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Tardigrade
Question
Mathematics
The distance of the point (1, 2, 1) from the line (x-1/2) = ( y-2/1)=( z-3/2) is
Q. The distance of the point (1, 2, 1) from the line
2
x
−
1
=
1
y
−
2
=
2
z
−
3
is
9232
109
KCET
KCET 2019
Three Dimensional Geometry
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A
3
5
12%
B
5
2
3
29%
C
3
20
19%
D
3
2
5
40%
Solution:
2
x
−
1
=
1
y
−
2
=
2
z
−
3
=
k
any point on the line = (2k + 1, k + 2, 2k + 3)
drs of AB : 2k, k, 2k + 2
drs of line : 2, 1, 2
A
B
⊥
l
in
e
⇒
4
k
+
k
+
4
k
+
4
=
0
9
k
=
−
4
k
=
9
−
4
D
i
s
t
an
ce
=
81
64
+
81
16
+
81
100
=
9
180
=
3
2
5