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Byju's Answer
Standard XII
Mathematics
Dot Product
The position ...
Question
The position vectors
→
a
,
→
b
,
→
c
of three points satisfy the relation
2
→
a
+
7
→
b
+
5
→
c
=
→
0
. Are these points collinear?
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Solution
2
→
a
+
7
→
b
+
5
→
c
=
0
∴
7
→
b
=
−
5
→
c
−
2
→
a
∴
→
b
=
5
→
c
−
2
→
a
7
∴
→
b
=
−
(
5
→
c
+
2
→
a
)
7
Above indicates
→
b
divides
→
c
and
→
a
internally
in ratio
5
:
2
Hence points are collinear.
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Similar questions
Q.
Prove that points
A
,
B
,
C
having positions vectors
→
a
,
→
b
,
→
c
are collinear, if
[
→
b
×
→
c
+
→
c
×
→
a
+
→
a
×
→
b
]
=
→
0
Q.
The position vectors of three points are
2
→
a
−
→
b
+
3
→
c
,
→
a
−
2
→
b
+
λ
→
c
and
μ
→
a
−
5
→
b
where
→
a
,
→
b
,
→
c
are non coplanar vectors, then the points are collinear when
Q.
If
→
a
,
→
b
and
→
c
are unit vectors satisfying
|
→
a
−
→
b
|
2
+
|
→
b
−
→
c
|
2
+
|
→
c
−
→
a
|
2
=
9
, then
|
2
→
a
+
5
→
b
+
5
→
c
|
is
Q.
If
→
a
,
→
b
,
→
c
are non-coplanar vectors, then show that the four points
2
→
a
+
→
b
,
→
a
+
2
→
b
+
→
c
,
4
→
a
−
2
→
b
−
→
c
and
3
→
a
+
4
→
b
−
5
→
c
are coplanar.
Q.
Let
→
a
,
→
b
,
→
c
be three vectors in the
x
y
z
space such that
→
a
×
→
b
=
→
b
×
→
c
=
→
c
×
→
a
≠
→
0
.
If
A
,
B
,
C
are points with position vectors
→
a
,
→
b
,
→
c
respectively, then the number of possible positions of the centroid of triangle
A
B
C
is.
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