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CBSE Questions for Class 11 Engineering Maths Introduction To Three Dimensional Geometry Quiz 10 - MCQExams.com
CBSE
Class 11 Engineering Maths
Introduction To Three Dimensional Geometry
Quiz 10
The distance of P(2,-3) from the x-axis is......
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0%
2
0%
-3
0%
3
0%
√
13
If the distance between the points (2,-3) and (5,b) is 5 , then b=.........
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0%
1
0%
2
0%
7
0%
5
Find the centroid of a triangle, mid-points of whose sides are
(
!
,
2
,
−
3
)
,
(
3
,
0
,
1
)
and
(
−
1
,
1
,
−
4
)
Report Question
0%
(
−
1
,
−
1
,
−
2
)
0%
(
−
1
,
2
,
−
2
)
0%
(
−
1
,
1
,
−
2
)
0%
(
1
,
1
,
−
2
)
The point in the
x
y
−
plane which is equidistant from
(
2
,
0
,
3
)
,
(
0
,
3
,
2
)
and
(
0
,
0
,
1
)
is
Report Question
0%
(
1
,
2
,
3
)
0%
(
−
3
,
2
,
0
)
0%
(
3
,
−
2.
0
)
0%
(
3
,
2
,
0
)
0%
(
3
,
2
,
1
)
Explanation
Let the points are
A
(
2
,
0
,
3
)
,
B
(
0
,
3
,
2
)
and
D
(
0
,
0
,
1
)
We know that Z-coordinate of every point an xy-plane is zero so let
p
(
x
,
y
,
0
)
be a point on xy-plane such that
P
A
=
P
B
=
P
C
Now,
P
A
=
P
B
⇒
P
A
2
=
P
B
2
⇒
(
x
−
2
)
2
+
(
y
−
0
)
2
+
(
0
−
3
)
2
=
(
x
−
0
)
2
+
(
y
−
3
)
2
+
(
0
−
2
)
⇒
4
x
−
6
y
=
0
⇒
2
x
−
3
y
=
0
......(i)
and ,
P
B
=
P
C
⇒
P
B
2
=
P
C
2
⇒
(
x
−
0
)
2
+
(
y
−
3
)
2
(
0
−
2
)
2
=
(
x
−
0
)
2
+
(
y
−
0
)
2
+
(
0
−
1
)
2
⇒
−
6
y
+
12
=
0
⇒
y
=
2
......(ii)
Putting
y
=
2
in equation (i), we get
x
=
3
Hence, the required point is
(
3
,
2
,
0
)
The foot of the perpendicular from the point
A
(
7
,
14
,
5
)
to the plane
2
x
+
4
y
−
z
=
2
is?
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0%
(
3
,
1
,
8
)
0%
(
1
,
2
,
8
)
0%
(
3
,
−
3
,
5
)
0%
(
5
,
−
3
,
−
4
)
Explanation
Let N be the foot of the perpendicular drawn from the point
A
(
7
,
14
,
5
)
and perpendicular to the plane
2
x
+
4
y
−
z
=
2
Then, the equation of the line PN is
x
−
7
2
=
y
−
14
4
=
z
−
5
−
1
=
λ
(say)
Let the coordinates of N be
N
(
2
λ
+
7
,
4
λ
+
14
,
−
λ
+
5
)
Since N lies on the plane
2
x
+
4
y
−
z
=
2
, so
2
(
2
λ
+
7
)
+
4
(
4
λ
+
14
)
−
(
−
λ
+
5
)
=
2
⇒
21
λ
=
−
63
⇒
λ
=
−
3
∴
required foot of the perpendicular is
N
(
−
6
+
7
,
−
12
+
14
,
3
+
5
)
, i.e.,
N
(
1
,
2
,
8
)
.
