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CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 13 - MCQExams.com
CBSE
Class 12 Commerce Maths
Vector Algebra
Quiz 13
If
→
u
=
→
a
−
→
b
;
→
v
=
→
a
+
→
b
&
|
→
a
|
=
|
→
b
|
=
2
,
then
|
→
u
×
→
υ
|
is equal to:
Report Question
0%
√
2
(
4
−
(
→
a
.
→
b
)
2
)
0%
2
√
(
16
+
(
→
a
.
→
b
)
2
)
0%
2
√
(
4
−
(
→
a
.
→
b
)
2
)
0%
2
√
(
16
−
(
→
a
.
→
b
)
2
)
→
r
.
ˆ
i
=
2
→
r
.
ˆ
j
=
4
→
r
.
ˆ
k
and
|
→
r
|
=
√
84
, then
|
→
r
.
(
2
ˆ
i
−
3
ˆ
j
+
ˆ
k
)
|
is equal to
Report Question
0%
0
0%
2
0%
4
0%
6
If
ˉ
a
=
ˆ
i
+
ˆ
j
−
2
ˆ
k
,
ˉ
b
=
2
ˆ
i
−
0
ˆ
j
+
ˆ
k
,
ˉ
c
=
3
ˆ
i
−
ˆ
k
and
ˉ
c
=
m
ˉ
a
+
n
ˉ
b
then m + n = ....
Report Question
0%
0
0%
1
0%
2
0%
-1
If the position vectors of
A
,
B
,
C
,
D
are
3
ˆ
i
+
2
ˆ
j
+
ˆ
k
,
4
ˆ
i
+
5
ˆ
j
+
5
ˆ
k
,
4
ˆ
i
+
2
ˆ
j
−
2
ˆ
k
,
6
ˆ
i
+
5
ˆ
j
−
ˆ
k
respectively then the position vector of the point of intersection of
¯
A
B
and
¯
C
D
is
Report Question
0%
2
ˆ
i
+
ˆ
j
−
3
ˆ
k
0%
2
ˆ
i
−
ˆ
j
+
3
ˆ
k
0%
2
ˆ
i
+
ˆ
j
+
3
ˆ
k
0%
2
ˆ
i
−
ˆ
j
−
3
ˆ
k
Explanation
→
A
B
=
→
b
−
→
a
=
(
4
i
+
5
ˆ
ȷ
+
5
ˆ
k
)
−
(
3
ˆ
i
+
2
ˆ
ȷ
+
ˆ
k
)
=
ˆ
ı
+
3
ˆ
ȷ
+
4
ˆ
k
,
a
n
d
→
C
D
=
→
d
−
→
c
=
(
6
ˆ
ı
+
5
ˆ
ȷ
−
ˆ
k
)
−
(
4
ˆ
ı
+
2
ˆ
ȷ
−
2
ˆ
k
)
=
→
2
ˆ
ı
+
3
ˆ
ȷ
+
ˆ
k
Let
→
A
B
and
→
C
D
intersect at
→
P
.
Line
A
B
=
(
3
i
+
2
j
+
ˆ
k
)
+
λ
(
i
+
3
ˆ
j
+
4
ˆ
k
)
and
Line
C
D
=
(
4
i
+
2
ˆ
ı
−
2
ˆ
k
)
+
μ
(
2
ˆ
i
+
3
ˆ
j
+
ˆ
k
)
W.r.t. line
A
0
,
P
=
(
3
+
λ
,
2
+
3
λ
,
1
+
4
λ
)
and
w
⋅
r
⋅
t
⋅
line
C
D
,
P
=
(
4
+
2
μ
,
2
+
3
μ
,
−
2
+
μ
)
Comparing both points,
⇒
3
+
λ
′
=
4
+
2
μ
,
2
+
3
λ
=
2
+
3
μ
1
+
4
λ
=
−
2
+
μ
Solving these equs.,
∴
→
p
=
2
ˆ
ı
−
ˆ
ȷ
−
3
ˆ
k
∴
option
D
is correct.
