Vector equation for line
Find a vector equation for the line through the point P = (–4, –1, 1) and parallel to the
vector v = (1, 4, 3). Assume r(0) = –4i –1j +1k and that v is the velocity vector of the
line.
z
vt
Q
v
P
r(0)
r(t)
y
x
Solution
From the above figure, let t be a scaling factor such that the direction vector vt
(red line) will be the displacement from point P to any point Q on the line (blue line). Let
the vectors r(0) and r(t) be the position vectors of point P and Q, respectively. By using
vector addition, we can derive vector r(t) as
r(t) = r(0) + vt
(1)
Substitute r(0) = –4i – j + k and v = i + 4j + 3k into equation (1), we will have the vector
equation for the line through the point P = (–4, –1, 1) and parallel to the vector v = (1, 4,
3) as
r(t) = (–4i – j + k) + (i + 4j + 3k)t
= (t – 4)i + (4t – 1)j + (3t + 1)k
Note that if we differentiate r(t) with respect to t, we will have
dr t  d
 r 0   vt   v  velocity vector of the line
dt
dt
Ans