Derivation
Conversion from vector form to Cartesian form
We have:
=
x
+
y
+
z
,
=
x1 + y1 + z1 and
=
x2 + y2 + z2.Substituting these values in = + m( –
), we
get:
= + m( –
), we
get:x +
y +
z =
x1 + y1 + z1 +
m[(x2 - x1) + (y2 - y1) +
(z2 - z1)]
+
y +
z =
x1 + y1 + z1 +
m[(x2 - x1) + (y2 - y1) +
(z2 - z1)]Equating the coefficients of , and , we
get:
, and , we
get:x = x1 + m(x2 – x1); y = y1 +
m(y2 – y1); z = z1 + m(z2 –
z1) . . . (i)
These are parametric equations of the line. Eliminating the parameter 'm' from (i), we get:
.
This is the Cartesian equation of the line.
.
This is the Cartesian equation of the line.