Let
→
a
=
ˆ
i
−
ˆ
j
,
→
b
=
ˆ
i
−
ˆ
j
=
→
c
=
ˆ
i
−
ˆ
j
, if
→
d
is a unit vector such that
→
a
.
→
d
=
0
=
|
→
b
→
c
→
d
|
then
→
d
equals:
Report Question
0%
±
ˆ
i
+
ˆ
j
−
2
ˆ
k
√
6
0%
±
ˆ
i
+
ˆ
j
−
2
ˆ
k
√
3
0%
±
ˆ
i
+
ˆ
j
+
2
ˆ
k
√
3
0%
±
ˆ
k
If
a
=
i
−
j
,
b
=
i
+
j
,
c
=
i
+
3
j
+
5
k
and
n
is a unit vector such that
b
,
n
=
0
,
a
,
n
=
0
then the value of
|
c
,
n
|
is equal to
Report Question
0%
1
0%
3
0%
5
0%
2
Explanation
b
n
=
0
(
ˆ
i
+
ˆ
j
)
(
a
ˆ
i
+
ˆ
i
e
)
=
0
a
+
b
=
c
(
−
ˆ
j
)
(
a
ˆ
i
+
b
ˆ
j
)
=
0
a
−
b
=
0
a
=
0
=
b
(
ˉ
c
⋅
ˉ
n
)
(
ˆ
i
+
3
ˆ
j
+
5
ˆ
b
)
⋅
(
c
ˆ
k
)
=
5
c
=
√
a
2
+
b
2
+
c
2
=
1
=
5
The value of
|
→
a
×
ˆ
i
|
2
+
|
→
a
×
ˆ
j
|
2
+
|
→
a
×
ˆ
k
|
2
is
Report Question
0%
a
2
0%
2
a
2
0%
3
a
2
0%
none of these
The values of
λ
such that
(
x
,
y
,
z
)
≠
(
0
,
0
,
0
)
and
(
ˆ
i
+
ˆ
j
+
3
ˆ
k
)
x
+
(
3
ˆ
i
−
3
ˆ
j
+
ˆ
k
)
y
+
(
−
4
ˆ
i
+
5
ˆ
j
)
z
=
λ
(
x
ˆ
i
+
y
ˆ
j
+
z
ˆ
k
)
are
Report Question
0%
0
,
1
0%
0
,
−
1
0%
1
,
−
1
0%
0
,
1
,
−
1
Explanation
⇒
(
x
+
3
y
−
4
z
−
λ
x
)
ˆ
i
+
(
x
−
3
y
+
5
z
−
λ
y
)
ˆ
j
+
(
3
x
+
y
−
λ
z
)
ˆ
k
=
0
⇒
x
+
3
y
−
4
z
−
λ
x
=
0
x
−
3
y
+
5
z
−
λ
y
=
0
3
x
+
y
−
λ
z
=
0
[
1
−
λ
3
−
4
1
−
3
−
λ
5
3
1
−
λ
]
[
x
y
z
]
=
[
0
0
0
]
⇒
|
1
−
λ
3
−
4
1
−
3
−
λ
5
3
1
−
λ
|
=
0
{
∵
for non-trivial
soution of equation
A
x
=
y
|
A
|
=
0
}
⇒
(
1
−
λ
)
(
(
−
3
−
λ
)
(
−
λ
)
−
5
)
−
1
(
−
3
λ
+
4
)
+
3
(
15
+
4
(
−
3
−
λ
)
)
=
0
=
(
1
−
λ
)
(
3
λ
+
λ
2
−
5
)
−
4
+
3
λ
+
3
(
3
−
4
λ
)
=
0
=
(
1
−
λ
)
(
λ
2
+
3
λ
−
5
)
−
4
+
3
λ
+
9
−
12
λ
=
0
=
λ
2
+
3
λ
−
5
−
λ
3
−
3
λ
2
+
5
λ
−
4
+
3
λ
+
9
−
12
λ
=
0
=
−
λ
3
−
2
λ
2
−
λ
=
0
⇒
λ
3
+
2
λ
2
+
λ
2
=
0
λ
(
λ
2
+
2
λ
+
1
)
=
0
λ
=
0
,
λ
2
+
2
λ
+
1
=
0
(
λ
+
1
)
2
=
0
λ
=
−
1
⇒
λ
=
0
,
−
1
∴
option
B
is correct.
