Eliminating the parameter λ from (2),
we get
1
x – x
a
 =
1
1
y – y
=z – z
b
c (3)
This is the Cartesian equation of the line �Note   If l, m, n are the direction cosines of the line, the equation of the line is
1
x – x
l
 =
1
1
y – y
=z – z
m
n
Example 6 Find the vector and the Cartesian equations of the line through the point
(5, 2, – 4) and which is parallel to the vector 
ˆ
ˆ
ˆ
3
2
8
i
j
k
+
− Solution We have
ar  =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
4
and
3
2
8
i
j
k
b
i
j
k
+
−
=
+
−
r
Therefore, the vector equation of the line is
rr =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
4
( 3
2
8
)
i
j
k
i
j
k
+
−
+ λ
+
−
Now, rr is the position vector of any point P(x, y, z) on the line