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JEE Main 8 Apr 2024 Shift 1 Solved Paper
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© examsnet.com
Question : 4
Total: 90
Let
P
(
x
,
y
,
z
)
be a point in the first octant, whose projection in the xy-plane is the point
Q
. Let
OP
=
γ
; the angle between
OQ
and the positive
x
-axis be
θ
; and the angle between
O
P
and the positive
z
-axis be
φ
, where
O
is the origin. Then the distance of
P
from the
x
-axis is :
[8 Apr 2024 Shift 1]
γ
√
1
−
s
i
n
2
φ
cos
2
θ
γ
√
1
+
cos
2
θ
s
i
n
2
φ
γ
√
1
−
s
i
n
2
θ
cos
2
φ
γ
√
1
+
cos
2
φ
s
i
n
2
θ
Validate
Solution:
👈: Video Solution
P
(
x
,
y
,
z
)
,
Q
(
x
,
y
,
O
)
;
x
2
+
y
2
+
z
2
=
γ
2
O
Q
=
x
^
i
+
y
^
j
cos
θ
=
x
√
x
2
+
y
2
cos
φ
=
z
√
x
2
+
y
2
+
z
2
⇒
s
i
n
2
φ
=
x
2
+
y
2
x
2
+
y
2
+
z
2
distance of
P
from
x
-axis
√
y
2
+
z
2
⇒
√
γ
2
−
x
2
⇒
γ
√
1
−
x
2
γ
2
=
γ
√
1
−
cos
2
θ
s
i
n
2
φ
© examsnet.com
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