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KCET 2017 Math Solved Paper
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© examsnet.com
Question : 54
Total: 60
The distance o f the point (-2, 4, -5) from the line
x
+
3
3
=
y
−
4
5
=
z
+
8
6
is
√
37
10
√
37
10
37
√
10
37
10
Validate
Solution:
Given line,
x
+
3
3
=
y
−
4
5
=
z
+
8
6
→ (1)
and point P(−2, 4, −5)
Let
x
+
3
3
=
y
−
4
5
=
z
+
8
6
=
λ
So, point on line is
Q
(
3
λ
−
3
,
5
λ
+
4
,
6
λ
−
8
)
Since, PQ line is perpendicular to line of Eq. (1). SO,
l
1
l
2
+
m
1
m
2
+
n
1
n
2
=
0
⇒
(
3
λ
−
1
)
3
+
5
λ
(
5
+
0
)
+
(
6
λ
−
3
)
6
=
0
⇒
9
λ
−
3
+
25
λ
+
36
λ
−
18
=
0
⇒
70
λ
=
21
⇒
λ
=
3
10
Therefore,
Q
(
−
21
10
,
55
10
,
−
62
10
)
So,
P
Q
=
√
(
−
21
10
−
(
−
2
)
)
2
+
(
55
10
−
4
)
2
+
(
−
62
10
−
(
−
5
)
)
2
=
√
1
100
+
225
100
+
144
100
=
√
370
100
=
√
37
10
© examsnet.com
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