Let
→
u
,
→
v
,
→
w
be such that
|
→
u
|
= 1,
|
→
v
|
= 2,
|
→
w
|
= 3 . If the projection
→
v
along
→
u
is equal to that of
→
w
along
→
u
and
→
v
,
→
w
are perpendicular to each other , then
|
→
u
→
v
+
→
w
|
equals
Report Question
0%
√
14
0%
√
7
0%
2
0%
14
Explanation
Given,
|
→
u
|
=
1
and
|
→
v
|
=
2
,
|
→
w
|
=
3
and,
→
v
⋅
→
u
=
→
w
⋅
→
u
→
v
⋅
→
u
=
→
ω
⋅
→
u
or,
→
v
⋅
→
ω
=
0
Now,
|
→
u
−
→
v
+
→
w
|
Squaring,
|
→
u
|
2
+
→
v
|
2
+
|
→
w
|
2
−
2
→
u
⋅
→
v
−
2
→
v
⋅
→
ω
+
2
→
u
⋅
→
w
⇒
1
+
4
+
9
−
2
→
u
⋅
→
v
+
2
→
u
⋅
→
v
-0
⇒
14
|
→
u
−
→
v
+
→
w
|
=
√
14
option
A
=
√
14
Vector
→
x
satisfying the relation
→
A
.
¯
x
=
c
and
→
A
×
→
x
=
→
B
is
Report Question
0%
c
→
A
−
(
¯
A
×
→
B
)
|
→
A
|
0%
c
→
A
−
(
→
A
×
→
B
)
|
→
A
|
2
0%
c
→
A
+
(
→
A
×
→
B
)
|
→
A
|
2
0%
None
For any vector
→
a
, the value of
(
→
a
×
ˆ
i
)
2
+
(
→
a
×
ˆ
j
)
2
+
(
→
a
×
ˆ
k
)
2
is equal to
Report Question
0%
4
|
→
a
|
2
0%
2
|
→
a
|
2
0%
|
→
a
|
2
0%
3
|
→
a
|
2
let
ˉ
a
be a unit vector and
ˉ
b
be a non-zero vector not parallel to
ˉ
a
if two sides of a triangle are represented by the vectors
√
3
(
ˉ
a
×
ˉ
b
)
a
n
d
ˉ
b
−
(
ˉ
a
.
ˉ
b
)
ˉ
a
then the angles of triangle are
Report Question
0%
90
∘
,
60
∘
,
30
∘
0%
45
∘
,
45
∘
,
90
∘
0%
60
∘
,
60
∘
,
60
∘
0%
75
∘
,
45
∘
,
60
∘
In a triangle ABC, if
A
=
(
0
,
0
)
,
B
=
(
3
,
3
√
3
)
,
C
=
(
−
3
√
3
,
3
)
then the vector of magnitude
2
√
2
units directed along
¯
A
O
, where O is the circumcentre of triangle ABC is?
Report Question
0%
(
1
−
√
3
)
ˉ
i
+
(
1
+
√
3
)
ˉ
j
0%
√
3
ˉ
i
+
2
ˉ
j
0%
ˉ
i
−
√
3
ˉ
j
0%
ˉ
i
+
2
ˉ
j
In a parallelogram ABD,
|
_
A
→
B
|
=
a
,
|
_
A
→
D
|
=
b
and
|
_
→
A
C
|
=
c
,
,
_
→
A
B
.
