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Solution
a. Since the direction vectors are parallel—that is, 211, 2, 52 12, 4, 102—
this means that the two lines are parallel. To show that the equations are
equivalent, we must show that a point on one of the lines is also on the other
line. This is based on the logic that, if the lines are parallel and they share a
common point, then the two equations must represent the same line. To check
whether 14, 10, 212 is also on L 1 substitute into its vector equation.
14, 10, 212 11, 0, 42 s11, 2, 52
Using the x component, we find 4 1 s, or s 5. Substituting s 5
into the above equation, 11, 0, 42 511, 2, 52 14, 10, 212. This
verifies that the point 14, 10, 212 is also on L 1 .
Since the lines have the same direction, and a point on one line is also on the
second line, the two given equations represent the same line.
b. The first check is to compare the direction vectors of the two lines. Since
2Q 1 the lines must be parallel. As in part a,
2 , 1 2 , 1R 11, 1, 22,
13, 10, 122 must be a point on L 3 for the equations to be equivalent. Therefore,
13, 10, 122 11, 6, 12 l11, 1, 22 must give a consistent value of l for
each component. If we solve, this gives an inconsistent result since l 4 for
the x and y components, and 12 1 2l, or l 11 for the z component. This
2
verifies that the two equations are not equations for the same line.
Symmetric Equations of Lines in R 3
We introduce a new form for a line in R 3 , called its symmetric equation. The
symmetric equation of a line is derived from using its parametric equations and
solving for the parameter in each component, as shown below
x x 0 ta 4 t x x 0
, a 0
y y 0 tb 4 t y a
y 0
, b 0
z z 0 tc 4 t z b
z 0
, c 0
c
x x
Combining these statements gives 0
y y 0
z z 0, a, b, c 0.
a b c
These equations are called the symmetric equations of a line in R 3 .
R 3
Symmetric Equations of a Line in
x x 0
y y 0
z z 0
, a 0, b 0, c 0
a b c
where 1x 0 , y 0 , z 0 2 is the vector from the origin to a point on the line, and 1a, b, c2
is a direction vector of the line.
NEL
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