374 JEE Main Chapterwise Topicwise Mathematics

14
Vector Algebra
TOPIC 1 clockwise sense. If, with respect to Ans. (a)
new system, a has components Let OA be 3 $i + $j and OB be α $i + β$j.
Algebra and p + 1 and 10, then a value of p is
Modulus of Vector Y
equal to [2021, 18 March Shift-I]
B
5 4
(a) 1 (b) − (c) (d) − 1
01 Let P1 , P2 , ……, P15 be 15 points on a 4 5
Ans. (d) A
circle. The number of distinct
After counter clockwise (or 45º
triangles formed by points 30º
anti-clockwise) rotation, the length of X′
P$ , P$ and P $ , such that the vector a remains constant. O X
i j k
Y′
$i + $j + k$ ≠ 15 is
[2021, 01 Sep. Shift-II] As, we can notice in OA, 1 / 3 = tan30 º.
(a) 12 (b) 419 (c) 443 (d) 455 So, it makes an angle of 30º with the
(p+1, √10) X-axis.
Ans. (c)
$i + $j + k$ = 15 Now, when OA is rotated further by
a (3p, 1) 45ºa nticlockwise, the resultant vector
where, $i = 1, $j + k$ = 14 OB makes an angle of 75º with the
a
⇒ ($j = 2, k$ = 12), ($j = 3, k$ = 11), O
X-axis.
So, OB = | OA | (cos 75° $i + sin 75° $j)
($j = 4, k$ = 10),
($j = 5, k$ = 9) ($j = 6, k$ = 8) … 5 ways i.e. | a | at old position = | a | at new position
Y
$i = 2, $j + k$ = 13 ⇒ C (0, β) B (α, β)
(3p) 2 + (1) 2 = (p + 1) 2 + ( 10 ) 2
⇒ ($j = 3, k$ = 10), … , ($j = 6, k$ = 7) … 4
⇒ 9p2 + 1 = p2 + 1 + 2p + 10
ways ⇒ 8p − 2p − 10 = 0
2
$i = 3, $j + k$ = 12
⇒ 4p2 − p − 5 = 0
X′ X
⇒ ($j = 4, k$ = 8), ($j = 5, k$ = 7) … 2 ways ⇒ (p + 1) (4p − 5) = 0 O
5 Y′
$i = 4 ,$j + k$ = 11 ⇒ (j$ = 5, k$ = 6) … 1 way ⇒ p= ,− 1
4 Let ∆OBC be the required triangle whose
∴Total = 12 ways area we have to determine.
Then, number of possible triangles using
03 Let a vector α $i + β$j be obtained by Area of ∆OBC = (1 /2) × (Base) × (Height)
vertices
P , P , P such that $i + $j + k$ ≠ 15 is
$ $
rotating the vector 3 $i + $j by an = 1 /2 × β × α
i j k$
1
15
C 3 − 12 = 455 − 12 = 443 angle 45° about the origin in = (2 sin 75° ) (2 cos 75° )
2
counterclockwise direction in the
02 A vector a has components 3p and = 2 sin 75° cos 75°
first quadrant. Then the area of
1 with respect to rectangular triangle having vertices (α, β), (0, β) = sin150 °
cartesian system. This system is and (0,0) is equal to = sin30 °
rotated through a certain angle [2021, 16 March Shift-I] = 1 /2
about the origin in the counter 1 1 Hence, the area is 1/2 sq. unit.
(a) (b) 1 (c) (d) 2 2
2 2
04 If vectors a 1 = x $i − $j + k$ and 06 Letα = (λ − 2) a + b and Clearly, angle bisector divides the sides
AB in OA : OB, i.e., 2 :2 = 1 : 1
a = i$ + y$j + zk$ are collinear, then a β = (4λ − 2) a + 3b be two given [using angle bisector theorem]
2
vectors where vectorsa and b are
possible unit vector parallel to the So, D is the mid-point of AB and hence
non- collinear. The value of λ for
vector x $i + y$j + zk$ is  3 + 1 3 + 1
which vectorsα and β are collinear, coordinates of D are  , 
[2021, 26 Feb. Shift-II] is [2019, 10 Jan. Shift-II]  2 2 
1 1 $ $
(a) (− $j + k$ ) (b) (i − j ) (a) 4 (b) −3 Now, equation of bisector OD is
2 2 (d) −4
(c) 3  3+1 
1 $ $ $ 1 $ $ $  −0
(c) (i + j − k ) (d) (i − j + k ) Ans. (d)
3 3 (y − 0) =  2  (x − 0) ⇒ y = x
Two vectors c and d are said to be  3+1 
Ans. (d)  −0
collinear if we can write c = λb for some  2 
Given, a 1 = x $i − $j + k$ and a 2 = $i + y$j + zk$ non-zero scalar λ. ⇒ x−y =0
x −1 1 Let the vectors α = (λ − 2) a + b According to the problem,
are collinear, then = = = λ (Say)
1 y z and β = (4λ − 2) a + 3 b are 3 β − (1 − β)
collinear, where a and bare = CM =
1 1 2 2
This gives x = λ, y = − , z = non-collinear.
λ λ
∴ We can writeα = kβ, for some [Distance of a point P (x 1 , y 1 ) from the line
Then, unit vector parallel to vector
k ∈ R − {0 } ax + by 1 + c 
x $i + y$j + zk$ will be ax + by + c = 0 is 1 
⇒ (λ − 2) a + b = k [(4λ − 2) a + 3 b] a 2 + b 2 
  1  k$ 
±  (λ)$i −   $j +
1
   ⇒ [(λ − 2) − k (4λ − 2)] a+(1 − 3k) b = 0 ⇒ |2β − 1 | = 3 ⇒ 2β = ± 3 + 1
  λ  λ 
= Now, as a and b are non-collinear,
2 2 ⇒ 2β = 4, − 2 ⇒ β = 2, − 1
−1
(λ) 2 +   +  
1 therefore they are linearly independent
λ  λ Sum of 2 and −1 is 1.
and hence
(λ 2 $i − $j + k$) λ (λ 2 $i − $j + k$) (λ − 2) − k (4λ − 2) = 0 and 1 − 3k=0
=± =± 08 If the vectors AB = 3i + 4k and
⇒ λ − 2 = k(4λ − 2) and 3k = 1
λ λ4 + 2 λ4 + 2 AC = 5$i − 2$j + 4k$ are the sides of a
λ − 2 = (4λ − 2) Q3k = 1 ⇒ k = 
1 1
⇒ ∆ABC, then the length of the median
± ($i − $j + k$) 3  3 
Take, λ = 1 =
3 ⇒ 3λ − 6 = 4λ − 2 ⇒ λ = −4 through A is [JEE Main 2013, 2003]
π (a) 18 (b) 72 (c) 33 (d) 45
05 If a unit vector a makes angles 07 Let 3 $i + $j, $i + 3$j and β$i + (1 − β) $j
3 Ans. (c)
π $ respectively be the position
with $i, with $jand θ ∈(0, π) with k, We know that, the sum of three vectors
4 vectors of the points A, B and C with of a triangle is zero.
then a value of θ is respect to the origin O. If the A
[2019, 9 April Shift-II] distance of C from the bisector of
5π π 5π 2π the acute angle between OA and OB
(a) (b) (c) (d)
6 4 12 3 3
is , then the sum of all possible
Ans. (d) 2
π values of β is [2019, 11 Jan. Shift-II] B C
Given unit vector a makes an angle M
3 (a) 1
π ∴ AB + BC + CA = 0
with $i, with $j and θ ∈(0, π) with k.
$
(b) 3
4 ⇒ BC = AC − AB [QAC = − CA]
(c) 4
Now, we know that (d) 2 AC − AB
cos2 α + cos2 β + cos2 γ = 1 , whereα, β, γ ⇒ BM =
Ans. (a) 2
are angles made by the vectors
According to given information, we have [QM is a mid-point of BC]
with respectively $i, $j and k.
$
the following figure. Also, AB + BM + MA = 0
π π
∴ cos2   + cos2   + cos2 θ = 1 A(Ö3, 1) [by properties of a triangle]
3  4
1 1 AC − AB
⇒ + + cos2 θ = 1 +1 ⇒ AB + = AM [QAM = − MA]
4 2 Ö3 2
+1 , 2
1 1 Ö3
⇒ cos2 θ = ⇒ cosθ = ± 2 M AB + AC
⇒ AM =
4 2 2 D 2
π 2π
⇒ cosθ = cos   or cos   C (b, (1–b)) 3 i + 4k$ + 5$i − 2 $j + 4k$
$
3  3  =
π 2π 2
⇒ θ = or
3 3 = 4 $i − $j + 4k$
B(1, Ö3)
2π 2 ⇒ | AM | = 42 + 12 + 42 = 33
So, θ is , according to options. O
3
09 Let a, b and c be three non-zero 12 If a, b and c are three non-zero a a2 1 + a3
vectors which are pairwise vectors such that no two of these 14 If b b2 1 + b3 = 0 and vectors
non-collinear. If a + 3b is collinear are collinear. If the vector a + 2b is c c2 1+ c3
with c and b + 2 c is collinear with a, collinear with c and b + 3c is
then a + 3b + 6 c is [AIEEE 2011] collinear with a (λ being some (1, a , a 2 ), (1, b, b2 ) and (1, c, c 2 ) are
(a) a + c (b) a (c) c (d) 0 non-zero scalar), then a + 2b + 6c non- coplanar, then the product
Ans. (d) equal to [AIEEE 2004] abc equal to [AIEEE 2003]

As, a + 3 b is collinear withc. (a) λa (b) λb (a) 2 (b) –1 (c) 1 (d) 0

∴ a + 3 b = λc …(i) (c) λc (d) 0 Ans. (b)
Also, b + 2c is collinear witha. Ans. (d) Since,
⇒ b + 2c = µa …(ii) If a + 2b is collinear withc, then a a2 1 + a3 a a2 1
From Eq. (i), we get a + 2b = tc …(i) b b2 1 + b = b b2
3
1
a + 3 b + 6 c = (λ + 6) c …(iii) Also, b + 3 c is collinear witha, then c c2 1 + c3 c c2 1
From Eq. (ii), we get b + 3 c = λa
⇒ b = λa − 3 c …(ii) a a2 a3
a + 3 b + 6 c = (1 + 3 µ ) a …(iv)
From Eqs. (i) and (ii), we get + b b2 b3 = 0
From Eqs. (iii) and (iv), we get
a + 2 (λa − 3 c) = tc c c2 c3
∴ (λ + 6) c = (1 + 3 µ ) a
⇒ (a − 6c) = tc − 2λa
Since,a is not collinear withc. a a2 1 a a2 1
On comparing the coefficients of
⇒ λ + 6 = 1 + 3µ = 0 ⇒ b b2 1 + abc b b 2 1 =0
a and c, we get
From Eq. (iv), we get 1 c c2 1 c c2 1
1 = − 2λ ⇒ λ = −
a + 3b + 6c = 0 2
a a2 1
and −6 = t ⇒ t = −6
10 The non-zero vectors a, b and c are ⇒ (1 + abc) b b2 1 =0
From Eq. (i), we get
c c2 1
related by a = 8b and c = − 7 b. a + 2b = −6c
Then, the angle between a and c is ⇒ a + 2b + 6c = 0  a a2 1 
 