The coordinates of the point where the line through the points
A
(
5
,
1
,
6
)
and
B
(
3
,
4
,
1
)
crosses the yz-plane is
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0%
(
0
,
17
,
−
13
)
0%
(
0
,
−
17
2
,
13
2
)
0%
(
0
,
17
2
,
−
13
2
)
0%
none of these
Explanation
Equation of the line passing through the points
A
(
5
,
1
,
6
)
and
B
(
3
,
4
,
1
)
is
x
−
5
2
=
4
−
1
−
3
=
7
−
6
5
....(i)
Let the line (i) cross the yz plane at point
(
o
,
h
,
k
)
Then
(
o
,
h
,
k
)
point lie on the line (I)
So,
0
−
5
2
=
h
−
1
−
3
=
k
−
6
5
Then,
k
−
6
5
=
−
5
2
or,
k
−
6
=
−
25
2
or,
k
=
6
−
−
25
2
=
12
−
25
2
=
−
13
2
Again,
h
−
1
−
3
=
−
5
2
or,
h
−
1
=
15
2
or,
h
=
15
2
+
1
=
17
2
Therefore the crossing points is
(
0
,
17
2
,
−
13
2
)
The point on the line
x
−
2
1
=
y
+
3
−
2
=
z
+
5
−
2
at a distance of 6 from the point
(
2
,
−
3
,
−
5
)
is
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0%
(
3
,
−
5
,
−
3
)
0%
(
4
,
−
7
,
−
9
)
0%
(
0
,
2
,
−
1
)
0%
(
−
3
,
5
,
3
)
The distance between two points P and Q is d and the length of their projections of PQ on the coordinate planes are
d
1
,
d
2
,
d
3
. Then
d
2
1
+
d
2
2
+
d
2
3
=
k
d
2
where K is
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0%
1
0%
5
0%
3
0%
2
The shortest distance between the lines
x
−
3
2
=
y
+
15
−
7
=
z
−
9
5
and
x
+
1
2
=
y
−
1
1
=
z
−
9
−
3
is
Report Question
0%
2
√
3
0%
4
√
3
0%
3
√
6
0%
5
√
6
The ratio in which the plane
→
r
⋅
(
→
i
−
2
→
j
+
3
→
k
)
=
17
divides the line joining the points
−
2
→
i
+
4
→
j
+
7
→
k
and
3
→
i
−
5
→
j
+
8
→
k
Report Question
0%
1 : 5
0%
1 : 10
0%
3 : 5
0%
3 : 10
Explanation
Let the plane
→
r
⋅
(
→
i
−
2
→
j
+
3
→
k
)
=
17
divided the line joining the points
−
2
→
i
+
4
→
j
+
7
→
k
and
3
→
i
−
5
→
j
+
8
→
k
in the ratio
t
:
1
at point P.
Therefore, point P is
3
t
−
2
t
+
1
→
i
+
−
5
t
+
4
t
+
1
→
j
+
8
t
+
7
t
+
1
→
k
This lies on the given plane
∴
3
t
−
2
t
+
1
(
1
)
+
−
5
t
+
4
t
+
1
(
−
2
)
+
8
t
+
7
t
+
1
(
3
)
=
17
Solving, we get
t
=
3
10
The point on the line
x
−
2
1
=
y
+
3
−
2
=
z
+
5
−
2
at a distance of 6 from the point
(
2
,
−
3
,
−
5
)
is
Report Question
0%
(
3
,
−
5
,
−
3
)
0%
(
4
,
−
7
,
−
9
)
0%
(
0
,
2
,
−
1
)
0%
(
−
3
,
5
,
3
)
Explanation
b. Direction cosines of the given line are
1
3
,
−
2
3
,
−
2
3
Hence, the equation of line can be point in the form
x
−
2
1
/
3
=
y
+
3
−
2
/
3
=
−
z
+
5
−
2
/
3
=
r
Therefore, any point on the line is
(
2
+
r
3
,
−
3
−
2
r
3
,
−
5
−
2
r
3
)
, where
r
=
±
6
.
