A line is determined by a point and a direction.
Thus, to find an equation representing a line in three dimensions choose
a point P_0 on the line
and a non-zero vector **v** parallel
to
the line. Since any constant multiple of a vector still points in the same
direction,
it seems reasonable that a point on the line can be found be starting at
the
point P_0 on the line
and following a constant multiple of the vector **v** (see the figure
below).

If
**r_0** is
the position
vector of the point
P_0, then the line must
have the form

This is the vector equation of a line in three dimensions. By letting
**r**=<x,y,z>,
**r_0**=<x_0,y_0,z_0>, and
**v**=<a,b,c> we obtain the equation

which leads to the parametric equations of the line passing through the
point
P_0=(x_0,y_0,z_0) and parallel to the vector
**v**=<a,b,c>:

To find the parametric equations of the line passing through
the point (-1,2,3) and parallel to the vector <3,0,-1>, we first find
the vector equation of the line. Here,
**r_0**=<-1,2,3>
and
**v**=<3,0,-1>. Thus, the line has vector equation
**r**=<-1,2,3>+t<3,0,-1>. Hence, the parametric equations
of the line are x=-1+3t, y=2, and z=3-t.

It is important to note that the equation of a line in three dimensions is not unique. Choosing a different point and a multiple of the vector will yield a different equation. If we had taken the point to be (2,2,-4) and the vector to be <-6,0,2> in the previous example we would have found the parametric equations of the line to be x=2-6t, y=2, and z=-4+2t. This is representation of the line is equivalent to the one found above.

There is a third representation of a line in three dimensions. Consider solving each of the parametric equations for t and then setting them equal:

These are the symmetric equations of the line. Note that we must have a, b, and c non-zero to use this representation; the line in the example above would have symmetric equations

Where does the line in the previous example intersect
the *xy*-plane? Well, the line intersects the *xy*-plane
when z=0. From the parametric
equation for z, we see that we must have 0=-3-t which implies t=-3. Thus,
x=-1+3t=-10 and y=2. The line intersect the *xy*-plane at the point
(-10,2).

A plane is determined by a point
P_0 in the plane and
a vector **n**
(called
the normal vector) orthogonal to the plane. If P is any other point in the
plane and
**r_0** and **r** are
the position vectors of the points
P_0 and P, respectively,
then an equation of the plane is

since each vector in the plane must be orthogonal to the normal vector
**n** and the
vector
**r**-**r_0** is
a vector in the plane. Either of these equations
is called a vector equation of the plane. With the substitutions
**n**=<a,b,c>, **r**=<x,y,z>, and
**r_0**=<x_0,y_0,z_0> and some simplification we
have an equation
of the plane passing through the point
P_0(x_0,y_0,z_0)
with normal vector
<a,b,c>:

This is frequently written as ax+by+cz=d. Notice that the
normal vector
can be identified directly from the coefficients on the left hand side of
the equation:
**n**=<a,b,c>. As with equations of lines in three dimensions, it
should be noted that there is not a unique equation for a given plane. The
graph of the plane -2x-3y+z=2 is shown with its normal vector.

Find an equation of the plane passing through the points
P(1,-1,3),
Q(4,1,-2), and R(-1,-1,1). Since we are not given a normal vector, we must
find one. By taking the cross product
of
the vector **a** from P to Q
and the vector **b** from Q to R, we obtain a vector which is orthogonal
to each of the original vectors (and thus orthogonal to the plane). Hence,
**a**=<3,2,-5>, **b**=<-5,-2,3>, and

So, an equation of the plane is

-4(x-1)+16(y+1)+4(z-3)=0 which implies -4x+16y+4z=-8

Two planes are parallel if their normal vectors are parallel (constant
multiples of one another). It is easy to recognize parallel planes written
in the form ax+by+cz=d since a quick comparison of the normal vectors
**n**=<a,b,c> can be made.

Find the formula for the distance D from a point
P_0(x_0,y_0,z_0) to the plane ax+by+cz+d=0. Let
P_1(x_1,y_1,z_1) be any other point in the plane.
Let **r**
=<x_0-x_1,y_0-y_1,z_0-z_1> be the vector from
P_1 to P_0. Then, from
the figure above, the distance D from the point to the plane is the
scalar projection of the
vector **r** onto the normal vector **n**:

But, since the point P_1 lies in the plane, -ax_1-by_1-cz_1=d. Hence, the distance D from the point P_0 to the plane is

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