[AIEEE 2008] Q b b
2
1 ≠ 0
π π 13 Consider points A, B, C and D with  c c2 
(a) π (b) 0 (c) (d) 1
4 2  
Ans. (a)
position vectors 7 $i − 4$j + 7 k$ , ⇒ 1 + abc = 0
$i − 6 $j + 10 k$ , − $i − 3$j + 4k$ and ⇒ abc = − 1
Since,a = 8 b and c = −7 b
So, a is parallel tob and c is anti-parallel 5 $i − $j + 5 k$ , respectively. Then,
to b. 15 The vector $i + x $j + 3 k$ is rotated
⇒ a and c are anti-parallel. ABCD is a [AIEEE 2003]
through an angle θ and doubled in
So, the angle betweena and c is π. (a) square (b) rhombus
magnitude, then it becomes
(c) rectangle (d) None of these
4 $i + (4x − 2) $j + 2 k$ . The value of x
11 If C is the mid-point of AB and P is Ans. (d)
Given that, OA = 7 $i − 4 $j + 7 k$ , are [AIEEE 2002]
any point outside AB, then
(a) − , 2 (b)  , 2
2 1
[AIEEE 2005] OB = $i − 6 $j + 10 k$ ,
 3  3 
(a) PA + PB + PC = 0 OC = − $i − 3 $j + 4 k$
2 
(b) PA + PB + 2PC = 0 (c)  , 0 (d) {2, 7}
and OD = 5$i − $j + 5 k$ 3 
(c) PA + PB = PC
Now, AB = ( 7 − 1) 2 + (−4 + 6) 2 + ( 7 − 10) 2 Ans. (a)
(d) PA + PB = 2PC
Since, the vector $i + x $j + 3k$ is doubled
Ans. (d) = 36 + 4 + 9 = 49 = 7
in magnitude, then it becomes
Let P be the origin outside of AB and C is BC = (1 + 1) 2 + (−6 + 3) 2 + (10 − 4) 2 4 $i + (4x − 2) $j + 2k$
mid-point of AB, then
= 4 + 9 + 36 = 49 = 7 ∴ 2 | $i + x $j + 3k$ | = |4 $i + (4x − 2) $j + 2k$ |
C B
A
CD = (−1 − 5) 2 + (−3 + 1) 2 + (4 − 5) 2 ⇒ 2 1 + x 2 + 9 = 16 + (4x − 2) 2 + 4
= 36 + 4 + 1 = 41 ⇒ 40 + 4x 2 = 20 + (4x − 2) 2
⇒ 3x 2 − 4x − 4 = 0
and DA = (5 − 7) 2 + (−1 + 4) 2 + (5 − 7) 2
P ⇒ (x − 2)(3x + 2) = 0
PA + PB = 4 + 9 + 4 = 17 2
PC = ⇒ 2PC = PA + PB ⇒ x = 2, −
2 Hence, option (d) is correct. 3
TOPIC 2 E F From (i),
7 | a |2 − 15| a |2 + 16a ⋅ b = 0
Scalar or Dot Product of Two ⇒ 16a ⋅ b = 8| a |2
Vectors and Its Applications H
G ⇒ 16| a | | b | cos θ = 8 | a |2
8|a|
B ⇒ cos θ = [Q| a | = | b |]
16 Let a and b be two vectors such A 16 | a | | b |
that | 2a + 3b | = | 3a + b | and the 1
1 D C ⇒ cosθ =
angle between a and b is 60°. If a 2
8 AG = 10 $i + 10 $j + hk$ ∴ θ = 60 °
is a unit vector, then | b | is equal to
BH = −10 $i + 10 $j + hk$
[2021, 31 Aug. Shift-I] 20 Let a, b, c be three mutually
(a) 4 (b) 6 Since, AG⋅BH = AG BH cosθ
(−100 + 100 + h2 )
perpendicular vectors of the same
(c) 5 (d) 8
magnitude and equally inclined at
= h2 + 200 ⋅ h2 + 200 .  
Ans. (c) 1
 5 an angle θ, with the vector a + b + c.
|2a + 3b | = |3a + b | Then, 36cos2 2θ is equal to ……… .
⇒ 5h2 = h2 + 200 ⇒ h2 = 50
⇒ |2a + 3b |2 = |3a + b |2 [2021, 20 July Shift-I]
⇒ h= 5 2
⇒4| a|2 + 9b|2 + 12a ⋅ b = 9| a|2 + |b|2 + 6a ⋅b Ans. (4)
⇒ 5 | a | 2 − 6 a ⋅ b − 8| b | 2 = 0
18 If the projection of the vector a ⊥ b⊥ c
a
is a unit vector. $i + 2$j + k$ on the sum of the two |a | = |b | = | c|
8
a ⋅ ( a + b + c) = | a | | a + b + c | cosθ
So, | a| = 8 vectors 2$i + 4$j − 5k$ and
a ⋅ a = | a | | a + b + c | cosθ
And, 5⋅64 − 6⋅ 8|b |   − 8 |b |2 = 0
1
−λ $i + 2$j + 3k$ is 1, then λ is equal to | a |2 = | a| | a + b + c | cos θ
 2
[2021, 26 Aug. Shift-II] | a | (| a | − | a + b + c | cos θ) = 0
⇒ |b |2 + 3|b | − 40 = 0
Ans. (5) As, | a | ≠ 0
⇒ (|b | + 8) (|b | − 5) = 0
Let a = $i + 2$j + k$, b = 2 $i + 4$j − 5k$ So, | a | = | a + b + c | cos θ
|b | = 5
Similarly, | b | = | a + b + c | cos θ
As, |b | = − 8 Not possible. c = −λ $i + 2$j + 3k$
| c | = | a + b + c | cosθ
Now, according to the questions,
Now, | a + b + c |2 = (a + b + c) ⋅ (a + b + c)
17 A hall has a square floor of a ⋅ (b + c)
=1 = a ⋅ a + b ⋅b + c ⋅ c
dimension 10m × 10m (see the | b + c|
= 3 | a |2
figure) and vertical walls. If the ⇒ a ⋅ b + a ⋅ c = | b + c|
∠GPH between the diagonals AG ∴| a + b + c | = 3 | a | = 3 | b | = 3 | c |
$
⇒ 5 + (− λ + 4 + 3) = | (2 − λ) $i + 6$j − 2k|
1 As, | a | = | a + b + c | cos θ
and BH is cos − 1 , then the height ⇒ (12 − λ ) 2 = (2 − λ ) 2 + 36 + 4
5 ⇒ | a | = 3 | a | cos θ
⇒ λ + 144 − 24λ = λ2 − 4λ + 4 + 40
2
1
of the hall (in m) is ⇒ cosθ =
⇒ λ=5
[2021, 26 Aug. Shift-II] 3
E F ∴ cos 2θ = 2 cos2 θ − 1
19 If (a + 3b) is perpendicular to
−1
= 2  − 1 =
1
H (7 a − 5b) and (a − 4b) is  3 3
G
perpendicular to (7 a − 2b), then the 1
So, 36 cos 2θ = 36 × = 4
2
angle between a and b (in degrees) 9
P is …… . [2021, 25 July Shift-II]
21 For p> 0, a vector V2 = 2i$ + (p + 1) $j is
B
Ans. (60)
A If (a + 3b) is perpendicular to (7 a − 5 b), obtained by rotating the vector
then (a + 3b) ⋅ (7 a − 5b) = 0 v 1 = 3p $i + $j by an angle θ about
D C
⇒ 7 | a |2 − 15 | b |2 + 16a ⋅ b = 0 ...(i) origin in counter clockwise
(a) 5 (b) 2 10 Also, (a − 4b) is perpendicular to (7 a − 2b). (α 3 − 2)
(c) 5 3 (d) 5 2 (a − 4b) ⋅ (7 a − 2b) = 0 direction. If tanθ = , then
(4 3 + 3)
Ans. (d) ⇒ 7 | a |2 + 8| b |2 − 30 a ⋅ b = 0 ...(ii)
Eq. (i) × 15 and Eq. (ii) × 8, the value of α is equal to
Let A = (0, 0, 0)
[2021, 20 July Shift-II]
B = (10, 0, 0) 105 | a |2 − 225 | b |2 + 240 a ⋅ b = 0
56 | a |2 + 64 | b |2 − 240 a ⋅ b = 0 Ans. (6)
G = (10, 10, h)
161 | a |2 − 161 | b |2 = 0 Since, V2 is obtained by rotating V1 .
H = (0, 10, h) ∴ | V1 |2 = | V2 |2
⇒ | a | = |b | ...(ii)
 ⇒ 3p2 + 1 = 4 + (p + 1) 2 23 In a triangle ABC, if 25 a, b and c be three unit vectors such
⇒ 3p2 + 1 = 4 + p2 + 1 + 2p |BC| = 8, |CA| = 7, | AB | = 10, then that | a − b | 2 + | a − c | 2 = 8. Then
⇒ 2p − 2p − 4 = 0 ⇒ p2 − p − 2 = 0
2
the projection of the vector AB on | a + 2b | 2 + | a − 2c| 2 is equal to …… .
⇒ (p + 1) (p − 2) = 0 ⇒ p = − 1 or p = 2 AC is equal to [2020, 2 Sep. Shift-I]
Since, p> 0 given, then p = − 1 is [2021, 18 March Shift-II] Ans. (2)
discarde(d) 25 85 127 115
V ⋅V (a) (b) (c) (d) Given, for unit vectors a, b and c
Now, cos θ= 1 2 4 14 20 16
| V1 | | V2 | | a − b |2 + | a − c |2 = 8
Ans. (b) ⇒ | a |2 + | b |2 − 2 a ⋅ b + | a |2 + | c |2
4 3+3 4 3+3 Projection of AB or AC = AB cos θ
= = −2a ⋅ c = 8
13 ⋅ 13 13 = 10 ⋅ cos θ ⇒ − 2 [a ⋅ b + a ⋅ c] = 4
(4 3 + 3) 2 B ⇒ a⋅ b + a⋅c = − 2 … (i)
sin θ = 1 − cos2 θ = 1 −
(13) 2 Now, | a + 2 b |2 + | a + 2c |2
= | a |2 + 4 | b |2 + 4 a ⋅ b + | a |2 + 4 | c |2
(13) 2 − (4 3 + 3) 2 10 8
= + 4a ⋅ c
13
= 10 + 4 [a ⋅ b + a ⋅ c]
112 − 24 3 θ
⇒ sinθ = A 7 C = 10 + 4 (− 2) [from Eq. (i)]
13 = 10 − 8 = 2
 10 + 7 − 82  85
2 2
sin θ 112 − 24 3 = 10 .  =
∴ tan θ = =  2 × 7 × 10  14
cos θ 4 3+3 26 Let the position vectors of points ‘A’
α 3 −2

 using cosθ =
c2 + b2 − a2 
 and ‘B ’ be $i + $j + k$ and 2$i + $j + 3k$ ,
Given, tanθ =  2bc  respectively. A point ‘ P ’ divides the
4 3+3
line segment AB internally in the ratio
∴ 112 − 24 3 = α 3 − 2 24 Let x be a vector in the plane λ :1(λ > 0). If O is the origin andOB ⋅ OP
⇒ 6 3 − 2 =α 3 − 2 containing vectors a = 2i$ − $j + k$ and − 3 | OA × OP| 2 = 6, then λ is equal to
∴ α =6 b = i$ + 2$j − k$ . If the vector x is ……… . [2020, 2 Sep. Shift-II]

22 In ∆ ABC, if | BC| = 3, | CA| = 5 and perpendicular to (3i$ + 2$j − k$ ) and its Ans. (0.80)
17 6 It is given that OA = $i + $j + k$
| BA| = 7, then the projection of the projection on a is , then the
vector BA on BC is equal to 2 and OB = 2$i + $j + 3k$

[2021, 20 July Shift-II] value of | x | 2 is equal to ………… . ∴Point ‘P ’ divides line segment AB
19 13 [2021, 17 March Shift-II] internally in the ratio λ:1, (λ > 0), then
(a) (b)
2 2 Ans. (486) (2λ + 1) i$ + (λ + 1) $j + (3λ + 1) k
$
11 15 OP =
(c) (d) Let x = λa + µb, where λ and µ are λ+ 1
2 2 1
scalars. ∴ OB⋅ OP = (14λ + 6)
Ans. (c) ⇒ x = λ (2$i − $j + k$) + µ ($i + 2$j − k$) λ+ 1
Given, ∆ABC, |BC | = 3, | CA | = 5, |BA | = 7
x = $i (2λ + µ ) + $j (2µ − λ) + k$ (λ − µ ) i$ i$ $
k
A and OA × OP = 1 1 1
Since, x is perpendicular to (3$i + 2$j − k$). 2λ + 1 3λ + 1
1
7 5 λ+ 1 λ+ 1
Then, x ⋅ (3$i + 2$j − k$ ) = 0
θ ⇒ 3λ + 8µ = 0 2λ $ λ $ λ $
B C … (i) = i− j− k
3
17 6 λ+ 1 λ+ 1 λ+ 1
Also, given projection of x on a is . λ
Now, projection of BA on 2 ∴ | OP × OP | = 6
BC= | BA | cos ∠ABC x ⋅ a 17 6 λ+ 1
∴ = 6λ2
(7) 2 + (3) 2 − (5) 2 |a| 2 ⇒ | OA × OP |2 =
Now, cos ∠ABC = (λ + 1) 2
2⋅ 7 ⋅3 ⇒ 2(2λ + µ ) + (λ − 2µ ) + (λ − µ ) = 51 It is given that,
49 + 9 − 25 ⇒ 6λ − µ = 51 … (ii) OB⋅ OP − 3 | OA × OP | 2 = 6
=
42 From Eqs. (i) and (ii), 14λ + 6 18λ2
33 11 ⇒ − =6
= = λ = 8, µ = − 3 λ+ 1 (λ + 1) 2
42 14
11 ∴ x = 13$i − 14$j + 11k$ ⇒14λ 2 + 20λ + 6 − 18λ2 = 6λ2 + 12λ + 6
∴Projection = |BA | ⋅ ⇒ 10 λ 2 − 8λ = 0 ⇒ λ = 0 or 0.8
14 ⇒ | x | = (13) 2 + (− 14) 2 + (11) 2
11 11 Q λ > 0, ∴ λ = 0.8
=7× = ∴ | x | = (13) 2 + (− 14) 2 + (11) 2 = 486
14 2 Hence, answer is 0.80
27 Let a,b,c ∈R be such that perpendicular to c, then the value Now, as (2x + λy) is perpendicular to y,
of | a + b − c | is ……… . so
a 2 + b2 + c 2 = 1. If (2x + λy) ⋅ y = 0
 2π  [2020, 5 Sep. Shift-II]
a cosθ = bcos θ +  ⇒ 2x ⋅ y + λ | y |2 = 0 ...(ii)
 3 Ans. (6.00)
From Eqs.(i) and (ii), we get
π
= ccosθ +  ,
4 It is given that projection of bon a is
equal to the projection of c on a, where λ=1
 3 | a|= 2, | b|= 4 and | c|= 4,
π a⋅ b a⋅c 31 A vector a = α i$ + 2$j + β k$ (α, β ∈ R)
where θ = , then the angle so = ⇒a ⋅ b = a ⋅ c
9 2 2 lies in the plane of the vectors,
$
between the vectors a$i + b$j + ck and bis perpendicular to c, so b⋅ c = 0 b = $i + $j and c = $i − $j + 4k$ . If a
and bi$ + c$j + ak$ is Now, | a + b − c |2 bisects the angle between b and c,
[2020, 3 Sep. Shift-II]
= | a |2 + | b |2 + | c |2 + 2a ⋅ b − 2 b⋅ c − 2a ⋅ c then [2020, 7 Jan. Shift-I]
π 2π π = 4 + 16 + 16 = 36 (a) a⋅ $i + 3 = 0 (b) a⋅k$ + 2 = 0
(a) (b) 0 (c) (d)
2 3 9 ∴ | a + b − c | = 6.
(c) a⋅ $i + 1 = 0 (d) a⋅k$ + 4 = 0
Ans. (a)
29 If a and b are unit vectors, then the Ans. (b)
It is given that, greatest value of 3 | a + b| + | a − b| Given vectors b = $i + $j and c = $i − $j + 4k$ .
2π  4π 
a cosθ = b cos  θ + 
 = c cos  θ +  is ....... [2020, 6 Sep. Shift-I] So, vector bisects the angle between b
 3   3 
Ans. (4.00) and cis
= k (let)
Let angle between unit vectors a and b  b c
⇒ a=
k
,b =
k a = λ ± 
cosθ  2π  is θ ∈[0, π].  | b| | c | 
cos  θ + 
 3  Then, | a + b|2 = | a |2 + | b|2 + 2a ⋅ b  $i + $j $i − $j + 4k$ 
= 1 + 1 + 2 cosθ = 2(1 + cosθ) ⇒ a = λ  ± 
18 
k
and c= …(i) θ  2
 4π  = 4 cos2
cos  θ +  λ
 3  2 = ((3$i + 3$j) ± ($i − $j + 4k$ ))
 θ 3 2
Now angle between vectorsa$i + b$j + ck $ ⇒ | a + b| = 2 cos  λ $
 2 = (4 i + 2$j + 4k$ )
and b$i + c$j + ak$ , wherea 2 + b 2 + c 2 = 1
3 2
  θ 
for a, b, c ∈ R is α = cos−1 | (ab + bc + ca)| Qθ ∈ [0, π] ⇒ cos 2  ≥ 0  =
λ $
(2 i + 4$j − 4k$ )
Q ab + bc + ca = k 2  
3 2
 Similarly, | a − b|2 = | a |2 + | b|2 − 2a ⋅ b
On comparing with given
 = 1 + 1 − 2cosθ

