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CBSE Questions for Class 12 Commerce Maths Three Dimensional Geometry Quiz 8 - MCQExams.com
CBSE
Class 12 Commerce Maths
Three Dimensional Geometry
Quiz 8
ABC is a triangle where
A
=
(
2
,
3
,
5
)
,
B
=
(
−
1
,
2
,
2
)
and
C
(
λ
,
5
,
μ
)
if the median through A is equally inclined to the positive axis then
λ
+
μ
is
Report Question
0%
7
0%
6
0%
15
0%
9
Explanation
A
D
is median
D
(
λ
−
1
2
,
7
2
,
μ
+
2
2
)
→
A
D
=
(
λ
−
1
2
−
2
,
7
2
−
3
,
μ
+
2
2
−
5
)
→
A
D
is equally in lined.
∴
paralle to
(
1
,
1
,
1
)
λ
−
5
2
=
1
2
=
μ
+
8
2
λ
=
6
μ
=
9
λ
+
μ
=
15
Then,
Option
C
is correct answer.
Projection of a vector on
3
coordinate axes are
6
,
−
3
,
2
respectively. Then DC's of vector are-
Report Question
0%
6
,
−
3
,
2
0%
6
5
,
−
3
5
,
2
5
0%
−
6
7
,
−
3
2
,
2
7
0%
6
7
,
−
3
7
,
2
7
Explanation
Projection of a vector on coordinate axis are
X
2
−
X
1
,
Y
2
−
Y
1
,
Z
2
−
Z
1
X
2
−
X
1
=
6
,
Y
2
−
Y
1
=
−
3
,
Z
2
−
Z
1
=
2
√
(
X
2
−
X
1
)
2
+
(
Y
2
−
Y
1
)
2
+
(
Z
2
−
Z
1
)
2
=
√
36
+
9
+
4
=
7
The Direction cosines of the vector are
6
7
,
−
3
7
,
2
7
If the foot of the perpendicular from
(
0
,
0
,
0
)
to a plane is
P
(
1
,
2
,
2
)
. Then, the equation of the plane is
Report Question
0%
−
x
+
2
y
+
8
z
−
9
=
0
0%
x
+
2
y
+
2
z
−
9
=
0
0%
x
+
y
+
z
−
5
=
0
0%
x
+
2
y
−
3
z
+
3
=
0
Explanation
Equation of normal
⇒
ˆ
i
+
2
ˆ
j
+
2
ˆ
k
equation of plane
⇒
(
x
−
1
)
1
+
(
x
y
−
2
)
2
+
(
z
−
2
)
2
=
0
x
−
1
+
2
y
−
4
+
2
z
−
4
=
0
x
+
2
y
+
2
z
−
9
=
0
P
(
1
,
1
,
1
)
and
Q
(
λ
,
λ
,
λ
)
are two points in the space such that
P
Q
=
√
27
,
the value of
λ
can be
Report Question
0%
−
4
0%
−
2
0%
2
0%
0
Explanation
P
Q
2
=
(
λ
−
1
)
2
+
(
λ
−
1
)
2
+
(
λ
−
1
)
2
⇒
3
(
λ
−
1
)
2
=
27
⇒
(
λ
−
1
)
2
=
9
⇒
λ
−
1
=
±
3
⇒
λ
=
1
±
3
⇒
λ
=
1
+
3
,
or
1
−
3
∴
λ
=
4
,
or
−
2
A
(
2
,
3
,
7
)
,
B
(
−
1
,
3
,
2
)
and
C
(
q
,
5
,
r
)
are the vertices of
Δ
A
B
C
. If the median through A is equally inclined to the coordinate axes then the coordinates of the vertex C is
Report Question
0%
(
7
,
5
,
14
)
0%
(
7
,
5
,
12
)
0%
(
7
,
5
,
10
)
0%
(
7
,
5
,
16
)
If the points
ˉ
a
+
ˉ
b
,
ˉ
a
−
ˉ
b
,
ˉ
a
+
k
ˉ
b
are collinear, then
Report Question
0%
k
has only one real value
0%
k
has two real value
0%
k
has no real values
0%
k
has infinite number of real values
Explanation
As the
3
point should be collinear area of triangle termed by then should be zero
considering
2
direction to be
(
ˉ
a
+
ˉ
b
)
−
(
ˉ
a
−
ˉ
b
)
=
2
ˉ
b
&
(
ˉ
a
+
ˉ
b
)
−
(
ˉ
a
+
k
ˉ
b
)
=
(
1
−
k
)
ˉ
b
(
2
ˉ
b
)
×
(
1
−
k
)
ˉ
b
=
0
this will be for any value of
k
as cross produced of
2
linear vector
=
0
A line
A
B
in three-dimensional space males angles
45
o
and
120
o
with the positive x-axis and the positive y-axis respectively. If
A
B
makes an acute angle
θ
with the positive z-axis, then
θ
equal
Report Question
0%
45
o
0%
60
o
0%
75
o
0%
30
o
Explanation
'l', 'm', 'n' be the Direction cosine of the line
l
=
c
o
s
α
=
c
o
s
45
∘
=
1
√
2
m
=
c
o
s
β
=
c
o
s
120
∘
=
−
1
2
n
=
c
o
s
θ
we know
l
2
+
m
2
+
n
2
=
1
⇒
d
1
2
+
1
4
+
c
o
s
2
θ
=
1
⇒
c
o
s
2
θ
=
1
−
1
2
−
1
4
⇒
c
o
s
2
θ
=
1
4
⇒
c
o
s
=
1
2
∴
θ
=
60
∘
If the points
(
α
,
−
1
)
,
(
2
,
1
)
and
(
4
,
5
)
are collinear, then find
α
by vector method.