Points are
(
4
,
−
7
,
−
9
)
and
(
0
,
1
,
−
1
)
The points
A
(
1
,
2
,
−
1
)
,
B
(
2
,
5
,
−
2
)
,
C
(
4
,
4
,
−
3
)
and
D
(
3
,
1
,
−
2
)
are
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0%
collinear
0%
vertices of a rectangle
0%
vertices of a square
0%
vertices of a rhombus
Explanation
AB
=
√
(
2
−
1
)
2
+
(
5
−
2
)
2
+
(
−
2
+
1
)
2
AB
=
√
(
1
)
2
+
(
3
)
2
+
(
−
1
)
2
AB
=
√
11
Similarly you find that BC
=
√
6
CD
=
√
11
and DA
=
√
6
Hence opposite sides of quadrilateral are equal, Now we check the diagonals
AC
=
√
(
4
−
1
)
2
+
(
4
−
2
)
2
+
(
−
3
+
1
)
2
AC
=
√
17
similarly BD
=
√
17
Diagonals are not equal
direction ratio of line passing through AB is (1,3,-1)
direction ratio of line passing through BC is (2,-1,-1), As the dot product dr of AB and BC are equal to 0 which means AB is perpendicular to BC,similarly check for others sides too
opposite sides are equal and diagonal are equal
hence it is rectangle
Let
A
(
2
ˆ
i
+
3
ˆ
j
+
5
ˆ
k
)
B
(
−
ˆ
i
+
3
ˆ
j
+
2
ˆ
k
)
and
C
(
λ
ˆ
i
+
5
ˆ
j
+
μ
ˆ
k
)
are vertices of a triangle and its median through
A
is equally inclined to the positive directions of the axes. The value of
λ
+
μ
is equal to
Report Question
0%
−
7
0%
2
0%
7
0%
17
Explanation
We have,
→
A
D
=
1
2
(
→
b
+
→
c
−
2
→
a
)
→
A
D
=
λ
−
5
2
ˆ
i
+
ˆ
j
+
μ
−
8
2
ˆ
k
The line is equally inclined to the axes. Hence, its direction ratios must be of the form
1
:
1
:
1
.
Hence,
λ
=
7
and
μ
=
10
If
(
0
,
b
,
0
)
is the centroid of the triangle formed by the points
(
4
,
2
,
−
3
)
,
(
a
,
−
5
,
1
)
and
(
2
,
−
6
,
2
)
. If
a
,
b
are the roots of the quadratic equation
x
2
+
p
x
+
q
=
0
, then
p
,
q
are
Report Question
0%
9
,
18
0%
−
9
,
−
18
0%
3
,
−
18
0%
−
3
,
18
Explanation
Since
a
,
b
are the roots of the equation
x
2
+
p
x
+
q
=
0
⇒
a
+
b
=
−
p
and
a
b
=
q
Centroid of triangle is
(
a
+
6
3
,
−
3
,
0
)
Given, centroid
(
0
,
b
,
0
)
Comparing, we get
b
=
−
3
and
a
+
6
3
=
0
⇒
a
=
−
6
Hence,
p
=
9
,
q
=
18
Hence, option A.
lf
O
A
B
C
is a tetrahedron such that the
O
A
2
+
B
C
2
=
O
B
2
+
C
A
2
=
O
C
2
+
A
B
2
, then which of the following is/are correct
Report Question
0%
A
B
⊥
O
C
0%
O
B
≠
C
A
0%
O
C
=
A
B
0%
A
B
⊥
B
C
Explanation
O
A
B
C
is a tetrahedron such that
O
A
2
+
B
C
2
=
O
B
2
+
C
A
2
=
O
C
2
+
A
B
2
Let
O
(
0
,
0
,
0
)
,
A
(
x
1
,
y
1
,
z
1
)
,
B
(
x
2
,
y
2
,
z
2
)
,
C
(
x
3
,
y
3
,
z
3
)
are coordinates of vertices.
O
A
2
+
B
C
2
=
(
x
2
1
+
y
2
1
+
z
2
1
)
+
(
(
x
3
−
x
2
)
2
+
(
y
3
−
y
2
)
2
+
(
z
3
−
z
2
)
2
)
=
(
x
2
1
+
y
2
1
+
z
2
1
)
+
(
x
2
2
+
y
2
2
+
z
2
2
)
+
(
x
2
3
+
y
2
3
+
z
2
3
)
−
2
(
x
2
x
3
+
y
2
y
3
+
z
2
z
3
)
Similarly,
O
B
2
+
C
A
2
=
(
x
2
1
+
y
2
1
+
z
2
1
)
+
(
x
2
2
+
y
2
2
+
z
2
2
)
+
(
x
2
3
+
y
2
3
+
z
2
3
)
−
2
(
x
1
x
3
+
y
1
y
3
+
z
1
z
3
)
O
C
2
+
A
B
2
=
(
x
2
1
+
y
2
1
+
z
2
1
)
+
(
x
2
2
+
y
2
2
+
z
2
2
)
+
(
x
2
3
+
y
2
3
+
z
2
3
)
−
2
(
x
1
x
2
+
y
1
y
2
+
z
1
z
2
)
∴
O
A
2
+
B
C
2
=
O
B
2
+
C
A
2
⇒
x
2
x
3
+
y
2
y
3
+
z
2
z
3
=
x
1
x
3
+
y
1
y
3
+
z
1
z
3
⇒
(
x
1
−
x
2
)
x
3
+
(
y
1
−
y
2
)
y
3
+
(
z
1
−
z
2
)
z
3
=
0
⇒
(
x
1
−
x
2
)
(
x
3
−
0
)
+
(
y
1
−
y
2
)
(
y
3
−
0
)
+
(
z
1
−
z
2
)
(
z
3
−
0
)
=
0
⇒
→
A
B
⋅
→
O
C
=
0
Hence,
O
C
is perpendicular to
A
B
.
