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Equations of
Lines

Equations of Lines
In this section we need to take a look at the equation of a line in 3
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This
doesn’t mean however that we can’t write down an equation for a line in 3-D space
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So, before we get into the equations of lines we first need to briefly look at vector functions
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At this point all that we need
to worry about is notational issues and how they can be used to give the equation of a curve
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So, consider the following vector function
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Note as well that a vector function can be a function of two or more variables
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The vector that the function gives can be a vector in whatever dimension we need it to be
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When we get to the real subject of this section,
equations of lines, we’ll be using a vector function that returns a vector in 3
Now, we want to determine the graph of the vector function above
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Recall that a position vector, say v = a,b , is a vector that starts at the
origin and ends at the point ( a,b )
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Here are some evaluations for our example
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The
points,

( −3,1)

( −1,1)

( 2,1)

( 5,1)

are all points that lie on the graph of our vector function
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In this sketch we’ve included the position vector (in gray and dashed) for several evaluations as
well as the t (above each point) we used for each evaluation
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Here’s another quick example
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In this case we get an ellipse
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We’ll be looking at lines in this section, but the graphs
of vector function do not have to be lines as the example above shows
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Imagine that a pencil/pen is attached to the end of the position vector and as we
increase the variable the resulting position vector moves and as it moves the pencil/pen on the end
sketches out the curve for the vector function
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We want to write down the
3
equation of a line in
and as suggested by the work above we will need a vector function to do
this
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In two dimensions we need the slope (m) and a point that was on the
line in order to write down the equation
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In this case we will need to acknowledge that a line can have a three
dimensional slope
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We already have a quantity that will do this for us
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So, let’s start with the following information
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Note, in all
likelihood, v will not be on the line itself
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Finally, let
P = ( x, y, z ) be any point on the line
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We’ll
do this with position vectors
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Also, for no apparent reason, let’s define a to be the vector with representation P0 P
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Now, we’ve shown the parallel vector, v , as a position vector but it doesn’t need to be a position
vector
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Next, notice that we can write r as follows,

r = r0 + a
If you’re not sure about this go back and check out the sketch for vector addition in the vector
arithmetic section
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Therefore there is a
number, t, such that

a =tv
We now have,

r = r0 + t v = x0 , y0 , z0 + t a, b, c
This is called the vector form of the equation of a line
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Notice that t v will be a vector that lies along the line and it tells us how far from
the original point that we should move
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As t varies over all possible values we will
completely cover the line
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There are several other forms of the equation of a line
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r = x0 , y0 , z0 + t a, b, c
x, y, z = x0 + ta, y0 + tb, z0 + tc
The only way for two vectors to be equal is for the components to be equal
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Notice as well that
this is really nothing more than an extension of the parametric equations we’ve seen previously
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To get a point on the line all we do is pick a t and plug into either form of the line
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There is one more form of the line that we want to look at
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We
can then set all of them equal to each other since t will be the same number in each
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If one of a, b, or c does happen to be zero we can still write down the symmetric equations
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In this case t will not exist in the parametric equation for y and
so we will only solve the parametric equations for x and z for t
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Example 1 Write down the equation of the line that passes through the points ( 2, −1,3) and

(1, 4, −3)
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Solution
To do this we need the vector v that will be parallel to the line
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In general, v won’t lie on the line itself
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All we need to do is let v be the vector that starts at the second point and ends at the first point
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So,

v = 1, −5, 6
Note that the order of the points was chosen to reduce the number of minus signs in the vector
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Once we’ve got v there really isn’t anything else to do
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We’ve got two and so we can use either one
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Here is
the vector form of the line
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Here are the parametric equations of the
line
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y = −1 − 5t
z = 3 + 6t

x − 2 y +1 z − 3
=
=
1
6
−5
Example 2 Determine if the line that passes through the point ( 0, −3,8 ) and is parallel to the
line given by x = 10 + 3t , y = 12 t and z = −3 − t passes through the xz-plane
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Solution
To answer this we will first need to write down the equation of the line
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We know that the new line must be parallel to the line given
by the parametric equations in the problem statement
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Now recall that in the parametric form of the line the numbers multiplied by t are the components
of the vector that is parallel to the line
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The equation of new line is then,

r = 0, −3,8 + t 3,12, −1 = 3t, −3 +12t,8 − t
If this line passes through the xz-plane then we know that the y-coordinate of that point must be
zero
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If
we can, this will give the value of t for which the point will pass through the xz-plane
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To get the complete coordinates of the point all we

1
need to do is plug t = into any of the equations
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4
1
3 31
⎛1⎞
⎛1⎞
r = 3 ⎜ ⎟ , −3 + 12 ⎜ ⎟ ,8 −
= , 0,
4
4
4
⎝4⎠
⎝4⎠
Recall that this vector is the position vector for the point on the line and so the coordinates of the

⎛ 3 31 ⎞
point where the line will pass through the xz-plane are ⎜ , 0, ⎟

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