Report Question
0%
4
0%
1
0%
8
0%
None of these
Explanation
If there points are collinear then vectors from one to another will have scalar triple produced
0
.Point
(
α
,
−
1
)
,
(
2
,
1
)
,
(
4
,
5
)
(
2
−
α
)
ˆ
i
+
2
ˆ
j
−
ˉ
A
2
ˆ
i
+
4
ˆ
j
−
ˉ
B
(
4
−
α
)
ˆ
i
+
6
ˆ
j
−
ˉ
C
ˉ
A
.
(
ˉ
B
×
ˉ
C
)
=
0
(
(
2
−
α
)
ˆ
i
+
2
ˆ
j
)
(
2
ˆ
i
+
4
ˆ
j
)
×
(
(
4
−
α
)
ˆ
i
+
6
ˆ
j
)
(
(
2
−
α
)
ˆ
i
+
2
ˆ
j
)
.
[
12
ˆ
k
−
16
ˆ
k
+
4
α
ˆ
k
]
=
0
4
α
=
4
α
=
1
Also the direction vector will be proportion
(
2
−
α
,
2
)
=
λ
(
4
−
2.5
−
1
)
(
2
−
α
,
2
)
=
λ
(
2
,
4
)
λ
=
1
2
as
2
=
4
λ
2
−
α
=
1
∴
α
=
1
The Cartesian equation of line
6
x
−
2
=
3
y
+
1
=
2
z
−
2
is given by
Report Question
0%
3
x
−
1
3
=
3
y
+
1
6
=
z
−
1
3
0%
3
x
+
1
3
=
3
y
−
1
6
=
z
−
1
3
0%
3
x
−
1
3
=
3
y
−
1
6
=
z
−
1
3
0%
3
x
−
1
6
=
3
y
−
1
3
=
z
−
1
3
Explanation
Solution -
6
x
−
2
=
3
y
+
1
=
2
z
−
2
6
x
−
2
6
=
3
y
+
1
6
=
2
z
−
2
6
x
−
1
3
1
=
y
+
1
3
2
=
z
−
1
3
DR
(
1
,
2
,
3
)
Point passing through line
(
1
/
3
,
−
1
/
3
,
1
)
3
x
−
1
3
=
3
y
+
1
6
=
z
−
1
3
A is correct.
The direction ratios of the joining
A
(
1
,
2
,
1
)
and
(
2
,
1
,
2
)
are
Report Question
0%
3
,
3
,
3
0%
−
1
,
1
,
−
1
0%
3
,
1
,
3
0%
1
√
3
,
1
√
3
,
1
√
3
Explanation
We have given points are:-
A
(
1
,
2
,
1
)
B
(
2
,
1
,
2
)
Now,
→
B
A
=
(
1
−
2
)
ˆ
i
+
(
2
−
1
)
ˆ
j
+
(
1
−
2
)
ˆ
k
=
−
ˆ
i
+
ˆ
j
−
ˆ
k
direction ratio are
a
=
−
1
b
=
1
c
=
−
1
Hence,
option
B
is correct.
The direction ratios of
A
B
are
−
2
,
2
,
1
. If coordinates of A are
(
4
,
1
,
5
)
and
l
(
A
B
)
=
6
, then coordinates of
B
?
Report Question
0%
(
0
,
5
,
−
7
)
0%
(
8
,
−
3
,
3
)
0%
(
0
,
7
,
5
)
0%
(
8
,
3
,
3
)
If the lines
L
1
a
n
d
L
2
are given by
ˉ
r
=
(
ˉ
i
+
2
ˉ
j
−
ˉ
k
)
+
t
(
¯
2
i
−
3
ˉ
j
+
ˉ
k
)
a
n
d
ˉ
r
=
(
ˉ
i
+
ˉ
j
+
ˉ
k
)
+
s
(
2
ˉ
i
+
ˉ
j
−
ˉ
k
)
, then
Report Question
0%
L
1
a
n
d
L
2
are perpendicular
0%
L
1
a
n
d
L
2
are parallel
0%
(
L
1
,
L
2
)
=
45
o
0%
(
L
1
,
L
2
)
=
60
o
The vector equation of line passing through the point
(
−
1
,
−
1
,
2
)
and parallel to the line
2
x
−
2
=
3
y
+
1
=
6
z
−
2
Report Question
0%
(
−
ˆ
i
−
ˆ
j
+
2
ˆ
k
)
+
λ
(
3
ˆ
i
+
2
ˆ
j
+
ˆ
k
)
0%
(
−
ˆ
i
−
ˆ
j
+
2
ˆ
k
)
+
λ
(
2
ˆ
i
+
3
ˆ
j
+
6
ˆ
k
)
0%
(
−
ˆ
i
−
ˆ
j
+
2
ˆ
k
)
+
λ
(
ˆ
i
+
2
ˆ
j
+
3
ˆ
k
)
0%
(
−
ˆ
i
−
ˆ
j
+
2
ˆ
k
)
+
λ
(
2
ˆ
i
+
3
ˆ
j
+
ˆ
k
)
Explanation
Given
L
1
:
2
x
−
2
=
3
y
+
1
=
6
z
−
2
⇒
2
(
x
−
1
)
=
3
(
y
+
1
3
)
=
6
(
z
−
1
3
)
⇒
x
−
1
3
=
y
+
1
3
2
=
z
−
1
3
1
Equation of line passing through
(
−
1
,
−
1
,
2
)
parallel to
L
1
x
+
1
3
=
y
+
1
2
=
z
−
2
1
vector equation is
(
−
ˆ
i
−
ˆ
j
+
2
ˆ
k
)
+
λ
(
3
ˆ
i
+
2
ˆ
j
+
ˆ
k
)
If
ˉ
a
,
ˉ
b
and
ˉ
c
are non-zero non collinear vectors and
θ
(
≠
0
,
π
)
is the angle between
ˉ
b
and
ˉ
c
if
(
ˉ
a
×
ˉ
b
)
×
ˉ
c
=
1
2
|
ˉ
b
|
ˉ
c
|
ˉ
a
. then
sin
θ
=
Report Question
0%
√
2
3
0%
√
3
2
0%
4
√
2
3
0%
2
√
2
3
Explanation
We have
(
→
a
×
→
b
)
×
→
c
=
1
2
|
→
b
|
|
→
c
|
→
a
→
c
×
(
→
a
×
→
b
)
=
1
2
|
→
b
|
|
→
c
|
→
a
−
[
(
→
c
.
