Finding a vector equation I need to find the directional derivative of a vector. However the direction of the line is given in the form $(x-1)/2 = (y-3)/-2 = z$  form. How do I find the vector equation for the direction from the above equation. The point where we need to find the directional derivative is $(1,1,1).$
See the question below 
QUESTION: Find the directional derivative of  M = 5*x^2*y -5*y^2*z + 2.5 z^2 x at the point p(1,1,1) in the direction of the line (x-1)/2 = (y-3)/-2 = z
Advanced Engineering Mathematics (B.S. Grewal) Vector calculus and it's application 
 A: $\frac{x-1}{2}=\frac{y-3}{-2}=z=t$, so $x=1+2t, y=3-2t, z=t$, so $r=(1,3,0)+t(2,-2,1)$.
A: Edit: In response to the new question.
You are given the function $M(x,y,z) = 5x^2y -5y^2z + 2.5 z^2 x$ and wish to know its directional derivative at the point $p = (1,1,1)$ in the direction of the line $(x-1)/2=(y-3)/(-2) = z$. As we discussed before, the points on this line are of the form $(1,3,0) + t(2,-2,1)$ for real values of $t$, and so the unit direction vectors for this line are $\pm u$ where
$$
    u = \frac{1}{\sqrt{1^2 + 3^2}}(1,3,0) = \frac{1}{\sqrt{10}}(1,3,0).
$$
We now compute $D_uM(1,1,1) = \nabla f(1,1,1) \cdot u$. First, note
$$
    \nabla f(x,y,z) = (f_x,f_y,f_z) = (10xy+2.5z^2,5x^2-10yz,-5y^2+5zx)
$$
so that
$$
    \nabla f(1,1,1) = (12.5,-5,0).
$$
Therefore
$$
    D_uM(1,1,1) = (12.5,-5,0)\cdot (1/\sqrt{10})(1,3,0) = (1/\sqrt{10})(12.5-15) = -2.5/\sqrt{10}
$$
or $-\sqrt{5}/2$. If they wished for you to compute the directional derivative along the opposite direction $-u$ for the given line, nothing above changes except the sign of the entries in $u$ in the last step, leaving you with $D_{-u}M(1,1,1) = \sqrt{5}/2$ instead.
