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KEAM 2010 Math Paper
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© examsnet.com
Question : 14
Total: 120
The equation of the line passing through the point (3,0,-4) and perpendicular to the plane
2
x
−
3
y
+
5
z
−
7
=
0
is
x
−
2
3
=
y
−
3
=
z
+
4
5
x
−
3
2
=
y
−
3
=
z
−
4
5
x
−
3
2
=
−
y
3
=
z
+
4
5
x
+
3
2
=
y
3
=
z
−
4
5
x
−
2
3
=
y
3
=
z
+
4
5
Validate
Solution:
Comparing given plane
2
x
−
3
y
+
5
z
−
7
=
0
with the plane
l
x
+
m
y
+
n
z
+
d
=
0
⇒
l
=
2
,
m
=
−
3
,
z
=
5
since, the required line is perpendicular to the given plane. Therefore, these are the direction cosine of the line.
∴ Equation of line is
x
−
3
2
=
y
−
0
−
3
=
z
+
4
5
⇒
x
−
3
2
=
−
y
3
=
z
+
4
5
© examsnet.com
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