→
b
)
→
a
−
(
→
c
.
→
a
)
→
b
]
=
1
2
|
→
b
|
|
→
c
|
→
a
(
→
c
.
→
a
)
→
b
−
(
→
c
.
→
b
)
→
a
=
1
2
|
→
b
|
|
→
c
|
→
a
→
c
.
→
a
=
0
→
c
.
→
a
=
−
1
2
|
→
b
|
|
→
c
|
cos
θ
=
−
1
2
⇒
θ
=
2
π
3
∴
sin
θ
=
√
3
2
Hence,
B
is the correct answer.
If
A
=
(
1
,
2
,
−
1
)
,
B
=
(
2
,
0
,
3
)
,
C
=
(
3
,
−
1
,
2
)
then the angle between
¯
A
B
and
¯
A
C
is
Report Question
0%
0
o
0%
90
o
0%
cos
−
1
(
20
√
21
√
22
)
0%
cos
−
1
(
15
√
21
√
11
)
Explanation
Given
A
(
1
,
2
,
−
1
)
,
B
=
(
2
,
0
,
3
)
,
C
=
(
3
,
−
1
,
2
)
Drs of
A
B
=
(
1
,
−
2
,
4
)
Drs of
A
C
=
(
2
,
−
3
,
3
)
cos
θ
=
2
+
6
+
12
√
1
+
4
+
16
√
4
+
9
+
9
⇒
cos
θ
=
20
√
21
√
22
⇒
θ
=
cos
−
1
(
20
√
21
√
22
)
A line d.c's proportional to
(
2
,
1
,
2
)
meets each of the lines
x
=
y
+
a
=
z
and
x
+
a
=
2
y
=
2
z
. Then the coordinates of each of the points of intersection are given by
Report Question
0%
(
3
a
,
2
a
,
3
a
)
;
(
a
,
a
,
2
a
)
0%
(
3
a
,
2
a
,
3
a
)
;
(
a
,
a
,
a
)
0%
(
3
a
,
3
a
,
3
a
)
;
(
a
,
a
,
a
)
0%
(
2
a
,
3
a
,
3
a
)
;
(
2
a
,
a
,
a
)
Explanation
L
1
:
x
1
=
y
+
a
1
=
z
1
L
2
:
x
+
a
2
=
y
1
=
z
1
Any point on
L
1
A
(
k
1
,
k
1
−
1
,
k
1
)
Any point on
L
2
B
(
2
k
2
−
1
,
k
2
,
k
2
)
Drs of
A
B
=
(
k
1
−
2
k
2
+
a
,
k
1
−
k
2
−
a
,
k
1
−
k
−
2
)
=
k
1
−
2
k
2
+
a
2
=
k
1
−
k
2
−
a
1
=
k
1
−
k
2
2
⇒
k
1
−
2
k
2
+
a
=
2
k
1
−
2
k
2
−
2
a
⇒
3
a
=
k
1
∴
2
k
1
−
2
k
2
−
2
a
=
k
1
−
k
2
⇒
2
k
1
−
k
1
−
k
2
=
2
a
⇒
k
1
−
k
2
=
2
a
⇒
k
2
=
a
∴
A
(
3
a
,
2
a
,
3
a
)
B
(
a
,
a
,
a
)
If
x
−
14
l
=
y
−
2
m
=
z
+
1
n
is the equation of the line through (1,2,-1) and (-1,0,1), then (l,m,n) is
Report Question
0%
(-1,0,1)
0%
(1,1,-1)
0%
(1,2,-1)
0%
(0,1,0)
If
A
=
(
1
,
2
,
3
)
,
B
=
(
2
,
10
,
1
)
,
Q
are collinear points and
Q
x
=
−
1
then
Q
z
is
Report Question
0%
−
3
0%
7
0%
−
14
0%
−
7
The direction cosines to two lines at right angles are (1,2,3) and (-2,
1
2
,
1
3
), then the direction cosine perpendicular to both given lines are:
Report Question
0%
√
25
2198
,
√
19
2198
,
√
729
2198
0%
√
24
2198
,
√
38
2198
,
√
730
2198
0%
1
3
,-2,
−
7
2
0%
None of the above
Explanation
Direction cosines of