_
→
D
B
has the value :
Report Question
0%
1
2
(
3
a
2
+
b
2
−
c
2
)
0%
1
2
(
a
2
−
b
2
+
c
2
)
0%
1
2
(
a
2
+
b
2
−
c
2
)
0%
1
3
(
b
2
+
c
2
−
a
2
)
If the position vectors of the vertices
A
,
B
and
C
of a
Δ
A
B
C
are respectively
4
ˆ
i
+
7
ˆ
j
+
8
ˆ
k
,
2
ˆ
i
+
3
ˆ
j
+
4
ˆ
k
and
2
ˆ
i
+
5
ˆ
j
+
7
ˆ
k
, then the position vector of the point, where the bisector of
∠
A
meets
B
C
is:
Report Question
0%
1
2
(
4
ˆ
i
+
8
ˆ
j
+
11
ˆ
k
)
0%
1
3
(
6
ˆ
i
+
13
ˆ
j
+
18
ˆ
k
)
0%
1
4
(
8
ˆ
i
+
14
ˆ
j
+
19
ˆ
k
)
0%
1
3
(
6
ˆ
i
+
8
ˆ
j
+
15
ˆ
k
)
I
f
|
→
a
|
=2 and
|
→
b
|
=3 and
|
→
a
|
.
|
→
b
|
=Then (a x (a x (a x (a x b)))) is equal to
Report Question
0%
4
ˆ
b
0%
−
4
ˆ
b
0%
4
ˆ
a
0%
−
4
ˆ
a
If C is the mid-point of AB and P is any point outside AB , then
Report Question
0%
→
P
A
+
→
P
B
+
→
P
C
= 0
0%
→
P
A
+
→
P
B
+ 2
→
P
C
=
→
0
0%
→
P
A
→
P
B
=
→
P
C
0%
→
P
A
→
P
B
= 2
→
P
C
If
→
a
and
→
b
are vectors such that
|
→
a
+
→
b
|
=
√
29
and
→
a
×
(
2
∧
i
+
3
∧
j
+
4
∧
k
)
=
(
2
∧
i
+
3
∧
j
+
4
∧
k
)
×
→
b
, then a possible value of
(
→
a
+
→
b
)
(
−
7
∧
i
+
2
∧
j
+
3
∧
k
)
is
Report Question
0%
0
0%
3
0%
4
0%
8
If
ˆ
u
and
ˆ
v
are unit vectors and
θ
is the acute angle between them , then 2
ˆ
u
3
ˆ
v
is a unit vector for
Report Question
0%
No value of
θ
0%
Exactly one value of
θ
0%
Exactly two values of
θ
0%
More than two values of
θ
Explanation
Given,
ˉ
u
and
→
v
are unit vector
|
→
u
|
=
1
|
→
v
|
=
1
Now,
⇒
|
2
→
u
|
|
→
3
v
|
sin
θ
=
1
|
→
2
u
×
→
→
3
v
|
[
∵
|
→
a
×
→
b
|
=
|
→
a
|
→
b
|
|
sin
θ
]
⇒
2
⋅
3
|
ˉ
u
|
∇
∣
sin
θ
=
1
sin
θ
=
1
6
There is only one value of
θ
for which
|
→
2
4
×
→
3
V
|
is a
unit vector.
option
B
Let
→
a
=
ˆ
i
ˆ
j
,
→
b
=
ˆ
j
ˆ
k
,
→
c
=
ˆ
k
ˆ
i
. If
→
d
is a unit vector such that
→
a
.
