�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 Therefore,
ˆ
ˆ
ˆ
xi
y j
z k
+
+
 =
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
5
2
4
( 3
2
8
)
i
j
k
i
j
k
+
−
+ λ
+
−
=
$
$
(5
3 )
(2
2 )
( 4
8 )
i
j
k
+ λ
+
+ λ
+ −
− λ
$
Eliminating λ , we get
5
3
x −
 =
2
4
2
8
y
z
−
+
=
−
which is the equation of the line in Cartesian form 11