L
1
=
(
1
,
2
,
3
)
Direction cosines of
L
2
=
(
−
2
,
1
2
,
1
3
)
Let direction ratios of required line be a,b,c then
a
+
2
b
+
3
c
=
0
→
1
and
−
2
a
+
b
2
+
c
3
=
0
⇒
−
12
a
+
3
b
+
2
c
=
0
→
2
Solving
1
and
2
a
+
2
b
+
3
c
=
0
−
12
a
+
3
b
+
2
c
=
0
⇒
a
4
−
9
=
b
−
36
−
2
=
c
3
+
24
⇒
a
+
5
=
b
+
38
=
c
−
27
⇒
a
=
+
5
,
b
=
+
38
,
c
=
−
27
∴
Direction cosines are
5
√
5
2
+
38
2
+
(
27
)
2
,
38
√
5
2
+
38
2
+
(
−
27
)
2
,
−
27
√
5
2
+
38
2
+
(
−
27
)
2
5
√
2198
,
38
√
2198
,
−
27
√
2198
The direction cosines of a vector
→
A
are
cos
α
=
4
5
√
2
,
cos
β
=
1
√
2
,
cos
γ
=
3
5
√
2
then, the vector
→
A
is
Report Question
0%
4
ˆ
i
+
ˆ
j
+
3
ˆ
k
0%
4
ˆ
i
+
5
ˆ
j
+
3
ˆ
k
0%
4
ˆ
i
−
5
ˆ
j
+
3
ˆ
k
0%
4
ˆ
i
−
ˆ
j
−
3
ˆ
k
If
ˉ
a
,
ˉ
b
,
ˉ
c
are non-coplaner vector , then the vectors
2
ˉ
a
−
4
ˉ
b
+
4
ˉ
c
,
ˉ
a
−
2
ˉ
b
+
4
ˉ
c
and
−
ˉ
a
+
2
ˉ
b
+
4
ˉ
c
are parellel.
Report Question
0%
True
0%
False
The angle between the lines
x
−
2
3
=
y
+
1
−
2
,
z
=
2
and
x
−
1
1
=
2
y
+
3
3
=
z
+
5
2
is equal to
Report Question
0%
π
/
2
0%
π
/
3
0%
π
/
6
0%
none of these
The direction ratios of the line joining the points
(
4
,
3
,
−
5
)
and
(
−
2
,
1
,
−
8
)
are
Report Question
0%
6
7
,
2
7
,
3
7
0%
6
,
2
,
3
0%
5
,
8
,
0
0%
3
,
7
,
9
Explanation
Given points
P
(
4
,
3
,
−
5
)
Q
(
−
2
,
1
,
−
8
)
Direction ratios of
P
Q
=
Position vector of
P
−
Position vector of
Q
=
4
−
(
−
2
)
,
3
−
1
,
−
5
−
(
−
8
)
=
6
,
2
,
3
If
(
1
2
,
1
3
,
n
)
are the direction cosines of a line then the value of
n
is
Report Question
0%
√
23
6
0%
23
6
0%
2
3
0%
3
2
Explanation
Given direction cosines are
1
2
,
1
3
,
n
we have that if
l
,
m
,
n
are direction cosines of a line then
l
2
+
m
2
+
n
2
=
1
1
4
+
1
9
+
n
2
=
1
⇒
n
2
=
1
−
(
13
36
)
=
23
36
n
=
√
23
6
The angle between the pair of lines with direction ratios (1, 1, 2) and
(
√
3
−
1
,
−
√
3
−
1
,
4
)
is
Report Question
0%
30
o
0%
45
o
0%
60
o
0%
90
o
Explanation
c
o
s
θ
=
1
(
√
3
−
1
)
+
1
(
−
√
3
−
1
)
+
2
(
4
)
√
1
2
+
1
2
+
2
2
√
(
√
3
−
1
)
2
+
(
−
√
3
−
1
)
2
+
4
2
=
√
3
−
1
−
√
3
−
1
+
8
√
6
.
√
4
−
2
√
3
+
4
+
2
√
3
+
16
=
6
√
6
.