The plane
a
x
+
b
y
+
c
z
+
(
−
3
)
=
0
meet the co-ordinate axes in
A
,
B
,
C
. The centroid of the triangle is
Report Question
0%
(
3
a
,
3
b
,
3
c
)
0%
(
3
a
.
3
b
,
3
c
)
0%
(
a
3
.
b
3
,
c
3
)
0%
(
1
a
.
1
b
,
1
c
)
Explanation
For finding the coordinates of the point where the plane
a
x
+
b
y
+
c
z
−
3
=
0
cuts the
x
axis,
we equate
y
and
z
to zero.
The point becomes
(
3
a
,
0
,
0
)
similarily, the point on y axis becomes
(
0
,
3
b
,
0
)
And that on z axis becomes
(
0
,
0
,
3
c
)
The centroid of the triangle formed by these points would be
(
3
a
+
0
+
0
3
,
0
+
3
a
+
0
3
,
0
+
0
+
3
c
3
)
∴
(
1
a
,
1
b
,
1
c
)
A
B
C
D
is a parallelogram.
L
is a point on
B
C
which divides
B
C
in the ratio
1
:
2
.
A
L
intersects
B
D
at
P
.
M
is a point on
D
C
which divides
D
C
in the ratio
1
:
2
and AM intersects
B
D
in
Q
.
Point
P
divides
A
L
in the ratio
Report Question
0%
1
:
2
0%
1
:
3
0%
3
:
1
0%
2
:
1
Explanation
Given,
B
L
:
L
C
=
1
:
2
and
D
M
:
M
C
=
1
:
2
Now,
B
L
:
B
C
=
1
:
3
Consider
△
A
P
D
and
△
L
P
B
∠
A
D
P
=
∠
L
B
P
∠
A
P
D
=
∠
B
P
L
By AA corollary,
△
A
P
D
∼
△
L
P
B
⇒
Corresponding parts are proportional.
⇒
A
P
L
P
=
A
D
B
L
⇒
A
P
L
P
=
B
C
B
L
(
∵
A
D
=
B
C
)
⇒
A
P
L
P
=
3
:
1
If the vertices of a triangle are
(
−
1
,
6
,
−
4
)
,
(
2
,
1
,
1
)
and
(
5
,
−
1
,
0
)
then the centroid of the triangle is
Report Question
0%
(
6
,
6
,
−
3
)
0%
(
2
,
2
,
−
1
)
0%
(
3
,
3
,
−
3
2
)
0%
none of these
Explanation
the centroid of the triangle is
(
x
1
+
x
2
+
x
3
3
,
y
1
+
y
2
+
y
3
3
,
z
1
+
z
2
+
z
3
3
)
Thus by substituting the vertices we get
=
(
−
1
+
2
+
5
3
,
6
+
1
−
1
3
,
−
4
+
1
+
0
3
)
=
(
6
3
,
6
3
,
−
3
3
)
∴
the centroid of the triangle is
(
2
,
2
,
−
1
)
The coordinates of a point which is equidistant from the point
(
0
,
0
,
0
)
,
(
a
,
0
,
0
)
,
(
0
,
b
,
0
)
and
(
0
,
0
,
c
)
are given by
Report Question
0%
(
a
2
,
b
2
,
c
2
)
0%
(
−
a
2
,
−
b
2
,
c
2
)
0%
(
a
2
,
−
b
2
,
−
c
2
)
0%
(
−
a
2
,
b
2
,
−
c
2
)
Explanation
Let
P
(
x
,
y
,
z
)
be the required point.
