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a b 2 a b 2 144 and a 4 then b is equal to
Question:
∣
a
⃗
×
b
⃗
∣
2
+
∣
a
⃗
.
b
⃗
∣
2
=
144
|\vec{a}\times\vec{b}|^2+|\vec{a}.\vec{b}|^2=144
∣
a
×
b
∣
2
+
∣
a
.
b
∣
2
=
144
and
∣
a
⃗
∣
=
4
|\vec{a}|=4
∣
a
∣
=
4
then
∣
b
⃗
∣
|\vec{b}|
∣
b
∣
is equal to
KCET - 2023
KCET
Updated On:
Apr 17, 2024
8
12
4
3
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The Correct Option is
D
Solution and Explanation
The correct answer is (D) : 4.
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Let
O
P
→
=
α
−
1
α
i
^
+
j
^
+
k
^
,
O
Q
→
=
i
^
+
β
−
1
β
j
^
+
k
^
\overrightarrow{OP}=\frac{\alpha-1}{\alpha}\hat{i}+\hat{j}+\hat{k},\overrightarrow{OQ}=\hat{i}+\frac{\beta-1}{\beta}\hat{j}+\hat{k}
OP
=
α
α
−
1
i
^
+
j
^
+
k
^
,
OQ
=
i
^
+
β
β
−
1
j
^
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k
^
and
O
R
→
=
i
^
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j
^
+
1
2
k
^
\overrightarrow{OR}=\hat{i}+\hat{j}+\frac{1}{2}\hat{k}
OR
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^
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j
^
+
2
1
k
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P
→
×
O
Q
→
)
.
O
R
→
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0
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OQ
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Let
a
⃗
=
i
^
+
j
^
+
k
^
,
b
⃗
=
−
i
^
−
8
j
^
+
2
k
^
,
and
c
⃗
=
4
i
^
+
c
2
j
^
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c
3
k
^
\vec{a} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{b} = -\hat{i} - 8\hat{j} + 2\hat{k}, \quad \text{and} \quad \vec{c} = 4\hat{i} + c_2\hat{j} + c_3\hat{k}
a
=
i
^
+
j
^
+
k
^
,
b
=
−
i
^
−
8
j
^
+
2
k
^
,
and
c
=
4
i
^
+
c
2
j
^
+
c
3
k
^
be three vectors such that
b
⃗
×
a
⃗
=
c
⃗
×
a
⃗
.
\vec{b} \times \vec{a} = \vec{c} \times \vec{a}.
b
×
a
=
c
×
a
.
If the angle between the vector
c
⃗
\vec{c}
c
and the vector
3
i
^
+
4
j
^
+
k
^
3\hat{i} + 4\hat{j} + \hat{k}
3
i
^
+
4
j
^
+
k
^
is
θ
\theta
θ
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tan
2
θ
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tan
2
θ
is:
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Mathematics
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Let
a
⃗
=
2
i
^
−
3
j
^
+
4
k
^
,
b
⃗
=
3
i
^
+
4
j
^
−
5
k
^
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a
=
2
i
^
−
3
j
^
+
4
k
^
,
b
=
3
i
^
+
4
j
^
−
5
k
^
, and a vector
c
⃗
\vec{c}
c
be such that
a
⃗
×
(
b
⃗
+
c
⃗
)
+
b
⃗
×
c
⃗
=
i
^
+
8
j
^
+
13
k
^
.
\vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times \vec{c} = \hat{i} + 8\hat{j} + 13\hat{k}.
a
×
(
b
+
c
)
+
b
×
c
=
i
^
+
8
j
^
+
13
k
^
.
If
a
⃗
⋅
c
⃗
=
13
\vec{a} \cdot \vec{c} = 13
a
⋅
c
=
13
, then
(
24
−
b
⃗
⋅
c
⃗
)
(24 - \vec{b} \cdot \vec{c})
(
24
−
b
⋅
c
)
is equal to ______.
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Mathematics
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View Solution
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a
⃗
=
2
i
^
+
j
^
−
k
^
\vec{a} = 2\hat{i} + \hat{j} - \hat{k}
a
=
2
i
^
+
j
^
−
k
^
,
b
⃗
=
(
(
a
⃗
×
(
i
^
+
j
^
)
)
×
i
^
)
×
i
^
\vec{b} = \left((\vec{a} \times (\hat{i} + \hat{j})) \times \hat{i}\right) \times \hat{i}
b
=
(
(
a
×
(
i
^
+
j
^
))
×
i
^
)
×
i
^
. Then the square of the projection of
a
⃗
\vec{a}
a
on
b
⃗
\vec{b}
b
is:
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Mathematics
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Let
a
⃗
=
6
i
^
+
j
^
−
k
^
\vec{a} = 6\hat{i} + \hat{j} - \hat{k}
a
=
6
i
^
+
j
^
−
k
^
and
b
⃗
=
i
^
+
j
^
\vec{b} = \hat{i} + \hat{j}
b
=
i
^
+
j
^
. If
c
⃗
\vec{c}
c
is a vector such that
∣
c
⃗
∣
≥
6
,
a
⃗
⋅
c
⃗
=
6
∣
c
⃗
∣
,
∣
c
⃗
−
a
⃗
∣
=
2
2
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∣
c
∣
≥
6
,
a
⋅
c
=
6∣
c
∣
,
∣
c
−
a
∣
=
2
2
and the angle between
a
⃗
×
b
⃗
\vec{a} \times \vec{b}
a
×
b
and
c
⃗
\vec{c}
c
is
6
0
∘
60^\circ
6
0
∘
, then
∣
(
a
⃗
×
b
⃗
)
×
c
⃗
∣
|(\vec{a} \times \vec{b}) \times \vec{c}|
∣
(
a
×
b
)
×
c
∣
is equal to:
JEE Main - 2024
Mathematics
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