√
24
=
6
12
=
1
2
θ
=
60
o
A line makes angles
α
,
β
,
γ
,
δ
with the four diagonals of a cube then
cos
2
α
+
cos
2
β
+
cos
2
γ
+
cos
2
δ
is equal to
Report Question
0%
1
0%
4
/
3
0%
3
/
4
0%
4
/
5
Explanation
REF.Image
Take 'O' as a corner
O
A
,
O
B
,
O
C
are 3 edges through
the axes
Let
O
A
=
O
B
=
O
C
=
a
coordinates of
O
=
(
o
,
o
,
o
)
A
(
a
,
o
,
o
)
B
(
o
,
a
,
o
)
C
(
o
,
o
,
a
)
P
(
a
,
a
,
o
)
L
(
o
,
a
,
a
)
M
(
a
,
o
,
a
)
N
(
a
,
a
,
o
)
The four diagonals
O
P
,
A
L
,
B
M
,
C
N
Direction cosine of
O
P
:
a
−
o
,
a
−
o
,
a
−
o
=
a
,
a
,
a
=
1
,
1
,
1
Direction cosine of
A
L
:
o
−
a
,
a
−
o
,
a
−
o
=
−
a
,
a
,
a
=
−
1
,
1
,
1
Direction cosine of
B
M
:
a
−
o
,
o
−
a
,
a
−
o
=
a
,
−
a
,
a
=
1
,
−
1
,
1
Direction cosine of
C
N
:
a
−
o
,
a
−
o
,
o
−
a
=
a
,
a
,
−
a
=
1
,
1
,
−
1
∴
DC's of OP are
1
√
3
,
1
√
3
,
1
√
3
DC's of AL are
−
1
√
3
,
1
√
3
,
1
√
3
DC's of BM are
1
√
3
,
−
1
√
3
,
1
√
3
DC's of CN are
1
√
3
,
1
√
3
,
−
1
√
3
Let l,m,n be dc's of line and line makes angle
α
with OP :-
cos
α
=
l
(
1
√
3
)
+
m
(
1
√
3
)
+
n
(
1
√
3
)
=
l
+
m
+
n
√
3
Similarly
cos
β
=
−
l
+
m
+
n
√
3
cos
δ
=
l
+
m
−
n
√
3
cos
γ
=
l
−
m
+
n
√
3
suaring and adding all the four
i.e ;
cos
2
α
+
cos
2
β
+
cos
2
γ
+
cos
2
δ
= \dfrac{1}{3}[(l+m+n)^{2}+(-l+m+n)^{2}+(l-m+n)^{2}+(l+m-n)^{2}]
=\dfrac{1}{3}[4l^{2}+4m^{2}+4n^{2}]=\dfrac{4}{3}(l^{2}+m^{2}+n^{2})
[\because l^{2}+m^{2}+n^{2}=1]=\dfrac{4}{3}
\therefore \cos^{2}\alpha +\cos^{2}\beta +\cos^{2}\gamma +\cos^{2}\delta =\dfrac{4}{3}
The direction ratios of the line, given by the planes x - y + z - 5 = 0, x - 3y - 6 = 0 are
Report Question
0%
(3, 1, -2)
0%
(2, -4, 1)
0%
(1,-1, 1)
0%
(0,2,1)
Explanation
Given,
x – y + z – 5 = 0 = x – 3y – 6
\Rightarrow x – y + z – 5 = 0
x – 3y – 6 = 0
\Rightarrow x – y + z – 5 = 0
(1)
x = 3y + 6
(2)
From (1) and (2) we get,
3y + 6 – y + z – 5 = 0
2y + z + 1 = 0
y =\dfrac{-z-1}{2}
Also,
y=\dfrac{x-6}{3}
From (2)
\therefore \dfrac{x-6}{3}=y=\dfrac{-z-1}{2}
So, the given equation can be re-written as
\dfrac{x-6}{3}=\dfrac{y}{1}=\dfrac{z+1}{-2}
Hence the direction ratios of the given line are proportional to
3,1,-2
If
\overline { O A } = 3 \overline { i } + \overline { j } - \overline { k }
,
| \overline { A B } | = 2 \sqrt { 6 }
and AB has the direction ratios 1, -1 , 2 then
| O B | =
Report Question
0%
\sqrt { 35 }
0%
\sqrt { 41 }
0%
\sqrt { 26 }
0%
\sqrt { 55 }
Explanation
\vec{OA}=3\hat{i}+\hat{j}-\hat{k}
|\vec{AB}|=2\sqrt{6}
has dr's
(1, -1, 2)
Unit vector with above dr's are
\vec{AB}=\left(\dfrac{1}{\sqrt{6}}\hat{i}-\dfrac{1}{\sqrt{6}}\hat{j}+\dfrac{2}{\sqrt{6}}\hat{k}\right)(2\sqrt{6})
\vec{AB}=2\hat{i}-2\hat{j}+4\hat{k}
\vec{AB}=\vec{OB}-\vec{OA}
\vec{OB}=(2\hat{i}-2\hat{j}+4\hat{k})+(3\hat{i}+\hat{j}-\hat{k})
\vec{OB}=(5\hat{i}-\hat{j}+3\hat{k})
|OB|=\sqrt{25+1+9}=\sqrt{35}
.
The direction cosines of a vector A are
\cos { \alpha } =\frac { 4 }{ 5\sqrt { 2 } } ,
cos \beta =\frac { 1 }{ \sqrt { 2 } } ,
and
cos \gamma = \frac{ 3 }{ 5\sqrt { 2 } } ,
then vector A is
Report Question
0%
4i+j+3k
0%
4i+5j+3k
0%
4i-5j-3k
0%
none
Explanation
\vec{A} = a\hat{i} + b\hat{j} + c\hat{k}
\cos \alpha = \dfrac{a}{\sqrt{a^2+ b^2+c^2}}
\cos \beta = \dfrac{b}{\sqrt{a^2 + b^2 + c^2}}
\cos \gamma = \dfrac{c}{\sqrt{a^2 + b^2 + c^2}}
\sqrt{4^2 + 5^2 + 3^2} = \sqrt{16 +25+ 9} = \sqrt{50} = 5\sqrt{2}
\vec{A} = 4\hat{i} + 5\hat{j} + 3\hat{k}
The vector
a = \alpha 1 + 2 j + \beta k
lies in the plane of the vectors
b = i + jt
and
c = j + k
and bisects the angle between
b
and
c
. Then which one of the following gives possible values
\alpha
and
\beta
.
