Derivation of cartesian form from vector form
We have
1
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
,
r
xi
y j
z k a
x i
y j
z k
= + +
= + +
r
r
and 
2
2
2 ˆ
ˆ
ˆ
,
b
x i
y j
z k
= + +
r
Substituting these values in (1), we get
$
$
$
$
$
$
1
1
1
2
1
2
1
2
1
[(
)
(
)
(
) ]
xi
y j
z k
x i
y j
z k
x
x i
y
y
j
z
z k
+
+
=
+
+
+ λ
−
+
−
+
−
$
$
$
Equating the like coefficients of 
jk
i
ˆ
,ˆ
,ˆ
, we get
x = x1 + λ (x2 – x1); y = y1 + λ (y2 – y1); z = z1 + λ (z2 – z1)
On eliminating  λ, we obtain
1
1
1
2
1
2
1
2
1
x
x
y
y
z
z
x
x
y
y
z
z
−
−
−
=
=
−
−
−
which is the equation of the line in Cartesian form Example 7 Find the vector equation for the line passing through the points (–1, 0, 2)
and (3, 4, 6) Solution Let  ar  and  b
r
 be the position vectors of the point A(– 1, 0, 2) and B(3, 4, 6) Then
ˆ
ˆ
2
a
i
=−+k
r
and
ˆ
ˆ
ˆ
3
4
6
b
i
j
k
= + +
r
Therefore
ˆ
ˆ
ˆ
4
4
4
b
a
i
j
k
−
=
+
+
r
r
Let rr  be the position vector of any point on the line