5 sqrt6` [given]

` => pm lamda/9 sqrt(54)= 5 sqrt6`

` => lamda = pm 15`

Hence, ` c = pm (15)/9 ( hat i - 7 hat j + 2 hat k)`

`= pm 5/3 ( hat i - 7 hat j + 2 hat k)`
Correct Answer is `=>` (A) ` 5/3 ( hat i - 7 hat j + 2 hat k)`
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 1615156960

`| a | =| b | = 5` and the angle between `a` and `b` is `pi/4 `.

The area of the triangle constructed on the

vectors `a - 2b` and `3a + 2b` is
GGSIPU 2014
(A)

`50`

(B)

`50sqrt(2)`

(C)

`(50)/sqrt(2)`

(D)

`100`

Solution:

We have, `| a | =| b | = 5` and angle between `a`

and `b` is `pi/4`.

Now, area of the triangle constructed on the

vectors is given by

`= 1/2 | (a - 2b) xx (3a + 2b) |`

` = 1/2 | 3 (a xx a) + 2 (a xx b) - 6 (b xx a) - 4 (b xx b) |`

`= 1/2 | 2(a xx b) + 6 (a xx b) |`

`[ a xx a = 0` and `a xx b = - b xx a]`

`= 1/2 | 8(a xx b) | = 8/2 | a xx b |`

`= 4 | a | | b | | sin( pi/4)| = 4 . 5 . 5 . 1/sqrt(2)`

`= 2 sqrt(2) xx 5 xx 5 = 50 sqrt(2)`
Correct Answer is `=>` (B) `50sqrt(2)`
Q 1778756606

Using vectors, find the area of the `triangleABC` with vertices `A(1, 2, 3), B(2, -1, 4)` and `C(4, 5, -1).`
NCERT Exemplar
Solution:

`vec(AB) = (2 - 1)hati + (-1 - 2)hatj + (4 - 3)hatk`

= `hati - 3hatj + hatk`

and `vec(AC) = (4 - 1)hati + (5 - 2)hatj + (-1 - 3)hatk`

= `3hati + 3hatj - 4hatk`

`therefore vec(AB) xx vec(AC) = [(hati, hatj, hatk) , (1 ,- 3 , 1) , (3 , 3 , -4)]`

= `hati (12 - 3) - hatj(-4 - 3) + hatk(3 + 9)`

= `9hati + 7hatj +12hatk`

and `|vec(AB) xx vec(AC)| = sqrt(9^2 + 7^2 + 12^2)`

= `sqrt(81 + 49 + 144)`

`therefore` Area of `triangleABC = 1/2 |vec(AB) xx vec(AC)|`

= `1/2 sqrt(274)` sq. units.
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 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 2436378272

In a `Delta ABC, AB= r i +j ,AC= s i-j` if the
area of triangle is of unit magnitude, then
UPSEE 2013
(A)

`| r-s | =2`

(B)

`| r+s | =1`

(C)

`| r +s | =2`

(D)

`|r-s | =1`

Solution:

Area of `Delta ABC =1/2 | vec (AB) xx vec (AC) | =1` (given)

`=> |vec (AB) xx vec (AC) | =2`

`=> | (r hat(i) + hat (j)) xx (s hat(i) - hat (j)) | =2`

`=> | (-r) (hat(i) xx hat (j)) +s (hat (j) xx hat (i)) | =2`

`=> | - (r +s) hat (k) | =2`

`=> | r+s | =2`
Correct Answer is `=>` (C) `| r +s | =2`
Q 2414301259

The area of a parallelogram whose adjacent
sides are represented by the vectors

`a= - hat i -2 hat j- 3 hat k` and `b=- hat i+2 hat j- 3 hat k` is
UPSEE 2011
(A)

`sqrt 14`

(B)

`sqrt 6`

(C)

`49/36`

(D)

`4 sqrt 10`

Solution:

`:.` Area of parallelogram `= | a xx b|`

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

`=| hat i (6+6) - hat j(3-3) + hat k (-2-2)|`

`=| 12 hat i -4 hat k|`

`=sqrt(12^2+4^2)`

`=sqrt(144+16)`

`=sqrt 160`

`=4 sqrt 10`
Correct Answer is `=>` (D) `4 sqrt 10`
Q 1877401386

Area of a rectangle having vertices

`A ( -i + 1/2 j +4k)`
` B (i + 1/2 j + 4k )`
`C (i - 1/2 j + 4k)`and
` D( -i - 1/2 j + 4k )` is
Class 12 Exercise 10.4 Q.No. 12
(A)

` 1/2` sq. unit

(B)

`1` sq. unit

(C)

`2` sq. unit

(D)

`4` sq. unit

Solution:

` vec (OA) =-hat i + 1/2 hat j + 4hat k , vec (OB) = hat i + 1/2 hat j + 4 hat k`

` vec (OC) = hat i - 1/2 hat j + 4 hat k , vec (OD) = - hat i - 1/2 hat j+ 4 hat k`

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

` vec (AC) = vec (OC ) - vec (OA) = 2 hat i - hat j`

` vec (AB) xx vec (AC) |( hat i , hat j , hat k ) , ( 2, 0, 0) ,( 2 , -1 ,0) | = hat k(-2) = -2 hat k`

`| vec (AB) xx vec (AC) |= sqrt (( -2)^2) = 2` sq. units

Hence area of rectangle `= 2` sq. units.
Correct Answer is `=>` (C) `2` sq. unit
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)`
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 2426378271