Report Question
0%
\alpha = 1 , \beta = 2
0%
\alpha = 2 , \beta = 1
0%
\alpha = 1 , \beta = 1
0%
\alpha = 2 , \beta = 2
Let
l_{1},\ m_{1},\ n_{1};\ l_{2},\ m_{2},\ n_{2};\ l_{3},\ m_{3},\ n_{3}
be the direction cosines of three mutually perpendicular line then
\begin{vmatrix} { l }_{ 1 } & m_{ 1 } & n_{ 1 } \\ { l }_{ 2 } & m_{ 2 } & n_{ 2 } \\ { l }_{ 3 } & m_{ 3 } & n_{ 3 } \end{vmatrix}
Report Question
0%
0
0%
\pm 1
0%
\pm 2
0%
\pm \dfrac{1}{2}
The point where
\vec{ x }
which is perpendicular to
(2,-3,1)
and
(1,-2,3)
and which satisfies the condition
\vec { x } \cdot ( \hat { i } + 2 \hat { j } - 7 \hat{ k } ) = 10
Report Question
0%
\left(3,5,1\right)
0%
\left(7,-5,1\right)
0%
\left(3,-5,1\right)
0%
\left(7,5,1\right)
Explanation
\bar{x} = (a , b, c)
\bar{x} \bot (2, -3, 1) \, \& \, x \bot (1, -2 , 3)
\Rightarrow x .(2, -3, 1) = 0
&
x. (1, -2, 3) = 0
\Rightarrow 2a - 3b + c = 0
__(1) &
a - 2b + 3c = 0
__(2)
\bar{x} . (\hat{i} + 2 \hat{j} - 7 \hat{k} ) = 10
\Rightarrow a + 2b - 7c = 10
__(3)
\left.\begin{matrix}(1) \Rightarrow c = 3b - 2a\\(2) \Rightarrow a = 2b - 3c \end{matrix}\right\}
Put in (3)
a + 2b - 7c = 10
2b - 3c + 2b - 7c = 10
\Rightarrow 4b - 10 c = 10
__(4)
(1)
\Rightarrow a = \dfrac{-c + 3b}{2}
a = \dfrac{3b - c}{2} = 2b - 3c
\Rightarrow 3b - c = 4b - 6c
\Rightarrow b = 5c
__(5)
from (4) & (5)
c = 1 , b = 5, a = 7
\Rightarrow \bar{x} = (7, 5 , 1)
The equation of the plane through
\left(0,-5,1\right)
which is perpendicular to the planes
2x+4y+2z+3=0
,
2x+5y+3z+4=0
is
Report Question
0%
x+y+z=6
0%
x-y+z=6
0%
x-y-z=6
0%
x+y+z+6=0
If
A(p,q,r)
and
B=(p\prime ,q\prime ,r\prime )
are two points on the line
\lambda x=\mu y=yz
such that
OA=3,OB=4
then
pp\prime +qq\prime +rr\prime
is equal to
Report Question
0%
7
0%
12
0%
5
0%
None
of
these
The angle between the lines whose de's satisfy the equation
l+m+m=0
and
l^2+m^{2}-n^{2}=0
is
Report Question
0%
\dfrac{\pi}{6}
0%
\dfrac{\pi}{2}
0%
\dfrac{\pi}{3}
0%
\dfrac{\pi}{4}
Explanation
Given that the equations
l+m+n=0
………..
(1)
l+m=-n
\Rightarrow -(l+m)=n
and
l^2+m^2+n^2=0
……….
(2)
Put the value of n in equation
(2)
l^2+m^2+n^2=0
\Rightarrow l^2+m^2-(-(l+m))^2=0
\Rightarrow l^2+m^2-(l^2+m^2-2ml)=0
\Rightarrow l^2+m^2-l^2-m^2+2ml=0
\Rightarrow 2ml0
\Rightarrow ml=0
\Rightarrow m=0, l=0
Let us put
m=0
in equation
(3)
l+o+n=0
l=-n
Hence, direction rates
(l, m, o)=(1, 0, -1)
Let us put
l=0
, we get
m=-n
Here, direction ratios
(l, m, n)=(0, 1, -1)
we know that,
\cos\theta =\dfrac{\vec{b_1}\cdot \vec{b_2}}{|\vec{b_1}||\vec{b_2}|}
=\dfrac{(1, 0, -1)\cdot (0, 1, -1)}{\sqrt{1^2+0^2+(-1)^2}\sqrt{0^2+1^2+(-1)^2}}
=\dfrac{1}{\sqrt{2}\sqrt{2}}
\cos\theta =\dfrac{1}{2}
\cos\theta =\cos\dfrac{\pi}{3}
\theta =\dfrac{\pi}{3}
Hence, this is the answer.
The angle between the lines, whose direction ratios are
1,1,2
and
\sqrt { 3 } - 1 , - \sqrt { 3 } - 1,4 ,
is
Report Question
0%
{45} ^ { \circ }
0%
{30} ^ { \circ }
0%
{60} ^ { \circ }
0%
{90} ^ { \circ }
Explanation
The direction ratios are
(1, 1, 2)
and
(\sqrt{3}-1, -\sqrt{3}-1, 4)
Angle between them is given by
\cos\theta =\dfrac{1(\sqrt{3}-1)+1(-\sqrt{3}-1)+2/4)}{\sqrt{1^2+1^2+2^2}\sqrt{(\sqrt{3}-1)^2+(-\sqrt{3}-1)^2+4^2}}
=\dfrac{\sqrt{3}-1-\sqrt{3}-1+8}{\sqrt{6}\sqrt{8+16}}
=\dfrac{6}{\sqrt{6}\sqrt{24}}=\dfrac{6}{\sqrt{144}}
=\dfrac{6}{12}
=\dfrac{1}{2}
\cos\theta =\dfrac{1}{2}
\Rightarrow \theta =\dfrac{\pi}{3}=60^o
.
