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JEE Main Math Class 12 Three Dimensional Geometry Part 3 Questions
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© examsnet.com
Question : 54
Total: 134
For
a
,
b
∈
Z
and
|
a
−
b
|
≤
10
, let the angle between the plane
P
:
ax
+
y
−
z
=
b
and the line
1
:
x
−
1
=
a
−
y
=
z
+1 be
cos
−
1
(
1
3
)
. If the distance of the point
(
6
,
−
6
,
4
)
from the plane
P
is
3
√
6
, then
a
4
+
b
2
is equal to
[8-Apr-2023 shift 2]
85
48
25
32
Validate
Solution:
θ
=
cos
−
1
1
3
∴
s
i
n
θ
=
√
1
−
1
9
=
2
√
2
3
s
i
n
θ
=
a
⋅
1
+
1
(
−
1
)
+
(
−
1
)
⋅
1
√
a
2
+
1
+
1
⋅
√
3
=
2
√
2
3
⇒
{
3
(
a
−
2
)
}
2
=
24
(
a
2
+
2
)
⇒
3
(
a
2
−
4
a
+
4
)
=
8
a
2
+
16
⇒
5
a
2
+
12
a
+
4
=
0
⇒
5
a
2
+
10
a
+
2
a
+
4
=
0
∴
a
=
−
2
,
−
2
5
∵
a
∈
z
∴
a
=
−
2
Distance of
(
6
,
−
6
,
4
)
from
−
2
x
+
y
−
z
−
b
=
0
is
3
√
6
∴
|
−
12
−
6
−
4
−
b
√
4
+
1
+
1
|
=
3
√
6
⇒
|
b
+
22
|
=
18
∴
b
=
−
40
,
−
4
∵
|
a
−
b
|
≤
10
∴
b
=
−
4
∴
a
4
+
b
2
=
32
Ans.
© examsnet.com
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