1
+
1 a = α$i + 2$j + βk,
$ (α, β ∈R)
π θ
 cosθ cos  θ + 2  cos  θ +  cos  θ + 4 π 
2 π = 2 (1 − cosθ) = 4 sin2 λ $
     2 and (4 i + 2$j + 4k$ ), we have λ = 3 2,
  3   3   3  3 2
 θ
 ⇒ | a − b| = 2 sin  so α = 4 and β = 4
1   2
+  which not satisfy the given options.
4π So, 3| a + b| + | a − b|
cos  + θ cosθ  Now, on comparing with
 3    θ θ 
= 2  3 cos   + sin    λ $
(2 i + 4$j − 4k$ )
  4π   2π     2  2  3 2
 cos  3 + θ + cos  3 + θ + cosθ 
= k2   having greatest value = 2 3 + 1 = 4 and a = α$i + 2$j + β k$ ,
 cosθ cos  2 π + θ cos  4 π + θ  [Qgreatest value of a cosθ + b sinθ is 3 2
  3   3   we have λ = , so α = 1 and β = −2
a +b ]
2 2 2
π
2 cos( π + θ) cos   + cosθ Now, since a⋅k$ = − 2 ⇒a⋅k$ + 2 = 0.
3
=k 2
30 If x and y be two non-zero vectors
2π 4π
cosθ cos  + θ cos  + θ
 3   3  such that | x + y| = | x| and 2x + λy is 32 Let A(3, 0, −1), B (2, 10, 6) and C(1, 2, 1) be
− cosθ + cosθ perpendicular to y, then the value
=k 2
=0 the vertices of a triangle and M be
2π 4π of λ is
cosθ cos  + θ cos  + θ [2020, 6 Sep. Shift-II]
the mid-point of AC. If G divides BM
 3   3  Ans. (1.00) in the ratio 2 : 1, then cos (∠GOA)
∴ α = cos−1 (0) = π /2 For two non-zero vectors x and y, it is
Hence, option (a) is correct.
(O being the origin) is equal to
given [2019, 10 April Shift-I]
|x+ y |=|x| 1 1
28 Let the vectors a, b, c be such that (a) (b)
On squaring both sides, we get 15 2 15
| a| = 2, | b| = 4 and | c| = 4. If the | x |2 + 2 ⋅ x ⋅ y + | y |2 = | x |2 1 1
projection of b on a is equal to the ⇒ | y |2 + 2x ⋅ y = 0 ...(i) (c) (d)
30 6 10
projection of c on a and b is
 Ans. (a) But projection of b on a = | a | Thus, a possible value of
b1 + b2 + 2 (λ 1 , λ 2 , λ 3 ) =  − , 4,0 
1
Key Idea Use the angle between two ∴ = 12 + 12 + ( 2) 2
2  2 
non-zero vectors a and bis given by
a⋅b b + b2 + 2
cosθ = and coordinates of the ⇒ 1 = 2 ⇒ b1 + b2 = 2 ...(i) 35 Let u be a vector coplanar with the
| a | | b| 2 vectors a = 2$i + 3$j − k$ and b = $j + k.
$
centroid i.e. Now,
 x 1 + x 2 + x 3 y 1 + y 2 + y 3 z1 + z2 + z3  a + b = ($i + $j + 2k$ ) + (b 1 $i + b 2 $j + 2k$ ) If u is perpendicular to a and
 , ,  u ⋅ b = 24, then | u| 2 is equal to
 3 3 3 
= (b + 1) $i + (b + 1) $j + 2 2k$
1 2 [JEE Main 2018]
of a triangle formed with vertices;
(x1 , y 1 , z1 ), (x2 , y 2 , z2 ) and (x3 , y 3 , z3 ). Q (a + b) ⊥ c, therefore (a + b) ⋅ c = 0 (a) 336 (b) 315 (c) 256 (d) 84
⇒{ (b 1 + 1) $i + (b 2 + 1) $j + 2 2k}
$ Ans. (a)
Given vertices of a ∆ABC are A(3, 0, − 1),
B(2, 10, 6) and C (1, 2, 1) and a point M is (5$i + $j + 2k$ ) = 0 Key idea If any vector x is coplanar with
mid-point of AC. An another pointG ⇒ 5(b 1 + 1) + 1(b 2 + 1) + 2 2 ( 2) = 0 the vector y and z, then x = λy + µz
divides BM in ratio ⇒ 5b 1 + b 2 = − 10 ...(ii) Here, u is coplanar with a and b.
2 : 1, soG is the centroid of ∆ABC. From Eqs. (i) and (ii),b 1 = − 3 and b 2 = 5 ∴ u = λa + µb
 3 + 2 + 1 0 + 10 + 2 −1 + 6 + 1  ⇒ b = − 3$i + 5$j + 2k$ Dot product with a, we get
∴ G , , 
 3 3 3  u⋅ a = λ (a ⋅ a) + µ (b ⋅ a)
⇒ | b | = (− 3) 2 + (5) 2 + ( 2) 2
= (2, 4, 2). ⇒ 0 = 14λ + 2µ …(i)
OG ⋅ OA = 36 = 6 [Qa = 2 i + 3 j − k, b = j + k, u ⋅ a = 0]
^ ^ ^ ^ ^
Now, cos(∠GOA) = , where O is
OG OA $,
34 Let a = 2i$ + λ 1 $j + 3k Dot product with b, we get
the origin. u⋅ b = λ (a ⋅ b) + µ (b ⋅ b)
$ and
b = 4i$ + (3 − λ ) $j + 6k
Q OG = 2i$ + 4$j + 2k
$ 2 24 = 2λ + 2µ …(ii)
⇒ OG = 4 + 16 + 4 = 24
$ be three
c = 3$i + 6$j + (λ 3 − 1) k [Qu⋅ b = 24]
and OA = 3$i − k
$
vectors such that b = 2a and a is Solving Eqs. (i) and (ii), we get
perpendicular to c. Then a possible λ = − 2, µ = 14
⇒ OA = 9 + 1 = 10
value of (λ 1 , λ 2 , λ 3 ) is Dot product with u, we get
and OG ⋅ OA = 6 − 2 = 4
4 1 [2019, 10 Jan. Shift-I] |u |2 = λ (u ⋅ a) + µ (u⋅ b)
∴ cos(∠GOA) = =
24 10 15 (a) (1, 3, 1) (b) (1, 5, 1) |u |2 = − 2(0) + 14(24)
 1  1  ⇒ |u |2 = 336
$, (c)  − , 4, 0 (d)  , 4, − 2
33 Let a = i$ + $j + 2 k  2  2 
Ans. (c) 36 Let a and b be two unit vectors. If
b = b $i + b $j + 2 k $ and
1 2
We have, a = 2i$ + λ 1 $j + 3k
$; the vectors c = a + 2b and
$ be three vectors
c = 5 i$ + $j + 2 k d = 5a − 4a are perpendicular to
b = 4i + (3 − λ ) $j + 6k
$ $
2
each other, then the angle between
such that the projection vector of b and c = 3$i + 6$j + (λ 3 − 1) k
$,
a and b is [AIEEE 2012]
on a is a. If a + b is perpendicular to π π
such that b = 2a (a) (b)
c, then | b | is equal to
Now, b = 2a 6 2
[2019, 9 Jan. Shift-II] π π
⇒ 4i + (3 − λ 2 ) $j + 6k
$ $ = 2 (2$i + λ $j + 3k
1
$) (c) (d)
(a) 6 (b) 4 3 4
(c) 22 (d) 32 ⇒ 4i$ + (3 − λ ) $j + 6k$ = 4i$ + 2λ $j + 6k$
2 1
Ans. (c)
Ans. (a) ⇒ (3 − 2λ 1 − λ 2 ) $j = 0 Given that,
According to given information, we have ⇒ 3 − 2λ 1 − λ 2 = 0 (i) a and b are unit vectors,
the following figure. ⇒ 2λ 1 + λ 2 = 3 ...(i) i.e., | a | = | b | = 1
a+b Also, as a is perpendicular to c, therefore (ii) c = a + 2b and d = 5a − 4b
a.c=0 (iii) c and d are perpendicular to each
b
c
1
$ ) ⋅ (3i$ + 6$j + (λ − 1) k
⇒ (2i$ + λ $j + 3k $) =0
3
other. i.e., c⋅ d = 0
θ ⇒ 6 + 6λ 1 + 3(λ 3 − 1) = 0 To find Angle between a and b.
a
⇒ 6λ 1 + 3λ 3 + 3 = 0 Now,
b⋅ a
Clearly, projection of b on a = ⇒ 2λ 1 + λ 3 = − 1 ... (ii) c ⋅d = 0 ⇒ (a + 2 b) ⋅ (5 a − 4 b) = 0
|a|
Now, from Eq. (i), λ 2 = 3 − 2λ 1 and from ⇒ 5 a ⋅a − 4 a ⋅b + 10 b ⋅a − 8 b ⋅b = 0
(b 1 $i + b 2 $j + 2 k$ ) ($i + $j + 2 k$ ) Eq. (ii)
= λ 3 = − 2λ 1 − 1 ⇒ 6 a ⋅b = 3
12 + 12 + ( 2) 2 1
∴ (λ 1 , λ 2 , λ 3 ) ≡ (λ 1 , 3 − 2λ 1 , − 2λ 1 − 1) ⇒ a ⋅b =
b1 + b2 + 2 b1 + b2 + 2 2
= = If
1
λ 1 = − , then λ 2 = 4, and λ 3 = 0 π
4 2 2 So, the angle betweena and b is .
3
37 Let ABCD be a parallelogram such Ans. (a) = 4 $i + 2 $j − 2 k$
that AB = q , AD = p and ∠BAD be Since, the given vectors are mutually ∴Work done = F ⋅ AB
an acute angle. If r is the vector orthogonal, therefore
= (7 $i + 2 $j − 4k$ ) ⋅ (4 $i + 2 $j − 2k$ )
that coincides with the altitude a ⋅b = 2 − 4 + 2 = 0
= 28 + 4 + 8 = 40 units
directed from the vertex B to the a ⋅c = λ − 1 + 2µ = 0 ...(i)
and b ⋅c = 2λ + 4 + µ = 0 ...(ii)
side AD, then r is given by 41 Let u, v, w be such that | u | = 1,
[AIEEE 2012] On solving Eqs. (i) and (ii), we get
3( p ⋅ q ) µ = 2 and λ = −3 | v | = 2, | w | = 3. If the projection v
(a) r = 3q p
(p ⋅ p ) Hence, (λ , µ ) = (−3, 2) along u is equal to that of w along
p⋅q u and v, w are perpendicular to
(b) r = − q +  p
 p⋅p 39 The value of a, for which the points, each other, then | u − v + w | equal
p⋅q A, B, C with position vectors to [AIEEE 2004]
(c) r = q −  p (a) 2 (b) 7 (c) 14 (d) 14
 p⋅p 2$i − $j + k$ , $i − 3 $j − 5 k$ and a $i − 3 $j + k$
3(p ⋅ q) Ans. (c)
(d) r = − 3q + p respectively are the vertices of a
(p ⋅ p) π Since, |u | = 1, | v | = 2, | w | = 3
right angled triangle with C = are v ⋅u
Ans. (b) 2 The projection of v along u =
[AIEEE 2006] |u |
Given
w ⋅u
(i) A parallelogram ABCD such that (a) –2 and –1 (b) –2 and 1 and the projection of w along u =
|u |
AB = q and AD = p. (c) 2 and –1 (d) 2 and 1
(ii) The altitude from vertex B to side Ans. (d) According to given condition,
v ⋅u w ⋅u
AD coincides with a vectorr. =
Since, position vectors of A, B, C are |u | |u |
To find The vectorr in terms of p and q.
2 $i − $j + k$ , $i − 3 $j − 5k$ and a $i − 3 $j + k$ ,
Let E be the foot of perpendicular from B ⇒ v ⋅u = w ⋅u …(i)
respectively. Since, v, w are perpendicular to each
to side A(d)
AE = Projection of vector q on Now, AC = (a $j − 3 $j + k$ ) − (2 $i − $j + k$ ) other.
∴ v ⋅w = 0 …(ii)
q⋅ p
p = q⋅p = = (a − 2) $i − 2 $j Now, |u − v + w |2 = |u |2 + | v |2 + | w |2
|p |
D and BC = (a $i − 3 $j + k$ ) − ($i − 3 $j − 5k$ ) −2u ⋅ v − 2 v ⋅ w + 2u ⋅ w
C
⇒ |u − v + w |2 = 1 + 4 + 9 − 2u ⋅ v + 2v ⋅u
= (a − 1) $i + 6k$
[from Eqs. (i) and (ii)]
E Since, the ∆ABC is right angled atC, then ⇒ |u − v + w |2 = 1 + 4 + 9
p r
AC ⋅BC = 0 ⇒ |u − v + w | = 14
A B ⇒ {(a − 2) i − 2 j } ⋅ {(a − 1) $i + 6k$ } = 0
$ $
q 42 a, b, c are three vectors, such that
⇒ (a − 2)(a − 1) = 0 a + b + c = 0,| a | = 1, | b| = 2, | c | = 3, then
AE = Vector along AE of length AE ∴ a = 1 and a = 2 a ⋅ b + b ⋅ c + c ⋅ a is equal to
q⋅p (q ⋅ p) p [AIEEE 2003]
= | AE | AE =   p= 40 A particle is acted upon by
 |p |  | p |2 (a) 0 (b) –7 (c) 7 (d) 1
constant forces 4i$ + $j − 3k$ and Ans. (b)
Now, applying triangles law in ∆ABE, we
get 3i$ + $j − k$ which displace it from a Given that, |a | = 1,|b | = 2,|c | = 3
and a + b + c = 0
AB + BE = AE point i$ + 2$j + 3k$ to the point
(q ⋅p) p Now, (a + b + c) 2 = |a |2 + |b |2 + |c |2
⇒ q+r=
| p |2
5$i + 4$j + k$ . The work done in + 2(a ⋅b + b ⋅c + c ⋅a )
standard units by the forces is ⇒ 0 = 12 + 22 + 32 + 2 (a ⋅b + b ⋅c + c ⋅a )
(q ⋅ p) p
⇒ r= −q given by [AIEEE 2004] ⇒ 2 (a ⋅b + b ⋅c + c ⋅a ) = − 14
| p |2 ⇒ a ⋅b + b ⋅c + c ⋅a = −7
(a) 40 units (b) 30 units
q⋅p
⇒ r = −q +  p (c) 25 units (d) 15 units 43 Given, two vectors are $i − $j and
p⋅p
Ans. (a) $i + 2$j, the unit vector coplanar with
38 If the vectors a = i$ − $j + 2 k$ , Total force, F = (4 $i + $j − 3k$ ) + (3 $i + $j − k$ )
the two vectors and perpendicular
b = 2 i$ + 4 $j + k$ and c = λ i$ + $j + µ k$ ∴ F = 7 $i + 2 $j − 4k$
to first is [AIEEE 2002]
are mutually orthogonal, then (λ , µ) The particle is displaced from
(a)
1 $ $
(i + j) (b)
1
(2 i$ + $j)
is equal to [AIEEE 2010] A ($i + 2 $j + 3k$ ) to B (5 $i + 4 $j + k$ ). 2 5
(a) (– 3, 2) (b) (2, – 3) Now, displacement, 1 $ $
(c) ± ( i + j) (d) None of these
(c) (– 2, 3) (d) (3, – 2) AB = ( 5$i + 4 $j + k ) − ( $i + 2 $j + 3 k$) 2
 Ans. (a) 50
1 Ans. (a)
Given two vectors lie in xy-plane. So, a 45 If ∑ tan − 1 = p, then the value of Given that, | a | = 2, | b | = 5 and | a × b | = 8
r =1 2r 2
vector coplanar with them is ⇒ | a | | b | sin θ= ± 8
a = x $i + y $j tan p is [2021, 26 Aug. Shift-II]
⇒ 10 sinθ = ± 8
101 50 4
Since, a ⊥ ($i − $j ) (a) (b) ⇒ sinθ = ±
102 51 5
⇒ (x $i + y $j ) ⋅ ($i − $j ) = 0 51 3
(c) 100 (d) So, cosθ = ± (Qsin2 θ + cos2 θ = 1)
⇒ x − y =0 50 5
⇒ x=y Ans. (b) Now, a ⋅ b = | a | | b | cosθ
∴ a = x $i + x $j Given, ∑ tan−1  2  = p
50 1
=2 × 5 × ± 
 3
and |a | = x + x = x 2
2 2 r=1  2r   5
Now, ∑ tan−1  
2 ⇒ a⋅b = ± 6
∴ Required unit vector 
a x ($i + $j ) 1 $ $  1 + 4r 2 − 1  ∴ | a⋅b | = 6
= = = (i + j )
|a | x 2 2  (2r + 1) − (2r − 1) 
= ∑ tan−1  
 1 + (2r + 1) (2r − 1)  48 Let a and b be two non-zero
−1 −1
= ∑ [tan (2r + 1) − tan (2r − 1)]
vectors perpendicular to each
TOPIC 3 −1 −1 −1 −1
= (tan 3 − tan 1) + (tan 5 − tan 3) + ...... +
other and | a| = | b|. If | a × b| = | a|,
Vector or Cross Product of −1 −1
tan 101 − tan 99 then the angle between the vectors
−1 −1
= tan (101) − tan 1 [a + b + (a × b)] and a is equal to
Two Vectors and Its [2021, 18 March Shift-II]
101 − 1  −1  50 
Applications = tan−1   = tan    1   1 
 1 + 101   51  −1
(a) sin   (b) cos−1  
 3  3
− 1 50 50
∴ = p ⇒ tanp =
44 Let a = i$ + 5$j + αk$ , b = $i + 3$j + βk$ tan
51 51  1 
(c) cos−1   (d) sin−1  
1
 2  6
and c = − i$ + 2$j − 3k$ be three
46 Let p = 2i$ + 3$j + k$ and q = $i + 2$j + k$ Ans. (b)
vectors such that, | b × c | = 5 3 and
a is perpendicular to b. Then, the be two vectors. If a vector Given, a ⊥ b … (i)
greatest amongst the values of | a | 2 r = (α$j + β$j + γk$ ) is perpendicular to | a| = |b| … (ii)
and | a × b| = | a|
is [2021, 27 Aug. Shift-I] each of the vectors (p + q) and
⇒ | a| |b| sin90° = | a| [from Eq. (i)]
Ans. (90) (p − q), and | r| = 3, then
⇒ |b| = 1 = | a| …(iii) [from Eq. (ii)]
Given, a = $i + 5$j + αk$ |α | + |β | + | γ | is equal to …… .
[2021, 25 July Shift-I] From Eq. (iii), we can say that
b = $i + 3$j + βk$ a × b are mutually perpendicular unit
Ans. (3)
and c = − $i + 2$j − 3k$ p = (2, 3,1), q = (1, 2,1)
vectors.
Let a = $i and b = $j
Q a ⊥ b ⇒a ⋅ b = 0 r is perpendicular top + q and p − q.
⇒ ($i + 5$j + αk$) ⋅ ($i + 3$j + βk$) = 0 p + q = (3, 5, 2) ∴ a × b = k$

