Lecture Notes, M261-004, Lines and Planes in Space
Sept 3, 2008
We continue to study the geometry of three dimensional space. This time, we will look at how
to form the equations for lines and planes, as well as computing some distances. We look at the
following topics
• The Equation for a Line
• The Distance from a Line to a Point
• The Equation for a Plane
• The Intersection of Two Planes
• The Distance from a Point to a Plane
1
The Equation for a Line
We have already seen how to parameterize a line in 3D space. A line is defined by a point it passes
through and a direction that it travels in. The direction is given by a vector. For example,
h1, 2, 3i + t h0, 1, −1i
passes through the point h1, 2, 3i and is spanned by the vector h0, 1, −1i. We could also write this
as a system

x = 1

y =t+2


z =3−t
For a line segment, the formula is simpler. The line segment between (x0 , y0 , z0 ) and (x1 , y1 , z1 )
is given by
n
t(x1 , y1 , z1 ) + (1 − t)(x0 , y0 , z0 ), t ∈ [0, 1] .
Example 1. Find the equation for the line passing through (1, −2, 0) and spanned by h−2, 1, 1i.
We get
y(t) = (1, −2, 0) + t h−2, 1, 1i
Example 2. Find the paramaterizations for both the line and the line segment joining (1, 1, 3) and
(2, −1, 1).
We get
t(2, −1, 1) + (1 − t)(1, 3, 3)
This gives the line. If we only want the line segment, we just restrict t to [0, 1] .
1
2
The Distance From a Line to a Point
We are often interested in the distance from a line to a point. We define this quantity as the
distance from the point to the closest point on the line. One way to do this would be to use
calculus to minimize the distance formula between the point and points on the line. It is much
easier, however, to think of it graphically. We draw a vector from the point that intersects the line
orthogonally. We are interested in the norm of this vector. Now, we draw a vector from the point
to any point on the line. We call this vector A. We can see using trigonometry that the distance
from the point to the line that we are looking for is given by
d=
kA × vk
kvk
where v is a vector spanning the line.
Example 3. Find the distance between the point (1, 0, −1) and the line given by t h2, 1, 3i
The line in question is spanned by h2, 1, 3i and passes through the origin. This means the
distance is given by
√
kh1, −5, 1ik
h1, 0, −1i × h2, 1, 3i
27
√
=√
=
d=
kh2, 1, 3ik
14
14
3
The Equation for a Plane
As we have seen before, every plane has a normal vector that is orthogonal to all vectors in the
plane. If we know this vector and a point in the plane, then we can get the equation for the plane.
If v is the normal vector and (x0 , y0 , z0 ) is the point in the plane we know, then for any other
point (x, y, z) in the plane, the vector hx − x0 , y − y0 , z − z0 i is orthogonal to v, or
v1 (x − x0 ) + v2 (y − y0 ) + v3 (z − z0 ) = 0
This is the general equation for a plane. Note that this means we can read off the normal vector
from a plane given by Ax + By + Cz = D. Its normal vector is hA, B, Ci.
Example 4. Find the equation for the plane containing h1, −2, 1i and normal to h7, −2, −3i.
We get
7(x − 1) − 2(y + 2) − 3(z − 1) = 0
Example 5. Find the equation for the plane defined by h1, 0, 0i, h9, 2, 3i and h2, 1, −5i.
Two vectors in the plane are h8, 2, 3i and h1, 1, −5i. Their cross product is
4
The Intersection of Two Planes
When two planes intersect, we get a line. We may be interested in parameterizing this line or in
finding the angle of intersection between the planes.
The line of intersection is orthogonal to the normal vectors of both planes. We can use this
information and a point in the intersection to get a formula for the line.
Example 6. Find the line of intersection between the planes x + 2y + z = 10 and 2x − y = 1.
2
The line is spanned by the cross product of the normal vectors of the planes:
h1, 2, 1i × h2, −1, 0i = h1, 2, −5i
We just need a point on the line. We can find this by substituting y = 2x − 1 into the equation
for the first plane and picking a point. I chose (0, −1, 12). The line is given by
(0, −1, 12) + t h1, 2, −5i .
The angle of intersection is given by the angle between the normal vectors of the planes.
Example 7. Find the angle of intersection of the two planes in the previous example.
We just take the angle of the normal vectors:
h1, 2, 1i · h2, −1, 0i
0
θ = cos−1
= cos−1
= 90◦
kh1, 2, 1ik kh2, −1, 0ik
kh1, 2, 1ik kh2, −1, 0ik
5
The distance from a Point to a Plane
We are often interested in the distance from a point to a plane. We define this as the shortest
possible distance from the point to all points on the plane. Graphically, we can see that this is
given by drawing a vector from the point to the plane that intersects the plane orthogonally. (Note
that this is parallel to the plane’s normal vector.) We can find this by taking the vector a between
the point and any point on the plane, and then taking the projection of a onto the normal vector
of the plane. Thus we get
n d = a ·
knk Example 8. Find the distance from the point (1, 1, 3) to the plane 3x + 2y + 6z = 6.
We just need a point on the plane. The plane’s normal vector is h3, 2, 6i. The point (0, 0, 1) is
on the plane, and the vector between that and (1, 1, 3) is given by h1, 1, 2i. Using the formula, the
distance is
h1, 1, 2i · h3, 2, 6i 17
=
d = kh3, 2, 6ik
7
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