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JEE Main Math Class 12 Three Dimensional Geometry Part 2 Questions
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© examsnet.com
Question : 3
Total: 100
Let
d
be the distance between the foot of perpendiculars of the points
P
(
1
,
2
,
−
1
)
and
Q
(
2
,
−
1
,
3
)
on the plane
−
x
+
y
+
z
=
1
. Then
d
2
is equal to___
[29-Jun-2022-Shift-1]
Your Answer:
Validate
Solution:
Foot of perpendicular from
P
x
−
1
−
1
=
y
−
2
1
=
z
+
1
1
=
−
(
−
1
+
2
−
1
−
1
)
3
⇒
p
′
≡
(
2
3
,
7
3
,
−
2
3
)
and foot of perpendicular from Q
x
−
2
−
1
=
y
+
1
1
=
z
−
3
1
=
−
(
−
2
−
1
+
3
−
1
)
3
⇒
Q
′
≡
(
5
3
,
−
2
3
,
10
3
)
P
′
Q
′
=
√
(
1
)
2
+
(
3
)
2
+
(
4
)
2
=
d
=
√
26
⇒
d
2
=
26
© examsnet.com
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