⇒ 1 + 15 + αβ = 0 p − q = (1, 1, 0) Now, [a + b + (a × b)] = ($i + $j + k$)

or αβ = − 16 ...(i) r is parallel to (p + q) × (p − q) ($i + $j + k$) ⋅ $i 1
i j k ∴ cosθ = =
$i $j k$ 3 1 3
⇒ 3 5 2 = (−2, 2, − 2)
Now, b × c = 1 3 β  1 
1 1 0 ∴ θ = cos−1  
− 1 2 −3  3
⇒ r = λ (−1, 1, −1)
= $i (− 9 − 2β) − $j (− 3 + β) + k$(2 + 3)
⇒ |r | = 3 49 Let a = 2i$ − 3$j + 4k$ and
b × c = $i (− 9 − 2β) + $j (3 − β) + 5k$
⇒ λ 1+ 1+ 1 = 3 b = 7 i$ + $j − 6k$ . If
Given, b× c = 5 3
⇒ λ=1 $ $ $
2 r × a = r × b,r. (i + 2 j + k) = − 3, then.
⇒ b× c = 75 ∴ r = (−1, 1, −1)
⇒ (− 9 − 2β)2 + (3 − β)2 + 25 = 75 α = −1, β = 1 and γ = − 1 r (2i$ − 3$j + k$ ) is equal to
⇒ β 2 + 6β + 8 = 0 ⇒ β = − 2, − 4 ∴ |α | + |β | + | γ | = 3 [2021, 17 March Shift-I]
(a) 12 (b) 8
From Eq. (i), we get 47 If | a| = 2,| b| = 5 and | a × b| = 8, then
For β = − 2, α = 8 (c) 13 (d) 10
| a. b| is equal to Ans. (a)
For β = − 4, α = 4 [2021, 25 July Shift-II]
a = 2$i − 3$j + 4k$
2
For maximum value of a , α = 8
(a) 6 (b) 4
b = 7 $i + $j − 6k$
2
∴ α = 1 + 25 + 64 = 90 (c) 3 (d) 5
 51 Let c be a vector perpendicular to Ans. (12)
If r × a = r × b the vectors a = $i + $j − k$ and Given, a = $i + 2$j − k$, b = $i − $j, c = $i − $j − k$
⇒ r × (a − b) = 0
b = i$ + 2$j + k$ r × a = c× a
⇒ r = λ (a − b) = λ (5$i + 4$j − 10k$) ⇒ r × a − c× a = 0
Now, r ⋅ ($i + 2$j + k$) = − 3 If c ⋅ (i$ + $j + 3k$ ) = 8 then the value of ⇒ (r − c) × a = 0
⇒ λ(5$i + 4$j − 10k$) ⋅ ($i + 2$j + k$) = − 3 c ⋅ (a × b) is equal to ……… . ∴ r − c= λ a
[2021, 16 March Shift-II] and r = λa + c
⇒ λ(5 + 8 − 10) = − 3
Ans. (28) ⇒r ⋅b = λa ⋅b + c⋅b (taking dot withb)
⇒ λ=−1
$ $ $ Since, cis perpendicular to a and b. ⇒ 0 = λa ⋅b + c⋅b [Qr ⋅b = 0]
∴r = − 5 i − 4 j + 10k
So, c = λ(a × b) ⇒ λ ($i + 2$j − k$) ⋅ ($i − $j) + ($i − $j − k$) ⋅ ($i − $j) = 0
So, r ⋅ (2$i − 3$j + k$)
a = $i + $j − k$ ⇒ λ(1 − 2) + 2 = 0 ⇒ λ = 2
= (− 5$i − 4$j + 10k$) (2$i − 3$j + k$) ∴ r = 2a + c
b = $i + 2$j + k$
= − 10 + 12 + 10 = 12 ⇒ r ⋅ a = 2a ⋅ a + c⋅ a [taking dot with a]
$i $j k$
= 2| a |2 + a ⋅ c
50 Let a = i$ + 2$j − 3k$ and Now, a × b = 1 1 − 1
= 2(1 + 4 + 1) + (1 − 2 + 1)
1 2 1
b = 2i$ − 3$j + 5k$ . If r × a = b × r, ⇒ r ⋅ a = 12
= (1 + 2) $i − (1 + 1 ) $j + (2 − 1)k$
r. (α i$ + 2$j + k$ ) = 3 and 54 Let a, b and c be three unit vectors
= 3$i − 2$j + k$
r. (2$i + 5$j − ak$ ) = − 1, α ∈R, then the such that a + b + c = 0. If
⇒ c = λ (3$i − 2$j + k$) λ = a ⋅ b + b ⋅ c + c ⋅ a and
value of α + | r| 2 is equal to
[2021, 16 March Shift-II] Now, c⋅ ($i + $j + 3k$) = 8 d = a × b + b × c + c × a, then the
(a) 9 (b) 15 λ(3$i − 2$j + k$ ) ($i + $j + 3k$ ) = 8 ordered pair, (λ , d) is equal to
[2020, 7 Jan. Shift-II]
(c) 13 (d) 11 ⇒ 4λ = 8
3   3 
Ans. (b) ∴ λ =2 (a)  , 3b × c (b)  − , 3c × b
2   2 
Given, a = ($i + 2$j − 3k$) So, c = 2(3$i − 2$j + k$)
3   3 
So, c⋅ (a × b) =2(3$i − 2$j + k$) (3$i − 2$j + k$)
(c)  , 3a × c (d)  − , 3a × b
and b = (2$i − 3$j + 5k$) 2   2 
If r × a = b × r, r ⋅ (α $i + 2$j + k$) = 3 =2(9 + 4 + 1) Ans. (d)
= 28
r ⋅ (2$i + 5$j − αk$) = − 1 Given three unit vectors a, b and csuch
that a + b + c = 0 …(i)
r × a =b×r 52 Let a = $i + α$j + 3k$ and b = 3i$ − α$j + k$ . If we do dot product in Eq. (i) with a,b
⇒ (r × a) = − (r × b) If the area of the parallelogram and c, then we get relations
⇒(r × a) + (r × b) = 0 whose adjacent sides are a⋅a + a⋅ b + a⋅c = 0
⇒ r × (a + b) = 0 represented by the vectors a and b ⇒ a ⋅ b + a ⋅ c = − 1, b⋅ a + b⋅ c = − 1
⇒ r = λ(a + b) is 8 3 square units, then a ⋅ b is and c⋅ a + c⋅ b = − 1
⇒ r = λ [(1 + 2) $i + (2 − 3) $j + (− 3 + 5)k$] equal to ………… . ∴On adding above three obtained
relations,
⇒ r = λ (3$i − $j + 2k$) [2021, 25 Feb. Shift-II]
Ans. (2) we get
r ⋅ (α $i + 2$j + k$) = 3 2(a ⋅ b + b⋅ c + c⋅ a) = − 3 (Qa ⋅ b = b⋅ a)
Area of parallelogram = |a × b|
⇒ λ (3$i − $j + 2k$) (α $i + 2$j + k$) = 3 ⇒ λ = a ⋅ b + b⋅ c + c⋅ a = −
3
= | ($i + α$j + 3k$) × (3$i −α$j + k$)| 2
⇒ λ (3α − 2 + 2) = 3
(64)(3) = 16α 2 + 64 + 16α 2 If we do cross product in Eq. (i) with a, b
⇒ αλ = 1
(given, area = 8 3) and c, then we get relation
r ⋅ (2$i + 5$j − αk$) = − 1 (squaring on both sides) 0 + a × b+ a × c= 0
⇒λ (3$i − $j + 2k$) (2$i + 5$j − αk$) = − 1 ⇒ α2 = 4 ⇒ a × b+ a × c= 0
⇒ λ (6 − 5 − 2α) = − 1 Now, a. b = 3 −α 2 + 3 and b× a + b× c= 0
⇒ λ (1 − 2α) = − 1 = 6 −α 2 = 6 − 4 = 2 and c× a + c× b=0
⇒ λ − 2αλ = − 1 ∴a × b + b × c + c × a = 3(a × b)
⇒ λ −2= − 1 ⇒ λ = 1 53 Let a = $i + 2$j − k$ , b = $i − $j and = 3( b × c) = 3(c × a)
So, α=1 c = i$ − $j − k$ be three given vectors. ∴ d = 3a × b = 3b × c = 3c × a
r = (3$i − $j + 2k$) If r is a vector such that r × a = c × a From the given options the ordered pair,
⇒ | r |2 = 9 + 1 + 4 = 14 and r ⋅ b = 0, then r ⋅ a is equal to  3 
(λ, d) =  − , 3a × b
∴ α + | r |2 = 1 + 14 = 15 ……… . [2021, 25 Feb. Shift-I]  2 
 r r
55 Let a = i$ − 2$j + k$ and b = i$ − $j + k$ be $ If
57 →Let→α = 3→$i + $jand β→= 2$i − $j + 3k. Ans. (d)
r Let point P whose position vector is
two vectors. If c is a vector such that β = β 1 − β 2 , where β 1 is parallel to α (− i$ + 2$j + 6k
$ ) and a straight line passing
→ r
b × c = b × a and c ⋅ a = 0, then c ⋅ b is and β 2 is perpendicular to α, then through Q (2, 3, − 4) parallel to the vector
→ →
equal to [2020, 8 Jan. Shift-II]
β 1 × β 2 is equal to n = 6$i + 3$j − 4k $.
1 3 1
(a) (b) − (c) − (d) − 1 [2019, 9 April Shift-I] P(–1,2,6)
2 2 2
1 $ ) (b) 1 (−3i$ + 9 $j + 5 k
Ans. (c) (a) (3$i − 9 $j + 5 k $)
2 2 d
Given vectors, a = i$ − 2 $j + k$ and (c) −3$i + 9 $j + 5 k
$ (d) 3$i − 9 $j − 5 k $)
b = i$ − $j + k$ and c, such that b × c = b × a
Ans. (b)
⇒ b × (c − a) = 0 Q(2,3,–4)
Given vectors α = 3$i + $j and
→
n=6^
i+3^
j–4^
k
⇒ b || c − a →
→ $ and →
β = 2$i − $j+ 3k
→ →
β = β 1 − β 2 such that β 1
⇒ c − a = λb Q Required distanced = Projection of line
→ →
⇒ c = a + λb is parallel toα andβ 2 is perpendicular toα segment PQ perpendicular to vectorn.
⇒ c = (1 + λ) i$ − (2 + λ) $j + (1 + λ) k$ | PQ × n |
So, β = λα = λ(3$i + $j)
→ =
1 |n |
Q c⋅ a = 0 (given)
Now, β 2 = β 1 − β = λ(3$i + $j) − (2$i − $j + 3k
→ → → $)
⇒ (1 + λ) + 2(2 + λ) + (1 + λ) = 0 Now, PQ = 3$i + $j − 10 k
$ , so
$ $
= (3λ − 2) i + (λ + 1) j – 3k $
⇒ 4λ + 6 = 0 ⇒ λ = −
3 i$ $j k
$
→ →
2 Q β 2 is perpendicular toα, so β 2 ⋅α = 0 PQ × n = 3 1 − 10 = 26$i − 48$j + 3k
$
1 1 1
∴ c = − i$ − $j − k$ [since if non-zero vectors a and bare 6 3 −4
2 2 2
1 1 1 1 perpendicular to each other, then (26) 2 + (48) 2 + (3) 2
So c⋅ b = − + − = − a ⋅ b = 0] So, d=
2 2 2 2 (6) 2 + (3) 2 + (4) 2
∴ (3λ − 2)(3) + (λ + 1)(1) = 0
Hence, option (c) is correct.
⇒ 9λ − 6 + λ + 1 = 0 676 + 2304 + 9 2989
1 = =
$ and b = i$ − $j + k
$, ⇒ 10 λ = 5 ⇒ λ = 36 + 9 + 16 61
56 Let a = 3i$ + 2$j + xk 2
→ 3$ 1$ = 49 = 7 units
for some real x. Then | a × b| = r is So, β1 = i+ j
2 2 $ and
59 Let a = 3i$ + 2$j + 2k
possible if [2019, 8 April Shift-II]
→ 3  $ be two vectors. If a
b = i$ + 2$j − 2k
β 2 =  − 2 $i +  + 1 $j − 3k
3 3 3 1 $
(a) 0 < r ≤ <r≤3 (b) and
2 2 2 2  2 
vector perpendicular to both the
1 3
(c) 3
3
< r <5
3
(d) r ≥ 5
3 = − $i + $j − 3k
$ vectorsa + b anda − b has the
2 2 2 2 2
magnitude 12, then one such vector is
Ans. (d) $i $j [2019, 12 April Shift-I]
$
Given vectors are a = 3i$ + 2$j + xk
$
→ →
k
(a) 4 (2 $i + 2 $j + k
$) (b) 4 (2 $i − 2 $j − k
$)
3 1
$ $
b= i − j+ k$ ∴ β 1× β 2 = 0
and 2 2 (c) 4 (2 $i + 2 $j − k
$) (d) 4 (− 2 $i − 2 $j + k$)
−1 −3
$i $j k $ 3
 