Then
O
P
=
P
A
=
P
B
=
P
C
Now
O
P
=
P
A
⟹
O
P
2
+
P
A
2
⟹
x
2
+
y
2
+
z
2
=
(
x
−
a
)
2
+
(
y
−
0
)
2
+
(
z
−
0
)
2
⟹
x
=
a
2
Similarily,
O
P
=
P
B
=
y
=
b
2
and
O
P
=
P
C
=
z
=
c
2
hence the coordinate of the required point are
(
a
2
,
b
2
,
c
2
)
If
A
=
(
0
,
0
,
2
)
,
B
=
(
√
2
,
√
2
,
2
)
,
C
=
(
√
2
,
√
2
,
0
)
and
D
=
(
8
√
2
−
20
17
,
12
√
2
+
4
17
,
20
−
8
√
2
17
)
, then
A
B
C
D
is a
Report Question
0%
rhombus
0%
square
0%
parallelogram
0%
none of these
Explanation
Given points are,
A
=
(
0
,
0
,
2
)
,
B
=
(
√
2
,
√
2
,
2
)
,
C
=
(
√
2
,
√
2
,
0
)
and
D
=
(
8
√
2
−
20
17
,
12
√
2
+
4
17
,
20
−
8
√
2
17
)
Length
A
B
=
√
√
2
2
+
√
2
2
+
(
2
−
2
)
2
=
2
Length
B
C
=
√
(
√
2
−
√
2
)
2
+
(
√
2
−
√
2
)
2
+
2
2
=
2
Length
C
D
=
√
(
8
√
2
−
20
17
−
√
2
)
2
+
(
12
√
2
+
4
17
−
√
2
)
2
+
(
20
−
8
√
2
17
)
2
=
2
Length
A
D
=
√
(
8
√
2
−
20
17
)
2
+
(
12
√
2
+
4
17
)
2
+
(
20
−
8
√
2
17
−
2
)
2
=
2
Angle between two vectors
¯
A
B
=
p
1
ˆ
i
+
q
1
ˆ
j
+
r
1
ˆ
k
=
√
2
ˆ
i
+
√
2
ˆ
j
+
0
ˆ
k
and
¯
B
C
=
p
2
ˆ
i
+
q
2
ˆ
j
+
r
2
ˆ
k
=
0
ˆ
i
+
0
ˆ
j
+
2
ˆ
k
is
c
o
s
θ
=
0
⇒
θ
=
90
o
All sides are equal and angle between
A
B
and
B
C
is
90
o
, Hence, its a square.
The equation of motion of a rocket are:
x
=
2
t
,
y
=
−
4
t
,
z
=
4
t
,
where the time
t
is given in seconds and the coordinate of a moving point in kilometers. At what distance will the rocket be from the starting point
O
(
0
,
0
,
0
)
in
10
seconds ?
Report Question
0%
60
km
0%
30
km
0%
45
km
0%
None of these
Explanation
Eliminating t from the given equation, we get the equation of the path
x
2
=
y
−
4
=
z
4
=
t
Thus the path of the Rocket represents a straight line passing through the origin for
t
=
10
s
e
c
.
we have
x
=
20
,
y
=
−
40
,
z
=
40
Let
→
r
=
x
→
i
+
y
→
j
+
z
→
k
⟹
|
→
r
|
=
√
x
2
+
y
2
+
z
2
=
√
400
+
1600
+
1600
=
60
k
m
What is the distance in space between
(
1
,
0
,
5
)
and
(
−
3
,
6
,
3
)
?
