Examples of Finding an Equation of a Plane
Example 1. Find an equation of the plane that passes through the point (1, 2, 3) and is parallel to the xy-plane.
We are given a point in the plane. The normal vector must be perpendicular to the xy-plane, so we can use the
direction vector for the z-axis, ~n = h 0, 0, 1 i. Thus, an equation of this plane is
0(x − 1) + 0(y − 2) + 1(z − 3) = 0
or
z−3=0
Example 2. Find an equation of the plane that contains the y-axis and makes an angle of
π
6
with the positive x-axis.
We can use any point on the y-axis, so (0, 0, 0) is as good as any. If the plane makes an angle of π6 with the positive
with the positive x-axis. The vector from the
x-axis, then the normal vector must make an angle of π6 + π2 = 2π
√
√ 3
origin to the point (−1, 0, 3) is such a vector, so ~n = h −1, 0, 3 i. Thus, an equation of this plane is
√
−1(x − 0) + 0(y − 0) + 3(z − 0) = 0
or
−x +
√
3z = 0
Example 3. Find an equation of the plane that passes through the points (1, −2, −1) and (2, 5, 6) and is parallel to
the x-axis.
We can use either of the given points. The vector between the points, ~u = h 2 − 1, 5 − (−2), 6 − (−1) i = h 1, 7, 7 i, is
in the plane, and a direction vector for the x-axis, ~v = h 1, 0, 0 i is also in the plane. So a normal vector for the plane
is
~n = ~u × ~v
ı̂ ̂
= 1 7
1 0
k̂
7
7 = ı̂ 0
0
1
7
− ̂ 0
1
1
7
+ k̂ 1
0
7
0
= ı̂(7 · 0 − 0 · 7) − ̂(1 · 0 − 1 · 7) + k̂(1 · 0 − 1 · 7) = h 0, 7, −7 i
Thus an equation of this plane is
0(x − 1) + 7(y + 2) − 7(z + 1) = 0
or
7y − 7z + 7 = 0
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