Ans. (b)
∴ a × b= 3 2 x 2 2
  Given vectors are
 1 −1 1   3 
− $j − − 0 
$ 9
 
= i  − − 0 a = 3i$ + 2$j + 2k
$
 2   2 
$ $
= i (2 + x) − j(3 − x) + k$ (−3 − 2) and b = i$ + 2$j − 2k
$
$ 9 + 1
+k   Now, vectors a + b = 4i$ + 4$j
$ $
= (x + 2) i + (x − 3) j − 5k$  4 4
3 9 5$ a − b = 2$i + 4k
= − $i + $j + k and $
⇒ | a × b | = (x + 2) 2 + (x − 3) 2 + 25 2 2 2
∴A vector which is perpendicular to both
1
= 2x 2 − 2x + 4 + 9 + 25 = (−3i$ + 9$j + 5k$) the vectors a + b and a − b is
2 $i $j k$

= 2 x 2 − x +
1 1 (a + b) × (a − b) = 4 4 0
 − + 38 58 The distance of the point having
 4 2 $ from
2
position vector − $i + 2$j + 6k 2 0 4
= 2 x −  +
1 75
the straight line passing through = $i (16) − $j(16) + k
$ (− 8)
 2 2
the point (2, 3, − 4) and parallel to the = 8(2$i − 2$j − k$)
75 1 $ is
vector, 6$i + 3$j − 4k
So, | a × b | ≥ [at x = , | a × b | is Then, the required vector along
2 2 [2019, 10 April Shift-II] (a + b) × (a − b) having magnitude 12 is
minimum] 8(2$i − 2$j − k$)
(a) 2 13 (b) 4 3 ± 12 ×
⇒ r≥ 5
3 8× 4+ 4+ 1
(c) 6 (d) 7
2 = ± 4(2$i − 2$j − k $)
 a⋅ c  b⋅ c 
60 Let a = 2$i + $j − 2k, (a) c +  (b) b + 
$ b = i$ + $j and c be Ans. (d)
b c
 a⋅ b  a⋅ b Given that, u = $i + $j , v = $i − $j ,
a vector such that | c − a | = 3,
a⋅ c  b⋅ c 
| (a × b) × c | = 3 and the angle (c) c −  b (d) b −  c w = $i + 2 $j + 3k$ ,
 a⋅ b  a⋅ b
between c and a × b is 30°. Then, u⋅n = 0 and v ⋅n = 0
a ⋅ c is equal to [JEE Main 2017] Ans. (c) u× v
i. e., n=
(a)
25
(b) 2 (c) 5 (d)
1 Given, a ⋅b ≠ 0, a ⋅ d = 0 …(i) |u × v |
8 8 and b ×c =b ×d
⇒ b × (c − d) = 0 $i $j k$
Ans. (b)
∴ b | | (c − d) Now, u× v = 1 1 0
We have, a = 2i$ + $j − 2k
$
⇒ c − d = λb
⇒ d = c − λb …(ii) 1 −1 0
⇒ |a | = 4 + 1 + 4 = 3
Taking dot product witha, we get
b = i$ + $j ⇒ | b | = 1 + 1 = 2 = 0 $i − 0 $j − 2k$ = − 2k$
and a ⋅ d = a ⋅ c − λa ⋅ b
Now, | c − a | = 3 ⇒| c − a |2 = 9 ⇒ 0 = a ⋅ c − λ (a ⋅ b) | w ⋅ u × v | | − 6k |
a ⋅c ∴ | w ⋅ n| = = =3
⇒ (c − a) ⋅ (c − a) = 9 ∴ λ= …(iii) |u × v | | − 2k |
a ⋅b
⇒ | c |2 + | a |2 − 2 c⋅ a = 9 …(i) (a ⋅c) [Qw ⋅ (u × w) = ($i + 2$j + 3k$ ) ⋅ (− 2k$ ) = − 6k$ ]
Again, | (a × b) × c | = 3 ∴ d =c − b
(a ⋅b) Hence, | w ⋅n | = 3
6
⇒ | a × b | | c | sin30 ° = 3 ⇒ | c | =
|a × b | 63 If u and v are unit vectors and θ is 66 A tetrahedron has vertices at
$i $j k
$ the acute angle between them, O(0, 0, 0), A(1, 2, 1), B (2, 1, 3) and C(−1, 1, 2).
But a × b = 2 1 − 2 = 2$i − 2$j + k
$ then 2 u × 3 v is a unit vector for Then, the angle between the faces
[AIEEE 2007]
1 1 0 OAB and ABC will be [AIEEE 2003]
(a) exactly two values of θ
6
∴ | c| = =2 …(ii) (b) more than two values of θ  19   17 
4+ 4+ 1 (a) cos−1   (b) cos−1  
(c) no value of θ  35   31
From Eqs. (i) and (ii), we get (d) exactly one value of θ (c) 30° (d) 90°
(2) 2 + (3) 2 − 2c⋅ a = 9 ⇒ 4 + 9 − 2c⋅ a = 9 Ans. (d) Ans. (a)
⇒ c⋅ a = 2 Since, (2u × 3v) is a unit vector.
Vector perpendicular to face OAB is n1 .
⇒ |2u × 3v | = 1
1
61 If a = (3i$ + k$ ) and ⇒ 6 |u | | v | | sin θ | = 1 Y
10 1
1 ⇒ sinθ = [Q|u | = | v | = 1] B(2, 1, 3)
b = (2i$ + 3$j − 6k$ ), then value of 6
7 Since, θ is an acute angle, then there is
(2 − b) ⋅ [(a × b) × (a + 2b)] is exactly one value of θ for which (2u × 3v) is
(a) −3 (b) 5 [AIEEE 2011] a unit vector.
X
(c) 3 (d) −5 64 For any vector a, the value of O A (1, 2, 1)
Ans. (d) (a × $i) 2 + (a × $j) 2 + (a × k$ ) 2 is equal to
1
a= (3$i + k$ ) [AIEEE 2005] C (–1, 1, 2)
10 (a) 4a 2 (b) 2a 2 (c) a 2 (d) 3a 2 Z
1
and b = (2$i + 3$j − 6 k$ ) Ans. (b) $i $j k$
7
Let a = a 1 $i + a 2 $j + a 3 k$
∴ (2a − b) ⋅ {(a × b) × (a + 2b)} = OA × OB = 1 2 1
= (2a − b) ⋅ {(a × b) × a + (a × b) × 2b} Then, a × $i = − a 2 k$ + a 3 $j 2 1 3
= (2a − b) ⋅ {(a ⋅ a ) b − (b ⋅ a ) a
a × $j = a 1 k$ − a 3 $i = 5 $i − $j − 3k$
+ 2 (a ⋅ b) b − 2 (b ⋅ b) a }
= (2a − b) ⋅ { 1 (b) − (0) a + 2 (0) b − 2 (1) a } a × k = − a1 j + a2 i Vector perpendicular to face ABC is n2
[as a ⋅ b = 0 and a ⋅ a = b ⋅ b = 1] ∴ (a × i) 2 + (a × j) 2 + (a × k ) 2 $i $j k$
= (2a − b) (b − 2a ) = a 22 + a 23 + a 21 + a 23 + a 21 + a 22
= AB × AC = 1 −1 2
= − (4 |a |2 − 4 a ⋅b + |b |2 ) = − {4 − 0 + 1 } = 2 (a 21 + a 22 + a 23 ) = 2a 2
=−5 −2 −1 1

62 The vectors a and b are not 65 Let u = i$ + $j , v = i$ − $j and = $i − 5 $j − 3k$

perpendicular and c and d are two w = i$ + 2$j + 3k$ . If n is a unit vector Since, angle between faces is equal to the
vectors satisfying b × c = b × d and such that u ⋅ n = 0 and v ⋅ n = 0, then
angle between their normals.
n ⋅n
a ⋅ b = 0. Then, the vector d is equal | w ⋅ n | is equal to [AIEEE 2003] ∴ cos θ = 1 2
|n1 | |n2 |
to [AIEEE 2011] (a) 0 (b) 1 (c) 2 (d) 3
 5 × 1 + (−1) × (−5) + (−3) × (−3) Ans. (c) Vector product
=
52 + (−1) 2 + (−3) 2 12 + (−5) 2 + (−3) 2 a × [(r − b) × a] + b × [(r − c) × b] + c (a + b) × [(a × (a − b) × b) × b]
5+ 5+9 19 × [(r − a) × c] = 0 = (a + b) × [(a × (a × b − b × b) × b]
= = ⇒a ⋅ a (r − b) − (a ⋅ (r − b)) a + b ⋅ b (r − c)
35 35 35 = (a + b) × [(a × (a × b)) × b]
− (b ⋅ (r − c)) b + c ⋅ c (r − a) − (c ⋅ (r − a)) c = 0
[Qb × b = 0]
θ = cos−1  
19
⇒ ⇒ | a |2 (r − b) − (r ⋅ a) a + | b |2 (r − c) − (r ⋅ b) b
 35  = (a + b) × [[(a ⋅ b) a − (a ⋅ a) b] × b]
+ | c |2 (r − a) − (r ⋅ c) c = 0
= (a + b) × [(a ⋅ b) (a × b)]
67 If the vectors a, b and c form the [Qa, b, c are mutually perpendicular;
= (a ⋅ b) (a + b) × (a × b)
sides BC, CA and AB respectively of ∴ a ⋅ b = b ⋅ c = c ⋅ a = 0]
= (a ⋅ b) [[(a + b) ⋅ b] a − [(a + b) ⋅ a] b]
a ∆ABC, then ⇒| a |2 [3r − (a + b + c)] − [(r ⋅ a) a + (r ⋅ b) b
a ⋅ b = ($i + $j + 2k$) ⋅ (− $i + 2$j + 3k$)
(a) a ⋅ b = b⋅ c = c ⋅ b = 0 [AIEEE 2002] + (r ⋅ c) c] = 0
= − 1+ 2+ 6= 7 ...(i)
(b) a × b = b× c = c× a [Q | a | = | b | = | c |]
(a + b) = 0 $i + 3$j + 5k$ ...(ii)
(c) a ⋅ b = b⋅ c = c ⋅ a = 0 ⇒ | a |2 [3r − (a + b + c) − r] = 0
∴ 3r − (a + b + c) − r = 0 (a + b) ⋅ a = (3$j + 5k$) ⋅ ($i + $j + 2k$) = 13 ...(iii)
(d) a ×a + a ×c+ a ×a =0
a + b+ c
Ans. (b) ⇒ r= (a + b) ⋅ b = (3$j + 5k$) ⋅ (− $i + 2$j + 3k$) = 21 .(iv)
2
Since, a +b+c=0 Substituting the values from Eqs. (i), (ii),
(iii) and (iv), we get
⇒ a +b = −c 70 Let a = $i + $j + k$ and b = $j − k$ . If c is a
⇒ (a + b) × c = − c × c (a ⋅b) [(a + b) ⋅b] a − [(a + b)⋅ ab]
vector such that a × c = b and = 7 [21a − 13b]
⇒ b ×c =c×a
a ⋅ c = 3, then a ⋅ (b × c) is equal to = 7 [21($i + $j + 2k$) − 13(− $i + 2$j + 3k$)]
Similarly, a ×b = b × c [2021, 26 Aug. Shift-I]
Hence, a ×b = b × c = c × a (a) −2 (b) − 6 = 7 (34$i − 5$j + 3k$)
(c) 6 (d) 2
68 If the vectors c, a = x i$ + y $j + z k$ and Ans. (a) 72 Let a = i$ + $j + k$ , b and c = $j − k$ be
b = $j are such that a, c and b form a Given, a × c = b three vectors such that a × b = c
right handed system, then c is a × (a × c) = a × b and a ⋅ b = 1. If the length of
[AIEEE 2002] (a. c) a − (a. a) c = a × b projection vector of the vector b on
(a) z $i − x k$ (b) 0 We have, a = (1, 1, 1), b = (0, 1, − 1), a ⋅ c = 3 the vector a × c is l, then the value
(c) y $j (d) − z $i + x k$ $i $j k$ of 3I 2 is equal to …………
[2021, 27 July Shift-I]
Ans. (a) a × b = 1 1 1 = −2$i + $j + k$
Ans. (2)
Since, the vectorsa = x $i + y $j + z k$ and 0 1 −1
Given, a = $i + $j + k$,
b = $j are such thata , c and b form a right So, 3a − 3c = (−2$i + $j + k$) c = $j − k$
handed system.
$i $j k$ ⇒ (3 $i + 3$j + 3k$) − 3 c = (−2$i + $j + k)
$
Given, a × b = c and a ⋅ b = 1
⇒ 3c = (5$i + 2$j + 2k$) Projection of bon a × c is
∴ c = b × a = 0 1 0 = z $i − x k$
Now, a ⋅ (b × c) l = |b | cos α
x y z
1 1 1
=   0 1 −1 = (4 − 5 − 5) = −2
1 1 b
TOPIC 4  3
5 2 2
3
α
Scalar and Vector a×c
and b ⋅ (a × c) = | b | | a × c | cosα
Triple Product 71 Let a = i$ + $j + 2k$ and (a × c)
∴ l = b⋅
b = − i$ + 2$j + 3k$ . Then the vector | a × c|
69 Let a, b and c be three vectors product As, we know that,
mutually perpendicular to each (a + b) × [(a × {(a − b) × b)} × b] is [a b c] = [b a c] = [c a b]
other and have same magnitude. If equal to [2021, 27 July Shift-I] So, if a ×b= c
a vector r satisfies. (a) 5(34$i − 5$j + 3k$ ) Then, c ⋅ (a × b) = c ⋅ c
a × {(r − b) × a} + b × {(r − c) × b} (b) 7 (34$i − 5$j + 3k$ ) [c a b] = | c |2
+ c × {(r − a) × c} = 0 , then r is equal (c) 7 (30 $i − 5$j + 7 k$ ) l=
[b a c] [c a b]
= =
| c |2
to [2021, 31 Aug. Shift-II] | a × c| | a × c| | a × c|
(d) 5(30 i$ − 5$j + 7 k$ )
1 1 $i $j k$
(a) (a + b + c) (b) (2a + b − c) Ans. (b)
3 3 a × c= 1 1 1 = − 2$i + $j + k$
1 1 Given, a = $i + $j + 2k$
(c) (a + b + c) (d) (a + b + 2c) 0 1 −1
2 2 b = − $i + 2$j + 3k$
 ∴ | a × c | = 22 + 12 + 12 = 6 α and β are integers, so possible values Ans. (d)
of α and β are a ×b = c
∴ c = $j − k$
If α = 1 ⇒β = 2 b× c = a
| c | = 12 + 12 = 2 If α = 2 ⇒β = 1 and | a| = 2
( 2) 2 2 If α = − 1 ⇒β = − 2 a × [(b + c) × (b − c)] = 0
∴ l= = (a)
6 6 If α = − 2 ⇒β = − 1 ⇒ a × [b × (b − c) + c × (b − c)] = 0
and
4
3l 2 = 3 ⋅ = 2 Now, b⋅ c = 10 ⇒ a × [−b × c + c × b] = 0
6 ⇒ − 3α − 2β − α = 10 ⇒ a × (− a − a) = 0 (True)
⇒ 2α + β + 5 = 0 (b) Projection of a on b × c is 2.
73 Let a,b and c be three vectors such ∴Value of α = − 2 and β = − 1 satisfy this (b × c) a. a | a|2
a. = = =2
that a = b × (b × c). If magnitudes of equation.
|b × c| | a| | a|
the vectors a, b and c are 2, 1 and So, a = $i + 2$j − k$ ⇒ b = 3$i − $j + 2k$ (c) [abc] + [cab] = 8
2, respectively and the angle c = 2$i − 2$j + k$ [abc] + [abc] = 8
π
between b and c is θ 0 < θ <  , then 1 2 −1 ⇒ [abc] = 4 ⇒ a .(b × c) = 4
 2 ⇒ a. a = 4 ⇒ | a|2 = 4 (True)
[a b c] = 3 −1 2
the value of (1 + tanθ) is equal to 2 −2 1 (d) |3a + b − 2c| = 51
2