Report Question
0%
4
0%
6
0%
2
√
11
0%
2
√
14
0%
12
Explanation
The distance between the points
(
1
,
0
,
5
)
and
(
−
3
,
6
,
3
)
is given below:
D
=
√
(
3
−
5
)
2
+
(
6
−
0
)
2
+
(
−
3
−
1
)
2
=
√
(
−
2
)
2
+
(
6
)
2
+
(
−
4
)
2
=
√
4
+
36
+
16
=
√
56
=
2
√
14
A triangle
A
B
C
is placed so that the midpoints of its sides are on the
x
,
y
and
z
axes respectively. Lengths of the intercepts made by the plane containing the triangle on these axes are respectively
α
,
β
,
γ
, then the coordinates of the centroid of the triangle
A
B
C
are
Report Question
0%
(
−
α
3
,
β
3
,
γ
3
)
0%
(
α
3
,
−
β
3
,
γ
3
)
0%
(
α
3
,
β
3
,
−
γ
3
)
0%
(
α
3
,
β
3
,
γ
3
)
Explanation
Equation of the plane containing the triangle
A
B
C
is
x
α
+
y
β
+
z
γ
=
1
which meets the axes in
(
α
,
0
,
0
)
,
(
0
,
β
,
0
)
and
(
0
,
0
,
y
)
.
Let the coordinates of
A
be
(
x
1
,
y
1
,
z
1
)
Since the middle point of
A
B
lies on the
z
-axis it is
(
0
,
0
,
γ
)
and thus the coordinates of
B
are
(
−
x
1
,
−
y
1
,
2
γ
−
z
1
)
Similarly the coordinates of
C
are
(
−
x
1
,
2
β
−
y
1
,
−
z
1
)
So that the middle point of
B
C
=
(
−
x
1
,
β
−
y
1
,
γ
−
z
1
)
=
(
α
,
0
,
0
)
⇒
x
1
=
−
α
,
y
1
=
β
,
z
1
=
γ
.
And thus the coordinates of
A
are
(
−
α
,
β
,
γ
)
Similarly the coordinates of
B
are
(
α
,
−
β
,
γ
)
and those of
C
are
(
α
,
β
,
−
γ
)
.
Hence, the coordinates of the centroid of the triangle
A
B
C
are
(
α
3
,
β
3
,
γ
3
)
Point A is
a
+
2
b
,
and a divides AB in the ratio 2 :The position vector of B is
Report Question
0%
2
a
−
b
0%
b
−
2
a
0%
a
−
3
b
0%
b
Explanation
Let us consider x be the position vector of B,
then a divides AB in the ratio
2
:
3
a
=
2
x
+
3
(
a
+
2
b
)
2
+
3
⇒
x
=
a
−
3
b
D
(
2
,
1
,
0
)
,
E
(
2
,
0
,
0
)
,
F
(
0
,
1
,
0
)
are mid point of the sides
B
C
,
C
A
,
A
B
of
Δ
A
B
C
respectively, The the centroid of
Δ
ABC is
Report Question
0%
(
1
3
,
1
3
,
1
3
)
0%
(
4
3
,
2
3
,
0
)
0%
(
−
1
3
,
1
3
,
1
3
)
0%
(
2
3
,
1
3
,
1
3
)
Explanation
Centroid of triangle coincide with the centroid of triangle formed by joing the mid-point of sides of triangle
So, centroid of
△
A
B
C
=
centroid of
△
D
E
F
=
(
2
+
2
+
0
3
,
1
+
0
+
1
3
,
0
+
0
+
0
3
)
=
(
4
3
,
2
3
,
0
)
If
P
(
x
,
y
,
x
)
is a point on the line segment joining
Q
(
2
,
2
,
4
)
and
R
(
3
,
5
,
6
)
such that the projection of
¯
O
P
on the axis are
13
5
,
19
5
,
26
5
,
respectively, then
P
divides
Q
R
in the ratio
Report Question
0%
1
:
2
0%
3
:
2
0%
2
:
3
0%
1
:
3
Explanation
Since
O
P
has projection
13
5
,
19
5
and
26
5
on the coordinate axis,
therefore
O
P
=
13
5
→
i
+
19
5
→
j
+
26
5
→
k
.
Suppose
P
divides the join of
Q
(
2
,
2
,
4
)
and
R
(
3
,
5
,
6
)
in the ratio
λ
:
1
.
