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JEE Mains 29-July-2022 Shift 1 Solved Paper
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© examsnet.com
Question : 9
Total: 90
If the foot of the perpendicular from the point
A
(
−
1
,
4
,
3
)
on the plane
P
:
2
x
+
m
y
+
n
z
=
4
, is
(
−
2
,
7
2
,
3
2
)
, then the distance of the point
A
from the plane
P
, measured parallel to a line with direction ratios
3
,
−
1
,
−
4
, is equal to :
[29-Jul-2022-Shift-1]
1
√
26
2
√
2
√
14
Validate
Solution:
(
−
2
,
7
2
,
3
2
)
satisfies the plane
P
:
2
x
+
m
y
+
n
z
=
4
−
4
+
7
m
2
+
3
n
2
=
4
⇒
7
m
+
3
n
=
16
.
.
.
.
.
.
(i)
Line joining
A
(
−
1
,
4
,
3
)
and
(
−
2
,
7
2
,
3
2
)
is perpendicular to
P
:
2
x
+
m
y
+
n
z
=
4
1
2
=
1
2
m
=
3
2
n
⇒
m
=
1
&
n
=
3
Plane
P
:
2
x
+
y
+
3
z
=
4
Distance of
P
from
A
(
−
1
,
4
,
3
)
parallel to the line
x
+
1
3
=
y
−
4
−
1
=
z
−
3
−
4
:
L
for point of intersection of
P
&
L
2
(
3
r
−
1
)
+
(
−
r
+
4
)
+
3
(
−
4
r
+
3
)
=
4
⇒
r
=
1
Point of intersection:
(
2
,
3
,
−
1
)
Required distance
=
√
3
2
+
1
2
+
4
2
=
√
26
© examsnet.com
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