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Tardigrade
Question
Mathematics
The distance of the point (-1,9,-16) from the plane 2 x+3 y-z=5 measured parallel to the line (x+4/3)=(2-y/4)=(z-3/12) is
Q. The distance of the point
(
−
1
,
9
,
−
16
)
from the plane
2
x
+
3
y
−
z
=
5
measured parallel to the line
3
x
+
4
=
4
2
−
y
=
12
z
−
3
is
1189
128
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JEE Main 2023
Three Dimensional Geometry
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A
20
2
0%
B
31
0%
C
13
2
0%
D
26
100%
Solution:
Equation of line
3
x
+
1
=
−
4
y
−
9
=
12
z
+
16
G.P on line
(
3
λ
−
1
,
−
4
λ
+
9
,
12
λ
−
16
)
point of intersection of line \& plane
6
λ
−
2
−
12
λ
+
27
−
12
λ
+
16
=
5
λ
=
2
Point
(
5
,
1
,
8
)
Distance
=
36
+
64
+
576
=
26