Planes
A (nonzero) vector n = (a, b, c) is said to be normal
to a plane if it is perpendicular to the plane. A plane
in space is determined by a point P0 (x0 , y0 , z0 ) in the
plane and a normal vector n = (a, b, c).
Let P (x, y, z) be an arbitrary point in the plane.
−−→
Then the vector P0 P is orthogonal to n. Hence,
−−→
n · P0 P = 0.
n
P
P0
1
Equation of a Plane
−−→
We have P0 P = (x − x0 , y − y0 , z − z0 ). Thus,
−−→
n · P0 P = 0 is equivalent to
a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0.
This is an equation of the plane through the point
P0 (x0 , y0 , z0 ) with normal vector n = (a, b, c).
Conversely, if a plane is given by an equation
ax + by + cz = d,
then (a, b, c) is a normal vector to the plane.
Example. Find an equation for the plane through
(−5, 1, 2) and perpendicular to n = 3 i − 5 j + 2 k.
Solution. The plane has an equation
3(x + 5) − 5(y − 1) + 2(z − 2) = 0
or 3x − 5y + 2z = −16.
2
A Plane Determined by Three Points
Example. Find an equation of the plane through
three points
P (1, 0, 2),
Q(3, −1, 6),
R(5, 2, 4).
Solution. Let
−−→
u = P Q = (2, −1, 4)
−→
and v = P R = (4, 2, 2).
Then n = u × v is a vector normal to the plane. We
have
i
j k n = u × v = 2 −1 4 = −10 i + 12 j + 8 k.
4 2 2 Moreover, the plane passes through point P (1, 0, 2).
Hence, the plane has an equation
−10(x − 1) + 12y + 8(z − 2) = 0
or −10x + 12y + 8z = 6, i.e., 5x − 6y − 4z = −3.
3
Intersection Points
Example. Let L be the line given by parametric
equations
x = 2 + 3t, y = −4t, z = 5 + t,
−∞ < t < ∞.
Find the point at which the line intersects the plane
4x + 5y − 2z = 18.
Solution. We substitute the expressions for x, y,
and z from the parametric equations of the line into
the equation of the plane:
4(2 + 3t) + 5(−4t) − 2(5 + t) = 18.
This simplifies to −10t = 20, so t = −2. Then we
have x = 2 + 3(−2) = −4, y = (−4)(−2) = 8, and
z = 5 + (−2) = 3. Therefore, the intersection point
is (−4, 8, 3).
4
The Distance from a Point to a Plane
Let us find the distance d from a point Q to
a plane with normal vector n. We draw the line
passing through Q and perpendicular to the plane.
Let R be the intersection point of the line with the
plane. Then d = |QR|.
Pick a point P on the plane. Let θ =
6
P QR.
Since 6 P RQ = 90◦ , we have d = |QR| = |P Q| cos θ.
n
Q
θ
d
P
R
5
−−→
Since the normal vector n is parallel to QR, the
−−→
angle between P Q and n is either θ or 180◦ − θ.
Hence,
−−→
|P Q · n|
cos θ = −−→
.
|P Q||n|
This in connection with d = |P Q| cos θ gives
−−→
|P Q · n|
.
d=
|n|
Example. Let Q be the point (2, 8, 5). Find the
distance from Q to the plane x − 2y − 2z = 1.
Solution. Setting y = 0 and z = 0 in the equation
of the plane, we obtain x = 1. Thus, P (1, 0, 0) is a
−−→
point in the plane. It follows that P Q = (1, 8, 5).
Moreover, n = (1, −2, −2) is a normal vector to the
plane. Hence, the distance from Q to the plane is
−−→
|P Q · n|
| − 25|
25
=
=
.
d=
|n|
3
3
6
The Distance from a Point to a Line
Let us find the distance d from a point Q to
a line with a direction vector v. We draw the line
passing through Q and perpendicular to the line. Let
R be the intersection point of these two lines. Then
d = |QR|.
Pick a point P on the line. Let θ =
6
QP R.
Since 6 P RQ = 90◦ , we have d = |QR| = |P Q| sin θ.
Q
d
P
θ
R
v
7
−−→
−−→
We observe that |P Q × v| = |P Q||v| sin θ. This
in connection with d = |P Q| sin θ gives
−−→
|P Q × v|
d=
.
|v|
Example. Let Q be the point (1, 1, 5). Find the
distance from Q to the line
x = 1 + t, y = 3 − t, z = 2t,
−∞ < t < ∞.
Solution. Choosing t = 0, we see that P (1, 3, 0) is
−−→
a point on the line. It follows that P Q = −2j + 5k.
Moreover, v = i − j + 2k is a direction vector of the
line. We have
i
j k −−→
P Q × v = 0 −2 5 = i + 5 j + 2 k.
1 −1 2 Hence, the distance from Q to the line is
√
−−→
30 √
|P Q × v|
= √ = 5.
d=
|v|
6
8