→
d
= 0 =
|
→
b
→
c
→
d
|
then
→
d
equals :
Report Question
0%
ˆ
i
+
ˆ
j
2
ˆ
k
√
6
0%
ˆ
i
+
ˆ
j
ˆ
k
√
3
0%
ˆ
i
+
ˆ
j
+
ˆ
k
√
3
0%
ˆ
k
The value of
λ
for
(
x
,
y
,
z
)
≠
(
0
,
0
,
0
)
and
(
i
+
j
+
3
k
)
x
+
(
3
i
−
3
j
+
k
)
y
+
(
−
4
i
+
5
j
)
z
=
λ
(
x
i
+
y
j
+
z
k
)
are
Report Question
0%
0, -1
0%
0, 1
0%
-2, 0
0%
0, 2
If
→
a
,
→
b
,
→
c
are unit vectors such that
→
a
+
→
b
+
→
c
=
0
then the value of
→
a
.
→
b
+
→
b
.
→
c
+
→
c
.
→
a
.
is
Report Question
0%
1
0%
-1
0%
-3/2
0%
none of these
If
¯
a
=
−
2
¯
i
+
3
¯
j
+
4
¯
k
and
¯
b
=
−
2
¯
i
−
2
¯
j
+
3
¯
k
then
¯
a
.
¯
b
is
Report Question
0%
2
0%
-2
0%
6
0%
none of these
Let position vector of the orthocentre of
△
A
B
C
be
→
r
. then, which of the following statement(s) is\are correct (Given position vector of points
a
ˆ
i
,
b
ˆ
j
,
c
ˆ
k
and
a
b
c
=
0
)
Report Question
0%
→
r
.
ˉ
i
=
a
1
a
2
+
1
b
2
+
1
c
2
0%
→
r
.
ˉ
i
=
1
a
(
1
a
2
+
1
b
2
+
1
c
2
)
0%
→
r
.
ˉ
i
→
r
.
ˉ
j
+
→
r
.
ˉ
j
→
r
.
ˉ
k
+
→
r
.
ˉ
k
→
r
.
ˉ
i
=
a
b
+
b
c
+
c
a
0%
→
r
.
ˉ
i
→
r
.
ˉ
j
+
→
r
.
ˉ
j
→
r
.
ˉ
k
+
→
r
.
ˉ
k
→
r
.
ˉ
i
=
b
a
+
c
b
+
a
c
If C is the mid point of AB and P is any point outside AB , then
Report Question
0%
→
P
A
+
→
P
B
+
→
P
C
= 0
0%
→
P
A
+
→
P
B
+
2
→
P
C
=
→
0
0%
→
P
A
+
→
P
B
=
→
P
C
0%
→
P
A
+
→
P
B
=
2
→
P
C
A particle in a plane from A to E along the shown path. It is given that AB=BC=DE=10 metre. Then the magnitude of net displacement of particle is :
Report Question
0%
10 m
0%
15 m
0%
5 m
0%
20 m
Let
a
=
∑
i
<
j
(
1
n
C
i
+
1
n
C
j
)
and
b
=
∑
i
<
j
(
i
n
C
i
+
j
n
C
j
)
, then?
Report Question
0%
b
=
(
n
−
1
)
a
0%
b
=
(
n
+
1
)
a
0%
b
=
n
2
a
0%
b
=
n
a
In parallelogram ABCD,
|
¯
A
B
|
=
a
,
|
¯
A
D
|
=
b
and
|
¯
A
C
|
=
c
then
¯
D
B
,
¯
A
B
has the value
Report Question
0%
3
a
2
+
b
2
−
c
2
2
0%
3
b
2
+
c
2
−
a
2
2
0%
3
c
2
+
b
2
−
a
2
2
0%
a
2
+
b
2
+
c
2
2
If
n
∑
i
=
1
→
a
i
=
→
0
where
|
→
a
i
|
=
1
∀
i
,
then the value of
∑
1
≤
i
<
j
≤
n
→
a
i
⋅
→
a
j
is
Report Question
0%
−
n
/
2
0%
−
n
0%
n
/
2
0%
n
If the vectors
→
a
,
→
b
,
→
c
satisfying
→
a
+
→
b
+ 2
→
c
= 0 . If
|
→
a
|
= 1 ,
|
→
b
|
= 4 ,
|
→
c
|
= 2 , then
→
a
.
→
b
+
→
b
.