The directions cosines of the line which is perpedicular to the lines whose direction cosines are proportional to (1, -1, 2) and (2,-1,-1) are:
Report Question
0%
\dfrac { 1 }{ \sqrt { 35 } } ,-\dfrac { 5 }{ \sqrt { 35 } } \frac { 3 }{ \sqrt { 35 } }
0%
-\dfrac { 1 }{ \sqrt { 35 } } ,\frac { 5 }{ \sqrt { 35 } } \dfrac { 3 }{ \sqrt { 35 } }
0%
\dfrac { 1 }{ \sqrt { 35 } } ,\dfrac { 5 }{ \sqrt { 35 } } \frac { 3 }{ \sqrt { 35 } }
0%
None of these
If a plane passes through the point
(1, 1, 1)
and is perpendicular to the line
\dfrac{x-1}{3}=\dfrac{y-1}{0}=\dfrac{z-1}{4}
then its perpendicular distance from the origin is
Report Question
0%
\dfrac{3}{4}
0%
\dfrac{4}{3}
0%
\dfrac{7}{5}
0%
1
Explanation
\begin{array}{l} Let\, eq{ u^{ n } }\, of\, plane\, be\, :\, a\left( { x-1 } \right) +b\left( { y-1 } \right) +\left( { z-1 } \right) =0 \\ 3\left( { x-1 } \right) +0\left( { y-1 } \right) +4\left( { z-1 } \right) =0 \\ 3x+4z=7 \\ perpendicular\, \, dis\tan ce\, from\, \, origin\, \, d=\frac { 7 }{ 5 } \\ Hence,\, the\, option\, C\, is\, correct\, answer. \end{array}
The direction cosines of a line equally inclined to three mutually perpendicular lines having direction cosines as
l_{1},m_{1},n_{1};l_{2},m_{2},n_{2}
and
l_{3},m_{3},n_{3}
are
Report Question
0%
l_{1}+ l_{2}+ l_{3},m_{1}+m_{2}+m_{3},n_{1}+n_{2}+n_{3}
0%
\dfrac{l_{1}+l_{2}+l_{3}}{\sqrt{3}},\dfrac{m_{1}+m_{2}+m_{3}}{\sqrt{3}},\dfrac{n_{1}+n_{2}+n_{3}}{\sqrt{3}}
0%
\dfrac{l_{1}+l_{2}+l_{3}}{3},\dfrac{m_{1}+m_{2}+m_{3}}{3},\dfrac{n_{1}+n_{2}+n_{3}}{3}
0%
None\ of\ these
l=m=n=1
represent the direction cosines of the
Report Question
0%
x-
axis
0%
y-
axis
0%
z-
axis
0%
none\ of\ these
Explanation
l=m=n=1
(1,1,1)
represent direction ratio of line equally inclined to
x,y\, \&\,zaxis
Hence,
option
D
is correct answer.
If the points (p. 0), (0, q) and (1, 1) are collinear then
\dfrac { 1 }{ p } +\dfrac { 1 }{ q }
is equal to
Report Question
0%
-1
0%
1
0%
2
0%
0
Explanation
If the area of triangle is zero, then the points are collinear.
Points are collinear.
Point are
(p, o)(o, q)(i,q)
\Delta =\dfrac {1}{2}[p(q-1)-0(1-0)+1(0-q)]
\Rightarrow \dfrac{1}{2}[p(q-1)-q]
\Rightarrow \dfrac{1}{2}[p(q-1)-q]
\Rightarrow \dfrac{1}{2}[pq-(p+q)]=o
\Rightarrow pq=p+q\Rightarrow \dfrac {p+q}{pq}=1
\Rightarrow \dfrac{1}{p}+\dfrac{1}{q}=1
The direction ratios of the line
x-y+z-5=\quad 0\quad =\quad x-3y-6\quad are
Report Question
0%
3,1,-2
0%
2,-4,1
0%
\frac { 3 }{ \sqrt { 14 } } ,\frac { 1 }{ \sqrt { 14 } } ,\frac { -2 }{ \sqrt { 14 } }
0%
\frac { 2 }{ \sqrt { 41 } } ,\frac { -4 }{ \sqrt { 41 } } ,\frac { 1 }{ \sqrt { 41 } }
Explanation
x-y+z-5=0=x-3y-6
x-y+z-5=0
--(I)
x-3y-6=0
--(II)
The Direction ratios of the line is
(\hat{i} - \hat{j} + \hat{k})\times(\hat{i} -3\hat{j})
=
3\hat{i}+\hat{j}-2\hat{k}
Hence,the direction ratios is (3,1,-2).