The points, whose position vectors are
`60 i + 3j, 40i- 8j` and `ai- 52j` collinear, if
UPSEE 2013
(A)

`a = 40`

(B)

`a = - 40`

(C)

`a =20`

(D)

`a= -20`

Solution:

` ( 40 i - 8 j ) - (60 i + 3j) = lambda [ (ai - 52 j) - (40 i - 8 j) ]`

For some scalar `lambda`

`-20 = lambda (a-40)` and `-11 = - 44 lambda`

`=> a- 40 = -20/lambda` and `lambda = 1/4`

`=> a= 40 -20/1 xx 4 = -40`
Correct Answer is `=>` (B) `a = - 40`
Q 2485634567

Let `a = 2i + j + k, b = i + 2j - k`, and a unit
vector `c` be caplanar. If `c` is perpendicular to `a`,
then `c` is
UPSEE 2012
(A)

`1/sqrt(2) (-j + k)`

(B)

`1/sqry(3) (-i -j - k)`

(C)

`1/sqrt(5) (i -2j)`

(D)

`1/sqrt(3) (i- j - k)`

Solution:

`c` is coplanar with `a` and `b`.

`:. c=xa+yb`

`=> c = x ( 2i + j + k) + y(i + 2j- k)`

`=> c = (2x +y) i + (x+2y) j + (x-y) k`

`a* c =0`

`:. 2 (2x+y) +x + 2y +x -y =0`

`=> y = -2x`

`c=- 3xj +3xk = 3x (-j + k)`


`|c| =1`

`:. 9x^2 + 9x^2 =1`

`=> x= pm 1/(3 sqrt(2))`


`=> c =1/sqrt(2) (-j + k)`
Correct Answer is `=>` (A) `1/sqrt(2) (-j + k)`
Q 2328745601

Let a, b and c be the distinct non-negative
numbers. If the vectors `ai + aj + ck, i + k` and
`c i + cj + bk` lie on a plane, then which one of the
following is correct?
NDA Paper 1 2010
(A)

c is the arithmetic mean ot a and b

(B)

c is the geometric mean of a and b

(C)

c is the harmonic mean of a and b

(D)

c is equal to zero

Solution:

Since, vectors `ai + aj + ck, i + k` and `ci + cj + bk` lie
on a plane i.e, are coplanar



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




Expanding along `R_2`,


`-1(ab-c^2) -1(ac-ac) = 0`


`=> ab -c^2 = 0`

So, c is the geometric mean of a and b.
Correct Answer is `=>` (B) c is the geometric mean of a and b
Q 2484501457

Let `a, b` and `c` represent vector quantities.
Which of the following points are collinear?
UPSEE 2011

(This question may have multiple correct answers)

(A) `a-2b+3c,2a+3b 4c,-7b+10c`
(B) `-2a + 3b + 5c, a+ 2b + 3c, 7a- c`
(C) `a,b,3a-2b`
(D) `a + b- c, 3a - 4 c, 2a + b + 3c`
Solution:

(a) `Delta= |(1,-2,3),(2,3,-4),(0,-7,10)|`

`= 7( -4- 6) + 10(3 + 4)`

`= - 70 + 70 = 0`, collinear

(b) `Delta=|(-2,3,5),(1,2,3),(7,0,-1)|`

`=4(9-10)-1(-4-3)`

`= - 7 + 7 = 0`, collinear

(c) `Delta ne 0`

(d) `Delta ne 0`
Correct Answer is `=>` (A)
Q 2805091868

Let `a, b` and `c` be the distinct non-negative numbers. If the vectors `a hat i +a hat j + c hat k, hat i + hat k`, c hat i + c hat j + b hat k` lie on a plane, then which one of the following is correct?

(A)

c is the arithmetic mean of a and b

(B)

c is the geometric mean of a and b

(C)

c is the harmonic mean of a and b

(D)

c is equal to zero

Solution:

Since, vectors `a hat i + a hat j + c hat k , hat i + hat k`

and `c hat i + c hat j + b hat k` are coplanar.

`:. | (a , a, c),(1 , 0 ,1),(c , c,b) | = 0`

`=> 1 ( ab - c^2 ) + 1 ( ac - ac) = 0`

` => ab - c^2 = 0`

So, `c` is the geometric mean of `a` and `b`.
Correct Answer is `=>` (B) c is the geometric mean of a and b
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 2825291161

If `hat a, hat b, hat c` are pan-coplanar unit vectors such that `hat a xx (hat b + hat c) = ( hat b + hat c ) / sqrt2 , hat b ` and `hat c` are non-parallel, then the angle between `hat a` nd `hat b` is

(A)

`pi/2`

(B)

`pi/3`

(C)

`(2 pi)/3`

(D)

`(3 pi)/4`

Solution:

Given `(a· b)b -(a· b)c = (b + c) /sqrt2`

On comparing the coefficients of `b` and

`c` on both sides, we get

`a· c = 1/sqrt2` and `a · b = - 1/sqrt2`

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

`:. a· b = | a | | b | cos theta = - 1/sqrt2`

`=> cos theta =- 1/sqrt2 [ ∵ | a | = | b | = 1]`

` => theta = (3 pi)/4`
Correct Answer is `=>` (D) `(3 pi)/4`
 
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