→
c
+
→
c
.
→
a
=
Report Question
0%
−
7
2
0%
−
17
2
0%
17
2
0%
7
2
The projection of the join of the point (3, 4, 2), (5, 1, 8) on the line whose d.c.'s are
(
2
7
,
−
3
7
,
6
7
)
is
Report Question
0%
7
0%
(
46
13
)
0%
(
42
13
)
0%
(
38
13
)
If
a
,
b
and care three mutually perpendicular vectors, then the projection of the vector
|
a
|
a
|
+
m
b
|
b
|
+
n
(
a
×
b
)
|
a
×
b
|
along the angle bisector of the vector
a
and
b
is
Report Question
0%
l
2
+
m
2
√
l
2
+
m
2
−
n
2
0%
√
1
2
+
m
2
+
n
2
0%
√
1
2
+
m
2
√
l
2
−
m
2
+
n
2
0%
1
+
m
√
2
Let OAB be a regular triangle with side unity (o being otogin). Also M, N are the points of intersection of AB, M being closer to A and N closer to B. Position vectors of A, B, M and N are
→
a
,
→
b
,
→
m
and
→
n
respectively. Which of the following hold (s) good ?
Report Question
0%
→
m
=
x
→
a
+
y
→
b
⇒
2
3
and
y
=
1
3
0%
→
m
=
x
→
a
+
y
→
b
⇒
x
=
5
6
and
y
=
1
6
0%
→
m
.
→
n
equals
13
18
0%
→
m
.
→
n
equals
15
18
The position vector of A is
2
→
i
+
3
→
j
+
4
→
k
→
A
B
=
5
→
i
+
7
→
j
+
6
→
k
, then the position vector of B is
Report Question
0%
−
7
→
i
−
10
→
j
−
10
→
k
0%
7
→
i
−
10
→
j
+
10
→
k
0%
7
→
i
+
10
→
j
−
10
→
k
0%
7
→
i
+
10
→
j
+
10
→
k
The position vector of a point lying on the joining the points whose position vectors are
¯
i
+
¯
j
−
¯
k
and
¯
i
−
¯
j
+
¯
k
is
Report Question
0%
¯
j
0%
¯
i
0%
¯
k
0%
¯
0
Area of diagonals is, ..., where diagonals
are
a
=
2
ˆ
i
−
3
ˆ
j
+
5
ˆ
k
, and
b
=
−
ˆ
i
+
ˆ
j
+
ˆ
k
Report Question
0%
√
21.5
0%
√
31.5
0%
√
28.5
0%
√
38.5
If
ˉ
a
,
ˉ
b
,
ˉ
c
are position vectors of the points A,B,C respectively such that
9
ˉ
a
−
7
ˉ
b
−
2
ˉ
c
=
ˉ
0
then point B divides AC in the ratio.....
Report Question
0%
Internally 7:2
0%
Externally 9:2
0%
Internally 9:7
0%
Externally 2:7
A vector
A
=
→
l
=
→
x
j
=
3
→
k
is rotated through an angle and is also doubled in magnitude resulting in
→
B
=
4
→
l
+
(
4
x
−
2
)
→
j
+
2
→
k
. An acceptable value of x is
Report Question
0%
1
0%
2
0%
3
0%
4/3
A stone projected vertically upwards raises 's' feets in 't' seconds where
S
=
112
t
−
16
t
2
then the maximum height it reached is
Report Question
0%
195 ft
0%
194 ft
0%
196 ft
0%
216 ft
Explanation
s
=
112
t
−
16
t
2
for maximum height
d
s
d
t
=
0
d
(
112
t
−
16
t
2
)
d
t
=
0
d
(
112
t
)
d
t
−
d
(
16
t
2
)
d
t
=
0
112
−
16
(
2
)
t
2
−
1
=
0
112
−
32
t
=
0
32
t
=
112
t
=
112
32
=
28
8
=
7
2
∴
maximum height
=
112
(
7
2
)
−
16
(
7
2
)
2
=
56
×
7
−
4
×
49
=
392
−
196
=
196
f
t
Given
¯
a
=
x
ˆ
i
+
y
ˆ
j
+
2
ˆ
k
,
¯
b
=
ˆ
i
−
ˆ
j
+
ˆ
k
,
¯
c
=
ˆ
i
+
2
ˆ
j
;
(
¯
a
^
¯
b
)
=
π
/
2
,
¯
a
.