[2021, 27 July Shift-II] (3a + b − 2c) ⋅ (3a + b − 2c)

(a) 3 + 1 (b) 2 = 1 (−1 + 4) − 2(3 − 4) − 1(−6 + 2)
= 9| a|2 + |b|2 + 4| c|2
3+1 =3+ 2+ 4=9
(c) 1 (d) = (9 × 4) + 1 + (4 × 4)
3 = 36 + 1 + 16 = 53 (False)
75 Let the vectors
Ans. (b) So, (d) is the correct option.
(2 + a + b) i$ + (a + 2b + c) $j − (b + c)k,
$
Given that,
a = b × (b × c) (1 + b) i$ + 2b$j − bk$ and 77 Let a vector a be coplanar with
and | a | = 2, |b | = 1, | c | = 2 (2 + b) $i + 2b$j + (1 − b) k,
$ a, b, c ∈R be vectors b = 2$i + $j + k$ and
Now, a = b × (b × c) coplanar. [2021, 25 July Shift-I] c = $i − $j + k.
$ If a is perpendicular to
= (b ⋅ c) b − (b⋅ b) c Then, which of the following is d = 3i$ + 2$j + 6k$ and | a| = 10. Then a
= (1⋅2 cosθ) b − (1) c true? possible value of
[a ⋅b = | a | |b | cosθ, a ⋅ a = | a −1 |2 ] (a) 2b = a + c (b) 3 c = a + b [a b c] + [a b d] + [a c d] is equal to
⇒ a = 2 cosθb − c (c) a = b + 2 c (d) 2a = b + c [2021, 22 July Shift-II]
| a |2 = (2 cosθ) 2 + 22 − 2⋅2 cosθ (b⋅ c) Ans. (a) (a) −42 (b) −40
⇒ 2 = 4 cos2 θ + 4 − 4 cosθ (2 cosθ) Given vectors are coplanar (c) −29 (d) −38
⇒ − 2 = − 4 cos2 θ a + b + 2 a + 2b + c −b − c Ans. (a)
⇒
1
cos2 θ = ⇒ sec2 θ = 2 ∴ b+1 2b −b = 0 a = λb + µc b = (2, 1,1)
2 b+2 2b 1 −b a ⊥ d  Given, c = (1, − 1,1)
⇒ 1 + tan2 θ = 2 ⇒ tan2 θ = 1 Apply, R 3 → R 3 − R 2 , R 1 → R 1 − R 2 |a| = 10  d = (3, 2, 6)
π  π
⇒ θ= where,0 < θ <  So, a ⋅d = 0
4  2 a + 1 a + c −c
⇒ (λb + µc) ⋅d = 0 ⇒ λ.b⋅ d + µc.d = 0
Hence, the value of 1 + tanθ = 1 + 1 = 2. b + 1 2b −b = 0
⇒λ (6 + 2 + 6) + µ (3 − 2 + 6) = 0
1 0 1 ⇒ 14λ + 7µ = 0
74 Let a = i$ − α $j + βk$ , b = 3i$ + β$j − αk$ = (a + 1)2b − (a + c)(2b + 1) − c (−2b) ⇒ µ = − 2λ
and c = −α i$ − 2$j + k$ , where α and β = 2ab + 2b − 2ab − a − 2bc − c + 2bc a = (2λ, λ, λ) + (µ ,−µ, µ )
are integers. If a ⋅ b = − 1 and ⇒ 2b − a − c = 0 = (2λ + µ ) $i + (λ − µ ) $j + (λ + µ )k$
b ⋅ c = 10, then (a × b) ⋅ c is equal to = 0 $i + 3λ$j − λk$ = λ (3j$ − k$)
……… . [2021, 27 July Shift-II]
76 Let three vectors a, b and c be
such that a × b = c,b × c = a and Now, | a| = 10
Ans. (9) ⇒ λ 2 (32 + 12 ) = 10
| a| = 2. Then, which one of the
Let a = $i − α$j + βk$ following is not true ? λ2 = 1 ⇒ λ = ± 1
b = 3$i + β$j − α k$ [2021, 22 July Shift-II]
[abc] + [abd] + [acd] = [abc] [a, b + c, d]
(a) a × ((b + c) × (b − c)) = 0 a = ±1(3$j − k$) 
and c = −α $i − 2$j + k$ 
(b) Projection of a on (b × c) is 2 b = 2$i + $j + k$ , b + c = 3$i + 2k$
Given that,
(c) [a b c] + [c a b] = 8 c = $i − $j + `k$ 
a ⋅b = − 1 
⇒ 3 −αβ −αβ = − 1 ⇒ αβ = 2 (d) | 3a + b − 2c |2 = 51
d = 3$i + 2$j + 6k$
 As, a is coplanar withb and c. So, − β + 2α + 1 = − 3 ⇒14x 2 + 6 = 20 ⇒14x 2 = 14
[abc] = 0 β − 2α = 4 ⇒x 2 = 1 ⇒x = ± 1 but x must be positive
0 3 −1  − 2 as in question conditions i.e. x > 0.
⇒ β − 2  =4
± 3 0 2 = ± (−36 − 6) = ± 42  β  ∴ x=−1 (Rejected)
3 2 6 ⇒ β 2 + 4 = 4β Hence, x = 1
⇒ β − 4β + 4 = 0
2
∴ y = 2x = 2 × 1 = 2
78 Let a = 2$i + $j − 2k$ and b = $i + $j. If c is ⇒ (β − 2) 2 = 0 Now, OP, OQ and OR are coplanar.
⇒ β =2 ∴ [OP OQ OR] = 0
a vector such that αβ = −2 ⇒α⋅2 = − 2 x y −1
a ⋅ c = | c|,| c − a| = 2 2 and the angle ⇒ α=−1 ⇒ − 1 2 3x = 0
π 1
between (a × b) and c is , then the [(a × b) ⋅ c] 3 z −7
6 3
value of | (a × b) × c| is a = < − 1, 2, 3 > 1 2 −1
[2021, 20 July Shift-I] b = < − 2, 1, − 1 > ⇒ −1 2 3 =0
(a)
2
(b) 4 (c) 3 (d) 3/2 c = < 1, − 2, − 1 > 3 z −7
3 $i $j k$ ⇒ 1 (− 14 − 3z) − 2(7 − 9) − 1(− z − 6) = 0
Ans. (d) (a × b) = − 1 2 3 ⇒ z = −2
a = 2$i + $j − 2k$ −2 1 − 1 ∴ x2 + y2 + z2 = 1 + 4 + 4 = 9
⇒ |a |= 4+ 1+ 4 =3
b = $i + $j = − 5$i − 7 $j + 3k$ 81 If (1, 5, 35), (7, 5, 5), (1, λ, 7) and
a⋅ c= | c| (a × b) ⋅ c = (− 5$i − 7 $j + 3k$) ($i − 2$j − k$) (2λ , 1, 2) are coplanar, then the sum
| c− a | = 2 2 = − 5 + 14 − 3 = 6 of all possible values of λ is
1 1 [2021, 26 Feb. Shift-I]
⇒ | c − a |2 = 8 ∴ [(a × b) − c] = × 6 = 2 39 39 44 44
⇒ (c − a) ⋅ (c − a) = 8 3 3 (a) (b) − (c) (d) −
5 5 5 5
⇒ | c |2 − 2a ⋅ c + | a |2 = 8
80 Let O be the origin. Let Ans. (c)
⇒ | c |2 − 2| c | + 9 = 8
⇒ | c |2 − 2 | c | + 1 = 0
OP = x $i + y$j − k$ and Let P(1, 5, 35), Q (7, 5, 5), R (1, λ, 7), S (2λ, 1, 2)

⇒ (| c | − 1) 2 = 0 OQ = − $i + 2$j + 3xk$ , x, y ∈R, x > 0, be Given P, Q , R, S are coplanar. Then,PQ,

PR, PSlie on the same plane.
⇒ | c| = 1 such that | PQ | = 20 and the vector
π PQ = (7 − 1) $i + (5 − 5) $j + (5 − 35)k$
| (a × b) × c | = | a × b | | c | sin   OP is perpendicular to OQ. If
6 = 6$i − 30k$
OR = 3i$ + z$j − 7k$ , z ∈R, is coplanar
1
= (| a × b | | c |)
with OP and OQ, then the value of PR = (1 − 1) $i + (λ − 5) $j + (7 − 35)k$
2
$i $j k$ x 2 + y 2 + z 2 is equal to = (λ − 5) $j − 28k$
[2021, 17 March Shift-II]
a × b = 2 1 − 2 = 2$i − 2$j + k$ PS = (2λ − 1) $i + (1 − 5) $j + (2 − 35) k$
(a) 7 (b) 9
1 1 0 (c) 2 (d) 1 = (2λ − 1) $i − 4$j − 33k$
|a ×b|= 4+ 4+ 1 =3 Ans. (b) QPQ,PR and PS lie on same plane, then
3
∴ | (a × b) × c | = | c | =
3 Given, OP = x $i + y$j − k$ 6 0 − 30
2 2 OQ = − $i + 2$j + 3xk$ 0 λ − 5 − 28 = 0
2λ − 1 − 4 − 33
OR = 3$i + z$j − 7k$
79 If a = α $i + β$j + 3k$ , b = − β $i − α$j − $j
and | PQ | = 20 Expand along first row,
and c = $i − 2$j − k$ ,such that a. b = 1 6[− 33λ( − 5) − 112] + 30 [2λ − 1)(λ − 5] = 0
Now, | PQ | = | OQ − OP | = | OP − OQ |
1 6(− 33λ + 53) + 30 (2λ 2 − 11λ + 5) = 0
and b. c = − 3, then [(a × b) ⋅ c] is = (x + 1) $i + (y − 2) $j − (1 + 3x)k$
3 60 λ2 − 528λ + 468 = 0
⇒ | PQ |2 = ( 20 ) 2 = 20
equal to ……… . 10 λ2 − 88λ + 78 = 0
⇒ (x + 1) 2 + (y − 2) 2 + (1 + 3x) 2 = 20
[2021, 17 March Shift-I] 5λ − 44λ + 39 = 0
2
… (i)
⇒ (x + 1) 2 + (2x − 2) 2 + (1 + 3x) 2 = 20
Ans. (2) Possible value of λ are roots of Eq. (i).
 QOP ⊥ OQ 
a = < α, β,3 >  ∴OP⋅ OQ = 0  Then, sum of all possible values of λ
b = < −β , −α , − 1 >   = Sum of roots of Eq. (i)
 ⇒ − x + 2y − 3x = 0  − (−44) 44
c = < 1, − 2, − 1 >  ⇒ y = 2x  = =
a ⋅b = 1   5 5
−αβ −αβ − 3 = 1 ⇒ x + 1 + 2x + 4x + 4 − 8x + 1 + 9x + 6x
2 2 2
[Qax + bx + c = 0, sum of roots
2

