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Byju's Answer
Standard XII
Mathematics
Distance Formula
Show that the...
Question
Show that the points (1, 1, 1) and (−3, 0, 1) are equidistant from the plane 3x + 4y − 12z + 13 = 0.
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Solution
We know that the distance of the point
x
1
,
y
1
,
z
1
from the plane
a
x
+
b
y
+
c
z
+
d
=
0
is given by
a
x
1
+
b
y
1
+
c
z
1
+
d
a
2
+
b
2
+
c
2
Distance of the point (1, 1, 1) from the plane 3x + 4y - 12z + 13 = 0
The required distance
=
3
1
+
4
1
-
12
1
+
13
3
2
+
4
2
+
-
12
2
=
3
+
4
-
12
+
13
9
+
16
+
144
=
8
13
units ... (1)
Distance of the point (-3, 0, 1) from the plane 3x + 4y - 12z + 13 = 0
The required distance
=
3
-
3
+
4
0
-
12
1
+
13
3
2
+
4
2
+
-
12
2
=
-
9
+
0
-
12
+
13
9
+
16
+
144
=
8
13
units .... (2)
From (1) and (2), we can say that the given points are equidistant from the given plane.
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Similar questions
Q.
If the points
(
1
,
1
,
λ
)
and
(
−
3
,
0
,
1
)
are equidistant from the plane,
3
x
+
4
y
−
12
z
+
13
=
0
,
then
λ
satisfies the equation:
Q.
Find an equation for the set of all points that are equidistant from the planes 3x − 4y + 12z = 6 and 4x + 3z = 7.
Q.
For what value(s) of a will the two points (1, a, 1) and (-3, 0, a) lie on opposite sides of the plane
3
x
+
4
y
−
12
z
+
13
=
0
?
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