¯
c
=
4
then
Report Question
0%
[
¯
a
¯
b
¯
c
]
2
=
|
¯
a
|
0%
[
¯
a
¯
b
¯
c
]
=
|
¯
a
|
0%
[
¯
a
¯
b
¯
c
]
=
0
0%
none of these
If
→
a
and
→
b
are vectors such that
|
→
a
+
→
b
|
=
√
29
and
→
a
×
(
2
ˆ
i
+
3
ˆ
j
+
4
ˆ
k
)
=
(
2
ˆ
i
+
3
ˆ
j
+
4
ˆ
k
)
×
→
b
,
then
a possible value of
(
→
a
+
→
b
)
⋅
(
−
7
ˆ
i
+
2
ˆ
j
+
3
ˆ
k
)
is
Report Question
0%
0
0%
3
0%
4
0%
8
For any three
→
a
,
→
b
,
→
c
(
→
a
−
→
b
)
(
→
b
−
→
c
)
×
(
→
c
−
→
a
)
is equal to
Report Question
0%
→
b
⋅
(
→
c
×
→
a
)
0%
2
→
a
⋅
(
→
b
×
→
c
)
0%
0
0%
none of these
The point D,E,F divide BC, CA and Ab of the triangle ABC in the ratio 1 : 4, 3 : 2 and 3 : 7 respectively and the point K divides AB in the ratio 1 : 3 then
(
→
A
D
+
→
B
E
+
→
C
F
)
:
→
C
K
is equal to
Report Question
0%
5 : 2
0%
2:5
0%
1:1
0%
none of these
(
→
r
.
→
i
)
(
→
r
×
→
i
)
+
(
→
r
.
→
j
)
(
→
r
×
→
j
+
)
(
→
r
.
→
k
)
(
→
r
×
→
k
)
is equal to
Report Question
0%
3
→
r
0%
→
r
0%
→
0
0%
None of these
If u =
ˆ
j
+
4
ˆ
K
,
V
=
ˆ
i
−
3
ˆ
K
w
=
c
o
s
θ
ˆ
i
+
s
i
n
θ
ˆ
j
are vectors in 3- dimensional space, then the maximum possible value of
|
u
×
v
.
w
|
is
Report Question
0%
√
13
0%
√
14
0%
5
0%
7
If u =
ˆ
j
+
4
ˆ
k
,
V
=
ˆ
i
=
−
3
K
a
n
d
W
=
c
o
s
θ
i
+
s
i
n
θ
ˆ
i
are vectors in 3-dimension space, then the maximum possible value of
|
u
×
v
.
w
|
is
Report Question
0%
√
1
3
0%
√
1
4
0%
5
0%
7
If
ˆ
i
×
(
→
a
×
ˆ
i
)
+
ˆ
j
×
(
→
a
×
ˆ
j
)
+
ˆ
k
×
(
→
a
×
ˆ
k
)
=
.
.
.
.
.
{
(
→
a
.
ˆ
i
)
ˆ
i
+
(
→
a
.
ˆ
j
)
ˆ
j
+
(
→
a
.
