The question is to look for vector equation for the tangent line to the curve of intersection of two given cylinders at a point $(3,4,2)$.
$$x^2+y^2=25$$ $$z^2+y^2=20$$. I found the value of $t$ to be $3$. I found general parametric equation to be $$x=t$$ $$y=\sqrt{25-x^2}$$ $$z=\sqrt{t^2-5}$$ and took derivative of them. $$t=3$$ $$\frac{dx}{dt}=1$$ $$\frac{dy}{dt}=\frac{-t}{\sqrt{25-t^2}}=\frac{-3}{4}$$ $$\frac{dz}{dt}=\frac{t}{\sqrt{t^2-5}}=\frac{3}{2}$$
and added the $x,y,z$ points given in the qustion and found it to be $$r(t)=(3+t)i+(4-\frac{3}{4}t)j+(2+\frac{3}{2}t)k$$
But the solution gives out a solution like this. $$r(t)=(3-4t)i+(4+3t)j+(2-6t)k$$ I can see the vectors are multiple of $-4$. I understand getting rid of denominator is acceptable but the direction of progression of t is also changed by multiplying $(-)$. Is this acceptable? or did I something wrong in solving the quesiton?
Your answer is perfectly correct. $$r(t)=(3+t)i+(4-\frac{3}{4}t)j+(2+\frac{3}{2}t)k$$
They re-parametrized it to get rid of fractions.