Then, the position vector of
P
is
(
3
λ
+
2
λ
+
1
)
→
i
+
(
5
λ
+
2
λ
+
1
)
→
j
+
(
6
λ
+
4
λ
+
1
)
→
k
∴
13
5
→
i
+
19
5
→
j
+
26
5
→
k
=
(
3
λ
+
2
λ
+
1
)
→
i
+
(
5
λ
+
2
λ
+
1
)
→
j
+
(
6
λ
+
4
λ
+
1
)
→
k
⇒
3
λ
+
2
λ
+
1
=
13
5
;
5
λ
+
2
λ
+
1
=
19
5
;
6
λ
+
4
λ
+
1
=
26
5
⇒
2
λ
=
3
⇒
λ
=
3
2
There are three points with position vectors
−
2
a
+
3
b
+
5
c
,
a
+
2
b
+
3
c
and
7
a
−
c
. What is the relation between the three points?
Report Question
0%
Collinear
0%
Forms a triangle
0%
In different plane
0%
None of the above
Explanation
The relation between the three points are collinear
Thus option A is correct answer
The plane
a
x
+
b
y
+
c
z
+
d
=
0
divides the line joining the points
(
x
1
,
y
1
,
z
1
)
and
(
x
2
,
y
2
,
z
2
)
in the ratio
Report Question
0%
−
(
a
x
1
+
b
y
1
+
c
z
1
+
d
)
(
a
x
2
+
b
y
2
+
c
z
2
+
d
)
0%
(
a
x
1
+
b
y
1
+
c
z
1
+
d
)
(
a
x
2
+
b
y
2
+
c
z
2
+
d
)
0%
a
x
1
x
2
+
b
y
1
y
2
+
c
z
1
z
2
d
2
0%
None of these
Explanation
Let the given plane meet the line joining the given points in
(
x
3
,
y
3
,
z
3
)
.
Then
a
x
3
+
b
y
3
+
c
z
3
+
d
=
0
...(1)
Also let the point
(
x
3
,
y
3
,
z
3
)
divide the line joining the given points in the ratio
m
:
n
Then
x
3
=
m
x
1
+
n
x
2
m
+
n
,
y
3
=
m
y
1
+
n
y
2
m
+
n
and
z
3
=
m
z
1
+
n
z
2
m
+
n
Substituting these values in (1), we get
a
(
m
x
1
+
n
x
2
m
+
n
)
+
b
(
m
y
1
+
n
y
2
m
+
n
)
+
c
(
m
z
1
+
n
z
2
m
+
n
)
+
d
=
0
⇒
a
(
m
x
1
+
n
x
2
)
+
b
(
m
y
1
+
n
y
2
)
+
c
(
m
z
1
+
n
z
2
)
+
d
(
m
+
n
)
=
0
⇒
m
(
a
x
1
+
b
y
1
+
c
z
1
+
d
)
+
n
(
a
x
2
+
b
y
2
+
c
z
2
+
d
)
=
0
⇒
n
m
=
−
(
a
x
1
+
b
y
1
+
c
z
1
+
d
)
(
a
x
2
+
b
y
2
+
c
z
2
+
d
)
Hence, option A.
The set of points in space 4 inches from a given line and 4 inches from a given point on this line is ______ , if given point lies on the given line.
Report Question
0%
a set consisting of two points
0%
a set consisting of four points
0%
a set consisting of two circles
0%
the empty set
Let
a
,
b
,
c
ϵ
R
such that
a
b
c
=
p
and
q
a
−
b
=
0
, where
p
and
q
are fixed positive number, then minimum distance of the point
(
a
,
b
,
c
)
from origin in the three dimensional coordinate system is:
Report Question
0%
√
3
(
p
(
q
2
+
1
)
2
q
)
1
/
3
0%
√
3
(
p
(
q
2
+
1
)
q
)
1
/
3
0%
√
3
(
p
)
1
/
3
0%
√
2
(
p
q
)
1
/
2
Explanation
We know that the distance between point
(
a
,
b
,
c
)
and origin
(
0
,
0
,
0
)
is:
d
=
√
(
a
2
+
b
2
+
c
2
)
.
Using the inequality:
A
M
≥
G
M
, for terms
a
2
,
b
2
and
c
2
:
a
2
+
b
2
+
c
2
3
≥
3
√
(
a
b
c
)
2
.
a
b
c
=
p
(Given)
⟹
a
2
+
b
2
+
c
2
3
≥
(
p
)
2
3
.
Taking square root on both sides of the inequality:
√
a
2
+
b
2
+
c
2
≥
√
3
p
1
/
3
.
Hence, Option C is correct.
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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