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EXAMPLES ON SECTION 12.2: LINES IN SPACE
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Finding Equations of a Line Given a Point and a Vector
Find an equation of the line through the point (1, 5, 2) and parallel to the vector
where the line intersects the yz- plane.
Also, determine
Notice that we can write down the equations immediately.
The parametric equations for the line are:
x = 1+4t, y = 5+3t,
z = 2+7t.
We show the graph below. Note that the line intersects the yz - plane where x = 0. Setting x = 0 in the last
equation, we solve for y and z to obtain
Alternatively, observe that we could solve x= 4t+1 for t (again where x = 0) and substitute this into the
parametric equations for y and z. So, the line intersects the yz - plane at the point
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Finding Equations of a Line Given Two Points
Find an equation of the line passing through the points P(1, 2, -1) and Q(5, -3, 4).
First, we need to find a vector that is parallel to the given line. The obvious choice is
Picking either point will give us equations for the line. Here, we use P, so that parametric
equations for the line are:
x=4t+1, y=-5t+2, z=5t-1.
DEFINITIONS
Since we have specified a line by choosing a point on the line and a vector with the same
direction, the following definition should be transparent.
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 Let l1 and l2 be two lines, with vectors a and b, respectively, and let be the angle between a and
b. Then the angle between l1 and l2 is .
 Nonparallel, nonintersecting lines are called SKEW lines.
 Reminder: The angle between two vector is given, for example, by cos( ) = a . b/ a b
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Showing Two Lines Are Not Parallel but Do Not Intersect
Show that these two lines are not parallel, yet do not intersect.
L1: x= -t+2, y= 2t+1, z= 2t+5
and
L2: x= s+1, y= -s+2, z= 3s+1
Skew lines
You will notice immediately that we have used different letters (t and s) as parameters for the two lines.
In this setting, the parameter is a dummy variable, so the letter used is not significant. Notice from
the figure that the lines are most certainly not parallel, but it is unclear whether or not they intersect.
You can read from the parametric equations that a vector parallel to l1 is a1 =
, while a vector
parallel to
. Since a1 is not a scalar multiple of a2 (check this!) or since their cross
product is not Zero, the vectors are not parallel and so the lines l1 and l2 are not parallel.
You might expect that the lines must then intersect, but let's take a careful look at this. The lines will
intersect if there's a choice of the parameters s and t that produces the same point, that is, that produces
the same values for all of x, y and z.
Setting the x - values equal, we get
2-t = 1+s,
so that s = 1-t.
Setting the y - values equal and setting s = 1-t , we get 1+2t=2-s=2-2(1-t)=1+t
Solving this for t yields
t = 0, which further implies that s = 1.
Setting the z - components equal gives
5+2t=3s+1
But this is not satisfied when t = 0 and s = 1. So, l1 and l2 do not intersect.