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Byju's Answer
Standard XII
Physics
Multiplication with Vectors
a⃗ = 2 i - j ...
Question
→
a
=
2
ˆ
i
−
ˆ
j
+
ˆ
k
,
→
b
=
ˆ
i
+
2
ˆ
j
−
ˆ
k
and
→
c
=
ˆ
i
+
ˆ
j
−
2
ˆ
k
. A vector coplanar with
→
b
and
→
c
whose projection on
→
a
is magnitude
√
2
3
is
A
2
ˆ
i
+
3
ˆ
j
−
3
ˆ
k
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B
−
2
ˆ
i
−
ˆ
j
+
5
ˆ
k
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C
2
ˆ
i
+
3
ˆ
j
+
3
ˆ
k
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D
2
ˆ
i
+
ˆ
j
+
5
ˆ
k
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Solution
The correct option is
B
−
2
ˆ
i
−
ˆ
j
+
5
ˆ
k
Let the required vector be
→
r
, then
→
r
=
x
1
→
b
+
x
2
→
c
and
→
r
⋅
→
a
=
√
2
3
(
|
→
a
|
)
=
2
Now,
→
r
⋅
→
a
=
x
1
→
a
⋅
→
b
+
x
2
→
a
⋅
→
c
⇒
2
=
x
1
(
2
−
2
−
1
)
+
x
2
(
2
−
1
−
2
)
⇒
x
1
+
x
2
=
−
2
⇒
→
r
=
x
1
(
ˆ
i
+
2
ˆ
j
−
ˆ
k
)
+
x
2
(
ˆ
i
+
ˆ
j
−
2
ˆ
k
)
=
ˆ
i
(
x
1
+
x
2
)
+
ˆ
j
(
2
x
1
+
x
2
)
−
ˆ
k
(
2
x
2
+
x
1
)
=
−
2
ˆ
i
+
ˆ
j
(
x
1
−
2
)
−
ˆ
k
(
−
4
−
x
1
)
,
where
x
1
∈
R
Suggest Corrections
0
Similar questions
Q.
Assertion :Vector
→
c
=
−
5
ˆ
i
+
7
ˆ
j
+
2
ˆ
k
is along the bisector of angle between
→
a
=
ˆ
i
+
2
ˆ
j
+
2
ˆ
k
and
→
b
=
−
8
ˆ
i
+
ˆ
j
−
4
ˆ
k
.
Reason:
→
c
is equally inclined to
→
a
and
→
b
.
Q.
If
→
r
×
→
b
=
→
c
×
→
b
&
→
r
,
→
a
=
0
where
→
a
=
2
∧
i
+
3
∧
j
−
∧
k
,
→
b
=
3
∧
i
−
∧
j
−
k
,
→
c
=
∧
i
+
∧
j
+
∧
k
then
→
r
Q.
Let
→
a
=
2
i
−
j
+
k
,
→
b
=
^
i
+
2
^
j
−
^
k
and
→
c
=
^
i
+
^
j
−
2
^
k
be three vectors. A vector in the plane of
→
b
and
→
c
whose projection on
→
a
is of magnitude
√
2
3
is
Q.
If
→
a
satisfies
→
a
×
(
ˆ
i
+
2
ˆ
j
+
ˆ
k
)
=
ˆ
i
−
ˆ
k
, then
→
a
is equal to
Q.
If
→
a
=
−
i
+
j
+
k
and
→
b
=
2
i
+
k
, then the vector
→
c
satisfying the conditions.
(i) that it is coplanar with
→
a
and
→
b
(ii) that its projection on
→
b
is
0
.
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