αβ = − 2 = 20 = − b / a}
b⋅ c = − 3
82 If a and b are perpendicular, then 84 Let x 0 be the point of local maxima Ans. (b)
a × (a × (a × (a × b))) is equal to off (x) = a ⋅ (b × c), where and As we know, the volume of
[2021, 26 Feb. Shift-I] $ . Then the value of parallelopiped, where coterminus edges
1 c = 7 i$ − 2$j + xk are given by vectors
(a) 0 (c) | a |4 b
2 a ⋅ b + b⋅ c + c⋅ a at x = x 0 is a = i$ + $j + nk
$ , b = 2$i + 4$j − nk
$
(c) a × b [2020, 4 Sep. Shift-I]
(d) | a |4 b and c = i$ + n$j + 3k$ , (n≥ 0), is
Ans. (d) (a) 14 (b) −4 (c) − 22 (d) − 30
1 1 n 
Ans. (c) 2 4 − n= 158 [given]
a × [a × { a × (a × a)}]
Given vectors a = x$i − 2$j + 3k$  
= a × (a × [(a ⋅b) a − (a ⋅ a)b]) 1 n 3 
[Using, a × (b × c) = (a ⋅ cb) − (a ⋅b) c] b = − 2$i + x$j − k$
⇒1(12 + n2 ) − 1(6 + n) + n(2n − 4) = 158
= a × [a × ((a ⋅b) a − | a |2 b)] and c = 7 $i − 2$j + xk$
⇒ 3n2 − 5n + 6 = 158
= a × [(a × (a ⋅b) a) − | a |2 (a × b)] And, it is given that
 x −2 3  ⇒ 3n − 5n − 152 = 0
2

= a × [0 − | a |2 (a × b)] [Qa × a = 0] ⇒ 3n2 − 24n + 19n − 152 = 0

f (x) = a ⋅ (b × c) =−2 x −1
= − | a |2 [a × (a × b)]   ⇒ (3n + 19)(n − 8) = 0
= − | a |2 [(a ⋅b) a − (a ⋅ a)b]  7 −2 x 
⇒n = 8 as n≥ 0
= x (x 2 − 2) + 2(−2x + 7) + 3(4 − 7 x)
= − (a ⋅b) a | a |2 + | a |4 b [Q(a ⋅ a) = | a |2 ] ∴ $ , b = 2$i + 4$j − 8k
a = $i + $j + 8k $
= x 3 − 27 x + 26
= 0 + | a |4 b [Qa ⋅b = 0] and c = i$ + 8$j + 3k
$
It is also given thatf (x) has local maxima
= | a |4 b at x = x 0 . ∴ a ⋅ c = 1 + 8 + 24 = 33
So, f ′ (x 0 ) = 0 ⇒3x 02 − 27 = 0 ⇒ x 02 = 9 and b⋅ c = 2 + 32 − 24 = 10
83 Let three vectors a, b and c be
⇒x 0 = ± 3, but maximum at x 0 = − 3 .
such that c is coplanar with a and b, 87 Let the volume of a parallelopiped
Now, a ⋅ b + b ⋅ c + c⋅ a
a⋅c = 7 and b is perpendicular to c, whose conterminous edges are
= − 2x − 2x − 3 − 14 − 2x − x + 7 x + 4 + 3x
where a = − i$ + $j + k$ and b = 2i$ + k$ , $ , v = $i + $j + 3k
given by u = $i + $j + λk $
= 3x − 13
then the value of2 | a+ b+ c | 2 is ........ . ∴The value of a ⋅ b + b ⋅ c + c⋅ a at $ be 1 cu. unit. If θ
and w = 2$i + $j + k
[2021, 24 Feb. Shift-I] x = x 0 = − 3,is −22
Ans. (75) Hence, option (c) is correct. be the angle between the edges u
Given, c is co-planar with a and b and w, then cosθ can be
a⋅ c = 7 85 If a = 2i$ + $j + 2k$ , then the value of [2020, 8 Jan. Shift-I]
5 7
b ⊥ c ⇒ b⋅ c = 0 | i$ × (a × i$)| 2 + | $j × (a × $j)| 2 (a) (b)
a = − $i + $j + k$ + | k$ × (a × k$ )| 2 is equal to ……… .
3 3 6 3
7 5
b = 2$i + k$ (c) (d)
[2020, 4 Sep. Shift-II] 6 6 7
Now, c = λ[b × (a × b)] Ans. (18) Ans. (b)
[Qcis coplanar with a and b]
Given vector a =2i$ + $j + 2k $
Since, the volume of parallelopiped
= λ [(b⋅b) a − (b⋅ a) b] $ $ $ $
∴ | i × (a × i)| = | a –(a ⋅ i) i| 2
2
whose coterminous edges are
= λ [( 5) 2 (− $i + $j + k$) u = i$ + $j + λk$ , v = i$ + $j + 3k$ and
= | a –2$i| 2 = | $j + 2k
$ |2 = 5
− (− 2 + 1) (2i$ + k$)] Similarly,  1 1 λ
= λ [5(− $i + $j + k$) + 2 $i + k$] | $j × (a × $j)|2 = | a –(a ⋅ $j) $j |2 = | a – $j |2 w = 2i$ + $j + k $ is [u v w] = 1 1 3 = 1
 
= λ (− 3$i + 5$j + 6k$) = |2i$ + 2k$ |2 = 8 2 1 1 
$ × (a × k $ )|2 = | a –(a ⋅ k $ )k$ |2 cubic unit (given)
Now, c⋅ a = 7 and | k
⇒ λ(− 3$i + 5$j + 6k$) ⋅ (− $i + $j + k$) = 7 $ |2 = |2$i + $j |2 = 5 ⇒|1(1 − 3) − 1(1 − 6) + λ (1 − 2)| = 1
= | a –2k
⇒ |− 2 + 5 − λ | = 1
⇒ 3λ + 5λ + 6λ = 7 ∴ | $i × (a × $i)|2 + | $j × (a × $j)|2 + | k
$ × (a × k $ )|2
⇒ |λ − 3 | = 1 ⇒ λ − 3 = ± 1
7 1
⇒ λ= = = 5 + 8 + 5 = 18 ⇒ λ = 2, 4
14 2
86 If the volume of a parallelopiped, Since, angle between u and w is θ, so
∴2 | a + b + c |2 |2 + 1 + λ | |λ + 3 |
2 whose coterminus edges are given cosθ = =
−3  5  $,
− 1 + 2 $i +  + 1 $j + (3 + 1 + 1) k$ by the vectors a = i$ + $j + nk 1+ 1+ λ 4+ 1+ 1 λ2 + 2 6
2
=2 
 2  2 
$ and c = $i + n$j + 3k
b = 2i$ + 4$j − nk $ for λ = 2,cosθ =
5
[by putting a, b, c]
6
(n≥ 0), is 158 cu units, then
= 2 + + 25
1 49 7
4 4  [2020, 5 Sep. Shift-I] for λ = 4, cosθ =
6 3
= 25 + 50 = 75 (a) n = 9 (b) b⋅ c = 10
Hence, option (b) is correct.
∴ Required value is 75. (c) a ⋅ c = 17 (d) n = 7
88 If the vectors p= (a + 1) i$ + aj$ + ak$ , Ans. (30)
There are three vectors given a, b and c, 91 Let α ∈R and the three vectors
q = ai$ + (a + 1) $j + ak$ and such that | a | = 3,| b | = 5 and b⋅ c = 10 $ , b = 2i$ + $j − αk
a = αi$ + $j + 3k $
r = ai$ + aj$ + (a + 1) k$ (a ∈R) are π
So, | b | | c | ⋅ cos = 10
3 and c = α $i − 2$j + 3k $ . Then, the set
coplanar and
π
3( p. q) 2 − λ | r × q| 2 = 0, then the (Q it is given angle between band cis ) S = {α : a, b and care coplanar}
3
value of λ is ........ . [2019, 12 April Shift-II]
5 | c |   = 10 ⇒| c | = 4
1
[2020, 9 Jan. Shift-I] ⇒ (a) is singleton.
 2
Ans. (1) (b) is empty.
Now, as a is perpendicular to the vector
The given vectors are
b × c, so (c) contains exactly two positive
p = (a + 1) i$ + aj$ + ak$, |a|| b × c|sin π numbers.
| a × (b × c)| =
q = ai$ + (a + 1) $j + ak$  2 (d) contains exactly two numbers only
and r = ai$ + aj$ + (a + 1) k$, (a ∈ R) are π one of which is positive.
=|a||b × c|=|a|| b|| c|sin
coplanar, So, [ p q r] = 0 3 Ans. (b)
a+1 3
a a = ( 3) (5) (4) = 30 Given three vectors are
⇒ a a+1 a =0 2 a = α$i + $j + 3k $
Hence answer is 30. $ $
b = 2i + j − αk $
a a a+1
⇒(a + 1) [(a + 1) 2 − a 2 ] − a [a (a + 1) − a 2 ]
and c = α $i − 2$j + 3k$
90 If the volume of parallelopiped
+ a [a 2 − a (a + 1)] = 0 $, α 1 3
formed by the vectors i$ + λ$j + k
⇒ (a + 1) [2a + 1] − 2a [a] = 0 Clearly, [a b c] = 2 1 −α
$ and λi$ + k
$j + λk $ is minimum, then λ
⇒ 2a 2 + 3a + 1 − 2a 2 = 0 α −2 3
1 is equal to [2019, 12 April Shift-I]
⇒ a=− =α (3 − 2α ) − 1(6 +α ) +3 (− 4 −α )
2