Hence, Option A is correct
The direction cosines of a line equally inclined to three mutually perpendicular lines having direction cosines as
l_1
,
m_1
,
n_1
:
l_2
,
m_2
,
n_2
and
l_3
,
m_3
,
n_3
are
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l_1
+
l_2
+
l_3
,
m_1
+
m_2
+
m_3
,
n_1
+
n_2
+
n_3
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\dfrac{l_1+l_2+l_3}{\sqrt{3}}
,
\dfrac{m_1+m_2+m_3}{\sqrt{3}}
,
\dfrac{n_1+n_2+n_3}{\sqrt{3}}
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\dfrac{l_1+l_2+l_3}{3}
,
\dfrac{m_1+m_2+m_3}{3}
,
\dfrac{n_1+n_2+n_3}{3}
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None of these
If
A(3\hat { i } +2\hat { j } +3\hat { k } ),B(-\hat { i } -\hat { j } +8\hat { k } ),C(-4\hat { i } +4\hat { j } +6\hat { k } )
are the vertices of a triangle then the equation of the line passing through the circumcentre and parallel to
\vec { A B }
is
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\hat { r } =\left( -\dfrac { 4 }{ 3 } \hat { i } +\dfrac { 5 }{ 3 } \hat { j } +\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )
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\hat { r } =\left( \dfrac { 4 }{ 3 } \hat { i } +\dfrac { 5 }{ 3 } \hat { j } +\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )
0%
\hat { r } =\left( -\dfrac { 4 }{ 3 } \hat { i } +\dfrac { 5 }{ 3 } \hat { j } -\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )
0%
\hat { r } =\left( \dfrac { 4 }{ 3 } \hat { i } -\dfrac { 5 }{ 3 } \hat { j } +\dfrac { 17 }{ 3 } \hat { k } \right) +t(2\hat { i } +3\hat { j } -5\hat { k } )
The cartesian from of equation a line passing through the point position vector
2\hat{i}-\hat{j}+2\hat{k}
and is in the direction of
-2\hat{i}+\hat{j}+\hat{k}
, is
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\dfrac{x-2}{-2}=\dfrac{y+1}{1}=\dfrac{z-2}{1}
0%
\dfrac{x+4}{-2}=\dfrac{y-1}{1}=\dfrac{z+2}{1}
0%
\dfrac{x+2}{4}=\dfrac{y-1}{-1}=\dfrac{z-1}{2}
0%
None \ of \ these
Explanation
Equation of a line passing through a point with position vector
\vec{a}
and parallel to
\vec{b}
is
\vec{r}=\vec{a}+\lambda\vec{b}
Here
\vec{a}= 2\hat{i}−\hat{j}+2\hat{k}
and
\vec{b}=−2\hat{i}+\hat{j}+\hat{k}
So,
\vec{r}=2\hat{i}−\hat{j}+2\hat{k}+\lambda\left(−2\hat{i}+\hat{j}+\hat{k}\right)
Equation of the line in vector form is
2\hat{i}−\hat{j}+2\hat{k}+\lambda\left(−2\hat{i}+\hat{j}+\hat{k}\right)
Since the line passes through a point with position vector
2\hat{i}−\hat{j}+2\hat{k}
\therefore {x}_{1}=2,\,{y}_{1}=-1,\,{z}_{1}=2
Also, line is in the direction of
−2\hat{i}+\hat{j}+\hat{k}
Direction ratios :
a=-2,\,b=1,\,c=1
Equation of line in cartesian form is
\dfrac{x-2}{-2}=\dfrac{y+1}{1}=\dfrac{z-2}{1}
If
\cos { \alpha ,\quad \cos { \beta ,\quad \cos { \gamma } } }
are the direction cosine of a line, then find the value of
{ cos }^{ 2 }\alpha +\left( \cos { \beta +\sin { \gamma } } \right)
\left( \cos { \beta - \sin { \gamma } } \right)
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2
0%
0
0%
-1
0%
1
\dfrac { x - 2 } { 1 } = \dfrac { y - 3 } { 1 } = \dfrac { z - 4 } { - 1 }
&
\dfrac { x - 1 } { k } = \dfrac { y - 4 } { 2 } = \dfrac { z - 5 } { 2 }
are coplanar then k=?
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any value
0%
exactly one value
0%
exactly
2
values
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exactly
3
values
Explanation
If this two lines are co-planar
\begin{vmatrix} 1& 1&-1 \\ k & 2 & 2\\ 2-1&3-4&4-5\end{vmatrix}=0
\begin{vmatrix} 1&1&1 \\k&2 &2\\ 1&-1&-1\end{vmatrix}=0
1(-2+2)-1(-k-2)-1(-k-2)=0
k + 2+ k + 2 = 0
k = -2
The plane through (1, 1, 1) (1, -1, 1) and (-7, -3, -5) is
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0%
Parallel to x-axis.
0%
Parallel to y-axis.
0%
Perpendicular to y-axis.
0%
Perpendicular to x-axis.
Direction ratio of line given by
\dfrac { x-1 }{ 3 } =\dfrac { 6-2y }{ 10 } =\dfrac { 1-z }{ -7 }
are:
Report Question
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<3,10,-7>
0%
<3,-5,7>
0%
<3,5,7>
0%
<3,5,-7>
Explanation
\cfrac{x-1}{3} = \cfrac{6 - 2y}{10} = \cfrac{1 - z}{-7}
\cfrac{x-1}{3} = \cfrac{2y - 6}{-10} = \cfrac{z - 1}{- \left( -7 \right)}
\cfrac{x - 1}{3} = \cfrac{y - 3}{\left( \cfrac{-10}{2} \right)} = \cfrac{z - 1}{7}
\cfrac{x-1}{3} = \cfrac{y - 3}{-5} = \cfrac{z-1}{7}
Therefore,
Direction ratios are
3, -5, 7
A normal to the plane
x=2
is...
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0%
(0,1,1)
0%
(2,0,2)
0%
(1,0,0)
0%
(0,1,0)
Explanation
The plane
x=2
is perpendicular to
x axis
So the angle is
\dfrac{\pi}{2},\cos \dfrac{\pi}{2}=0
The plane
x=2
is parallel to both
y axis
and
z axis
So the angle is
0,\cos 0=1
So the Drs are
(0,1,1)
0:0:2
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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