ˆ
k
)
ˆ
k
}
Report Question
0%
−
1
0%
0
0%
2
0%
N
o
n
e
o
f
t
h
e
s
e
The magnitude of two vectors which can be represented in the form i+j+(2x)k is
√
18
. Then the unit vector that is perpendicular to these two vectors is
Report Question
0%
−
i
+
j
√
2
0%
i
−
j
8
√
2
0%
−
i
+
j
8
0%
−
i
+
j
2
√
2
The length of vector
→
A
G
is
Report Question
0%
√
17
0%
√
51
/
3
0%
3
/
√
6
0%
√
59
/
4
Explanation
Point
G
is
(
4
3
,
1
3
,
8
3
)
⋅
Therefore
|
→
A
G
|
2
=
(
5
3
)
2
+
1
9
+
(
5
3
)
2
=
51
9
⇒
|
→
A
G
|
=
√
51
3
→
A
B
=
−
4
ˆ
i
+
4
ˆ
j
+
0
ˆ
k
→
A
C
=
2
ˆ
i
+
2
ˆ
j
+
2
ˆ
k
∴
→
A
B
×
→
A
C
=
−
8
|
ˆ
i
ˆ
j
ˆ
k
1
−
1
0
1
1
1
|
=
8
(
ˆ
i
+
ˆ
j
−
2
ˆ
k
)
Area of
Δ
A
B
C
=
1
2
|
→
A
B
×
→
A
C
|
=
4
√
6
→
A
D
=
−
3
ˆ
i
−
5
ˆ
j
+
3
ˆ
k
The length of the perpendicular from the vertex
D
on the opposite face
=
∣
projection of
→
A
D
on
→
A
B
×
→
A
C
∣
=
|
(
−
3
ˆ
i
−
5
ˆ
j
+
3
ˆ
k
)
(
ˆ
i
+
ˆ
j
−
2
ˆ
k
)
√
6
|
=
|
−
3
−
5
−
6
√
6
|
=
14
√
6
For non-zero vectors
→
a
,
→
b
and
→
c
,
|
(
→
a
×
→
b
)
⋅
→
c
|
=
|
→
a
|
|
→
b
|
|
→
c
|
holds if and only if
Report Question
0%
→
a
⋅
→
b
=
0
,
→
b
⋅
→
c
=
0
0%
→
b
⋅
→
c
=
0
,
→
c
⋅
→
a
=
0
0%
→
c
⋅
→
a
=
0
,
→
a
⋅
→
b
=
0
0%
→
a
⋅
→
b
=
→
b
⋅
→
c
=
→
c
⋅
→
a
=
0
Explanation
Step 1: Simplifying given expression
Given that
|
(
→
a
×
→
b
)
.
→
c
|
=
|
→
a
|
|
→
b
|
|
→
c
|
⟹
|
|
→
a
|
|
→
b
|
s
i
n
θ
ˆ
n
.
→
c
|
=
|
→
a
|
|
→
b
|
|
→
c
|
Here
ˆ
n
is the resultant of vector
a
and
b
and
ˆ
n
.
→
c
=
|
ˆ
n
|
|
→
c
|
c
o
s
α
⟹
|
|
→
a
|
|
→
b
|
s
i
n
θ
|
ˆ
n
|
|
→
c
|
c
o
s
α
|
=
|
→
a
|
|
→
b
|
|
→
c
|
Since
ˆ
n
is a unit vector
|
ˆ
n
|
=
1
Step 2: Calculating
⟹
|
|
→
a
|
|
→
b
|
|
→
c
|
s
i
n
θ
.
c
o
s
α
|
=
|
→
a
|
|
→
b
|
|
→
c
|
⟹
|
s
i
n
θ
|
|
c
o
s
α
|
=
1
Step 3: Calculating result
If we assume that
θ
=
π
2
,
α
=
0
⟹
→
a
and
→
b
are perpendicular
⟹
→
a
.
→
b
=
0
As
α
=
0
,
→
c
is along the resultant of vector
a
and
b
⟹
→
c
is perpendicular to
→
a
and
→
b
⟹
→
c
.
→
a
=
0
and
→
c
.
→
b
=
0
Hence,
→
a
.
→
b
=
→
a
.
→
c
=
→
b
.
→
c
=
0
0:0:1
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0
Answered
1
Not Answered
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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