3 1 1
(a) − (b) = − 3 α 2 − 18 = − 3 (α 2 + 6)
2 1 1$ 1 2 1
So, p = i$ − $j − k, q = − i$ + $j − k$ 3 3
QThere is no value ofα for which
3 3 3 3 3 3 (c) 3 (d) − 3 − 3(α 2 + 6) becomes zero, so
1 1 2
and r = − i$ − $j + k$ Ans. (b) α 1 3
3 3 3
2 2 = 2 1 −α [a b c] ≠ 0
 3 Key Idea Volume of parallelopiped
So, ( p⋅ q) 2 =  − − +  =   =
2 2 1 1
formed by the vectors a, b and cis α −2 3
 9 9 9  9 9
V = [ a b c]. ⇒vectors a, b and care not coplanar
i$ $j k$ for any value α ∈R.
Given vectors are $i + λ$j + k
$ , $j + λk
$ and
and r × q = −
1
−
1 2 $ $
λi + k, which forms a parallelopipe(d) So, the set S = {α : a, b and c are
3 3 3 coplanar} is empty set.
1 2 1 ∴Volume of the parallelopiped is
− − 1 λ 1 $ and c be a
3 3 3 92 Let a = i$ − $j, b = $i + $j + k
V= 0 1 λ = 1 + λ3 − λ
= i$  −  − $j  +  + k$  − − 
1 4 1 2 2 1
 9 9  9 9  9 9
vector such that a × c + b = 0 and
λ 0 1
1 1 1
a ⋅ c = 4, then | c| 2 is equal to
= − i$ − $j − k$ ⇒ V = λ3 − λ + 1 [2019, 9 Jan. Shift-I]
3 3 3 19
1 1 1 1 On differentiating w.r.t. λ, we get (a) 8 (b)
∴ | r × q |2 = + + = dV 2
9 9 9 3 = 3 λ2 − 1 17
QIt is given that dλ (c) 9 (d)
dV 2
3( p⋅ q) 2 − λ | r × q |2 = 0 For maxima or minima, =0 Ans. (b)
dλ
3  − λ   = 0
1 1
⇒ 1 We have, (a × c) + b = 0
 9  3 ⇒ λ=±
3 ⇒ a × (a × c) + a × b = 0
⇒ λ=1
d 2V (taking cross product with a on both
Hence, answer is 1. and = 6λ
dλ2 sides)
2 3 > 0 , for λ = 1 $i $j k$
89 Let a, b and c be three vectors such 
= 3 ⇒ (a ⋅c) a − (a ⋅a ) c + 1 −1 0 = 0
that | a| = 3,| b| = 5, b ⋅ c = 10 and the 1
π 2 3 < 0 , for λ = − 1 1 1
angle between b and c is . If a is  3
3 d 2V 1 (Qa × (b × c) = (a . c) b − (a ⋅b) c)
Q 2 is positive for λ = , so volume ‘V ’
perpendicular to the vector b × c, dλ 3 ⇒ 4($i − $j ) − 2c + (− $i − $j + 2k$ ) = 0
then | a × (b × c)| is equal to ...... . is minimum for λ =
1
[2020, 9 Jan. Shift-II] 3 [Qa ⋅a = ($i − $j )($i − $j ) = 1 + 1 = 2 and a ⋅c = 4]
 ⇒ 2c = 4$i − 4$j − $i − $j + 2k$ ⇒ (µ − 1) [µ (µ + 1) − 1 − 1] = 0 3 $ $
and a$ × ( b$ × c$ ) = (b + c)
⇒ (µ − 1) [µ 2 + µ − 2] = 0 2
3$i − 5$j + 2k$
⇒ c= ⇒ (µ − 1) [(µ + 2) (µ − 1)] = 0 3 $ $
2 Now, consider a$ × ( b$ × c$ ) = (b + c)
⇒ µ = 1 or − 2 2
9 + 25 + 4 19
⇒ |c |2 = = So, sum of the distinct real values of 3 3
4 2 ⇒ (a$ ⋅c$ ) b$ − (a$ ⋅b$) c$ = b$ + c$
µ = 1 − 2 = − 1. 2 2
$ , b = i$ + λ$j + 4k
93 Let a = i$ + 2$j + 4k $ On comparing, we get
95 Let a, b and c be three unit vectors,
$ be 3 3
and c = 2i$ + 4$j + (λ 2 − 1) k out of which vectors b and c are a$ ⋅b$ = − ⇒ |a$ | |b$ | cos θ = −
2 2
coplanar vectors. Then, the non-parallel. If α and β are the
3
non-zero vector a × c is angles which vector a makes with ⇒ cosθ = − [Q |a$ | = |b$ | = 1]
2
[2019, 11 Jan. Shift-I] vectors b and c respectively and π 5π
1 ⇒ cos θ = cos  π −  ⇒ θ =
(a) − 10 $i + 5 $j (b) − 10 $i − 5 $j a × (b × c) = b, then | α − β | is equal  6 6
(c) − 14 $i − 5 $j (d) − 14 $i + 5 $j 2
Ans. (a) to [2019, 12 Jan. Shift-II]
97 Let a, band c be three non-zero
(a) 30º (b) 45º vectors such that no two of them
We know that, if a, b, c are coplanar
(c) 90º (d) 60º
vectors, then [a b c] = 0 are collinear and
Ans. (a) 1
1 2 4
1 (a × b) × c = | b|| c| a. If θ is the angle
∴ 1 λ 4 =0 Given, a × ( b × c) = b 3
2 between vectors band c, then a
2 4 λ2 − 1 1
⇒ (a ⋅ c) b − (a ⋅ b) c = b value of sinθ is [JEE Main 2015]
⇒ 1 { λ (λ2 − 1) − 16} − 2((λ2 − 1) − 8) + 4 2
(4 − 2λ) = 0 2 2 − 2 2 −2 3
[Qa × ( b × c) = (a ⋅ c) b − (a ⋅ b) c] (a) (b) (c) (d)
⇒ λ3 − λ − 16 − 2λ2 + 18 + 16 − 8λ = 0 3 3 3 3
On comparing both sides, we get
⇒ λ3 − 2λ2 − 9λ + 18 = 0 1 Ans. (a)
a⋅c = …(i) 1
⇒ λ2 (λ − 2) − 9 (λ − 2) = 0 2 Given, (a × b) × c = |b | |c |a
3
⇒ (λ − 2)(λ2 − 9) = 0 and a⋅ b = 0 …(ii) 1
⇒ (λ − 2) (λ + 3) (λ − 3) = 0 ⇒ − c × (a × b) = |b| |c |a
Qa, b and care unit vectors, and angle 3
∴ λ = 2, 3 or − 3 between a and bis α and angle between 1
a and cis β, so ⇒ − (c ⋅b) ⋅a + (c ⋅a )b = |b| |c |a
If λ = 2, then $ $ 3
i j k$ | a | | c| cosβ =
1
[from Eq. (i)]  1 |b| |c | + (c ⋅b)  a = (c ⋅a ) b
a× c= 1 2 4 2  3 
1
2 4 3 ⇒ cosβ = [Q| a | = 1 = | c | ] Since, a and b are not collinear.
2 1
π c ⋅b + |b| |c | = 0 and c ⋅a = 0
= $i (6 − 16) − $j (3 − 8) + k$ (4 − 4) ⇒ β= …(iii) 3
3 1
= − 10 $i + 5$j ⇒ |c | |b| cosθ + |b| |c | = 0
Qcos π = 1  3
 3 2  |b| |c |  cosθ +  = 0
$i $j k$ 1
⇒
 3
If λ = ± 3, then a × c = 1 2 4 = 0 and | a | | b| cosα = 0 [from Eq. (ii)]
1
π ⇒ cosθ + = 0 (Q|b| ≠ 0, |c | ≠ 0)
2 4 8 ⇒ α= …(iv) 3
2 1
(because last two rows are proportional). ⇒ cosθ = −
From Eqs. (iii) and (iv), we get 3
π π π 8 2 2
94 The sum of the distinct real values |α − β | = − = = 30 ° ⇒ sinθ= =
2 3 6 3 3
of µ, for which the vectors,
$ i$ + µ$j + k
µi$ + $j + k, $ are
$ , $i + $j + µk 96 Let a, b and c be three unit vectors 98 If [a × bb × cc × a] = λ [a b c] 2 , then
3 λ is equal to [JEE Main 2014]
coplanar, is [2019, 12 Jan. Shift-I] such that a × (b × c ) = (b + c). If b
(a) 2 (b) 0 (c) 1 (d) − 1 2 (a) 0 (b) 1
is not parallel to c, then the angle (c) 2 (d) 3
Ans. (d)
between a and b is [JEE Main 2016] Ans. (b)
Given vectors,µ$i + $j + k$ , $i + µ$j + k$ , 3π π
$i + $j + µk$ will be coplanar, if (a) (b) Use the formulae
4 2 a × (b × c) = (a ⋅ c)b − (a ⋅b) c,
µ 1 1 2π 5π
(c) (d) [a b c] = [b c a ] = [c a b]
1 µ 1 =0 3 6 and [a a b] = [a b b] = [ a c c] = 0
1 1 µ Ans. (d)
Further simplify it and get the result.
⇒ µ (µ 2 − 1) − 1 (µ − 1) + 1 (1 − µ ) = 0 Given, |a$ | = |b$ | = |c$ | = 1
 Now, [a × b b × c c × a ] (a) exactly two values of (p, q). 103 Let a = $i + $j + k$ , b = i$ − $j + 2k$ and
(b) more than two but not all values of
= a × b ⋅ ((b × c) × (c × a ))
(p, q). c = x$i + (x − 2) $j − k$ . If the vector c
= a × b ⋅ ((k × c × a )) [here, k = b × c] (c) all values of (p, q). lies in the plane of a and b, then x
= a × b ⋅ [(k ⋅a ) c − (k ⋅c) a ] (d) exactly one value of (p, q).
equal to [AIEEE 2007]
= (a × b) ⋅ ((b × c ⋅a ) c − (b × c ⋅ c) a ) Ans. (d)
(a) 0 (b) 1 (c) – 4 (d) –2
= (a × b) ⋅ ([b c a ] c) − 0 Since, [3u p v p w ] − [p v w q u]
Ans. (d)
[Q[b × c ⋅c] = 0] − [2w qv qu] = 0
Since, given vectorsa , b and c are
= a × b ⋅ c [b c a ] = [a b c] [b c a ] ∴3p2 [u⋅ (v × w)] − pq [v ⋅ (w × u)] coplanar.
= [a b c] 2 {Q[a b c] = [b c a ]} − 2q 2 [w ⋅ (v × u)] = 0 1 1 1
Hence, [a × b b × c c × a ] = λ [a b c] 2 ⇒ (3p2 − pq + 2q 2 ) [u ⋅ (v × w)] = 0 ∴ 1 −1 2 = 0
⇒ [a b c] 2 = λ [a b c] 2 But [u v w] ≠ 0 x x − 2 −1
⇒ λ=1 ⇒ 3p − pq + 2q = 0
2 2
⇒1 { 1 − 2 (x − 2)} − 1 (−1 − 2 x)
∴ p=q =0 + 1 (x − 2 + x) = 0
99 If the vectors p$i + $j + k$ , i$ + q $j + k$ ⇒ 1 − 2x + 4 + 1 + 2x + 2x − 2 = 0
and $i + $j + r k$ (where, p ≠ q ≠ r ≠ 1) 102 The vector a = α$i + 2$j + βk$ lies in ⇒ 2 x = − 4 ⇒ x = −2
are coplanar, then the value of the plane of the vectors b = i$ + $j
104 If (a × b) × c = a × (b × c), where
pqr − (p + q + r) is [AIEEE 2011]
and c = $j + k$ and bisects the angle a, b and c are any three vectors
(a) − 2 (b) 2 (c) 0 (d) −1
between b and c. Then, which one such that a ⋅ b ≠ 0, b ⋅ c ≠ 0, then a
Ans. (a) and c are [AIEEE 2006]
of the following gives possible
Given, a = p $i + $j + k,
$ b = $i + q$j + k$ and π
$ $ $
values of α and β? [AIEEE 2008] (a) inclined at an angle of between
c = i + j + rk are coplanar and 6
(a) α = 1, β = 1
(b) α = 2 , β = 2 them.
p ≠ q ≠ r ≠ 1.
(c) α = 1, β = 2 (b) perpendicular.
Since,a , b and c are coplanar.
(d) α = 2 , β = 1 (c) parallel.
p 1 1 π
(d) inclined at an angle of between
⇒ [a b c] = 0 ⇒ 1 q 1 = 0 Ans. (a)
them. 3
1 1 r Given that, b = $i + $j and c = $j + k$ . Ans. (c)
⇒ p (q r − 1) − 1 (r − 1) + 1 (1 − q) = 0 The equation of bisector ofb and c is
Since, (a × b) × c = a × (b × c)
 $i + $j $j + k$  ∴ (a ⋅c) b − (b ⋅c) a = (a ⋅c) b − (a ⋅b) c
⇒ pqr − p − r + 1 + 1 − q = 0 r = λ(b + c) = λ  + 
∴ pqr − (p + q + r) = − 2  2 2  ⇒ (b ⋅c) a = (a ⋅b) c
λ $ (a ⋅b)
= ( i + 2$j + k$ ) …(i) ⇒ a= ⋅c
(b ⋅c)
100 Let a = $j − k$ and a = i$ − $j − k$ . Then, 2
Hence, a is parallel toc.
Since, vectora lies in plane ofb and c.
the vector b satisfying a × b + c = 0
∴ a = b + µc
and a⋅ b = 3, is [AIEEE 2010]
λ $
105 Let a = $i − k$ , b = xi$ + $j + (1 − x) k$ and
(a) − i$ + $j − 2 k$ (b) 2 i$ − $j + 2 k$ ⇒ ( i + 2$j + k$ ) = ($i + $j ) + µ( $j + k$ )
2 c = y i$ + x $j + (1 + x − y) k$ . Then, [a b c]
(c) $i − $j −2 k$ (d) $i + $j −2 k$ On equating the coefficient ofi both depends on [AIEEE 2005]
Ans. (a) sides, we get (a) Neither x nor y (b) Both x and y
a ×b +c =0 λ
We have, =1 ⇒ λ= 2 (c) Only x (d) Only y
⇒ a × (a × b) + a × c = 0 2
⇒ (a ⋅b) a − (a ⋅a ) b + a × c = 0 On putting λ = 2 in Eq. (i), we get Ans. (a)
⇒ 3a − 2b + a × c = 0 r = $i + 2$j + k$ Given, vectors are
⇒ 2b = 3a + a × c Since, the given vectora represents the a = $i − k$ , b = x $i + $j + (1 − x) k$
⇒ 2b = 3$j − 3k$ − 2 $i − $j − k$ same bisector equationr. and c = y i + x $j + (1 + x − y) k$
$
= − 2 $i + 2 $j − 4k$ ∴ α = 1 and β = 1 1 0 −1
∴ b = − $i + $j − 2k$ Alternate Solution ∴ [a b c] = x 1 1− x
Since,a , b and c are coplanar. y x 1+ x − y
α 2 β
101 If u, v and w are non-coplanar
⇒ 1 1 0 =0 ApplyingC 3 → C 3 + C 1 , we get
vectors and p, q are real numbers,
0 1 1 1 0 0
then the equality
= x 1 1 = 1(1 + x) − x = 1
[3u p v p w] − [p v w q u] ⇒ α (1 − 0) − 2 (1 − 0) + β (1 − 0) = 0
y x 1+ x
− [2 w q v q u] = 0 holds for ⇒ α + β = 2, which is possible for
[AIEEE 2009] α = 1, β = 1. Thus, [a b c] depends upon neither x nor y.
106 Let a , b and c be distinct λ (a 1 + b 1 ) λ (a 2 + b 2 ) λ (a 3 + b 3 ) vectors b and c, then sinθ is equal
∴ λ2 b 1 λ2 b 2 λ2 b 3 to [AIEEE 2004]
non-negative numbers. If the
λc 1 λc 2 λc 3
vectors a $i + a $j + c k$ , i + k and (a)
1
(b)
2
(c)
2
(d)
2 2
a1 a2 a3 3 3 3 3
c $i + c $j + b k$ lie in a plane, then c is
= b1 + c 1 b2 + c 2 b3 + c 3 Ans. (d)
[AIEEE 2005] 1
b1 b2 b3 Given that, |b | |c |a = (a × b) × c
(a) the harmonic mean of a and b . 3
a1 a2 a3 a1 a2 a3
(b) equal to zero. We know that, (a × b) × c = (a ⋅c) b − (b ⋅c) a
⇒ λ b1
4
b2 b3 = − b1 b2 b3 1
(c) the arithmetic mean of a and b . ∴ |b | | c |a = (a ⋅ c) b − (b ⋅c) a
(d) the geometric mean of a and b . c1 c2 c3 c1 c2 c3 3
Ans. (d) ⇒ λ = −1 4 On comparing the coefficients ofa and b,
we get
Since, the given vectors lie in a plane. So, no real value of λ exists.
1
|b | | c | = −b ⋅ c and a ⋅ c = 0
a a c 3
108 If a, b,c are non-coplanar vectors
∴ 1 0 1 =0 ⇒
1
|b | |c | = − |b | |c | cos θ
c c b and λ is a real number, then the 3
vectors a + 2b + 3c, λb + 4c and ⇒
1
cosθ = − ⇒ 1 − sin2 θ =
1
ApplyingC 1 → C 1 − C 2 , we get (2λ − 1) c are non-coplanar for 3 9
0 a c (a) all values of λ. [AIEEE 2004] ⇒ sin2 θ =
8
1 0 1 =0 (b) all except one value of λ. 9
0 c b (c) all except two values of λ. ∴ sinθ =
2 2 Q 0 ≤ θ ≤ π 
(d) no value of λ. 3  2 
⇒ −1 (ab − c 2 ) = 0
Ans. (c)
⇒ c 2 = ab 110 If u, v and w are three
The three vectors (a + 2b + 3c),(λb + 4c)
Hence, c is GM of a and b.
and(2λ − 1) c are non-coplanar, if non-coplanar vectors, then
1 2 3 (u + v − w ) ⋅ [(u − v) × (v − w )]
107 If a, b, c are non-coplanar vectors
0 λ 4 ≠0 equal to [AIEEE 2003]
and λ is a real number, then (a) 0 (b) u ⋅ v × w
0 0 2λ − 1
[λ (a + b) λ2 b λc] = [a b + c b] for (c) u ⋅ w × v (d) 3u ⋅ v × w
1
[AIEEE 2005] ⇒ (2 λ − 1)(λ) ≠ 0 ⇒ λ ≠ 0,
2 Ans. (b)
(a) exactly two values of λ.
(b) exactly three values of λ.
So, these three vectors are non- (u + v − w) ⋅ [(u − v) × (v − w)]
coplanar for all except two values of λ. = (u + v − w) ⋅ [u × v − u × w − v × v + v × w]
(c) no value of λ.
(d) exactly one value of λ. = u⋅ (u × v ) − u⋅ (u × w ) + u⋅ (v × w ) + v ⋅ (u × v )
109 Let a, b and c be non-zero vectors − v ⋅ (u × w) + v ⋅ (v × w) − w ⋅ (u × v)
Ans. (c) 1
such that (a × b) × c = | b|| c | a. If θ is + w ⋅ (u × w) − w ⋅ (v × w)
Given that, 3 = u⋅ v × w − v ⋅u × w − w ⋅u × v {[a ,a b] = 0}
[λ (a + b) λ2 b λ c] = [a b + c b]
an acute angle between the = u⋅ v × w + w ⋅u × v − w ⋅u × v = u⋅ v × w