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Byju's Answer
Standard XII
Mathematics
Differentiability in an Interval
Given that ...
Question
Given that
→
A
×
→
B
=
→
B
×
→
C
=
→
0
if
→
A
→
B
→
C
are not null vectors, Find the value of
→
A
×
→
C
A
→
A
×
→
B
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B
→
0
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C
→
C
×
→
B
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D
→
C
×
→
A
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Solution
The correct option is
A
→
0
We have,
→
A
×
→
B
=
→
B
×
→
C
=
→
0
→
A
×
→
B
+
→
C
×
→
B
=
→
0
(
→
A
+
→
C
)
×
→
B
=
→
0
→
A
+
→
C
=
λ
→
B
→
A
×
→
C
+
0
=
λ
(
→
B
×
→
C
)
→
A
×
→
C
=
→
0
H
e
n
c
e
,
t
h
e
o
p
t
i
o
n
B
i
s
c
o
r
r
e
c
t
.
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0
Similar questions
Q.
If
→
a
,
→
b
,
→
c
are unit vectors such that
→
a
+
→
b
+
→
c
=
→
0
and
(
→
a
,
→
b
)
=
π
3
then
|
→
a
×
→
b
|
+
|
→
b
×
→
c
|
+
|
→
c
×
→
a
|
=
Q.
If
→
a
,
→
b
,
→
c
are unit vectors such that
→
a
+
→
b
+
→
c
=
→
0
and
(
→
a
,
→
b
)
=
π
3
, then
∣
∣
→
a
×
→
b
∣
∣
+
∣
∣
→
b
×
→
c
∣
∣
+
|
→
c
×
→
a
|
=
Q.
→
a
≠
→
0
,
→
b
≠
→
0
,
→
a
×
→
b
=
→
0
,
→
c
×
→
b
=
→
0
⇒
→
a
×
→
c
=
Q.
If
→
a
+
→
b
+
→
c
=
→
0
show that
→
a
×
→
b
=
→
b
×
→
c
=
→
c
×
→
a
.
Q.
Prove that points
A
,
B
,
C
having positions vectors
→
a
,
→
b
,
→
c
are collinear, if
[
→
b
×
→
c
+
→
c
×
→
a
+
→
a
×
→
b
]
=
→
0
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