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Byju's Answer
Standard XII
Mathematics
Point Form of Normal: Ellipse
Find the dist...
Question
Find the distance of the plane 2x − 3y + 4z − 6 = 0 from the origin.
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Solution
The given equation of the plane is
2
x
-
3
y
+
4
z
=
6
.
.
.
1
Now,
2
2
+
-
3
2
+
4
2
=
4
+
9
+
16
=
29
Dividing (1) by
29
, we get
2
29
x
-
3
29
y
+
4
29
z
=
6
29
,
which is the normal form of plane (1).
So
,
the
length of the
perpendicular from the origin to the plane
=
6
29
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Similar questions
Q.
Find the distance of the plane 2x − 3y + 4z − 6 = 0 from the origin.
Q.
What is the distance of the plane
2
x
−
3
y
+
4
z
=
6
from the origin?
Q.
Find the perpendicular distance from origin to the plane
2
x
+
3
y
+
4
z
+
5
=
0
Q.
If
p
1
,
p
2
,
p
3
denote the distance of the plane
2
x
−
3
y
+
4
z
+
2
=
0
from the planes
2
x
−
3
y
+
4
z
+
6
=
0
,
4
x
−
6
y
+
8
z
+
3
=
0
and
2
x
−
3
y
+
4
z
−
6
=
0
respectively, then
Q.
Find the equation of the plane which is at a distance of
6
√
29
from the origin and its normal vector from the origin is
2
^
i
−
3
^
j
+
4
^
k
