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