Cartesian equation of a straight line in space

Question

We learned that if the parametric equations of a straight line in space are $$x=at+x(A) ; y=bt+ y(A) ; z=ct+z(A)$$

Then the cartesian equation of this line is: $$(x-x(A))/a =(y-y(A))/b =(z-z(A))/c$$ Where $a$, $b$, and $c$ are the coordinates of its direction vector. But if one of them was zero, what to do? How would the equation be?

Thanks for answering.

Answer 1

If, say, $a=0$, then the Cartesian equations will be $x=x(A)$ and$$\frac{y-y(B)}b=\frac{z-z(A)}c.$$And if both $a$ and $c$ are $0$, then the Cartesian equations will be simply $x=x(A)$ and $z=z(A)$ ($y$ can be any real number).

Answer 2

If, for instance, $a = 0, b, c \neq 0$, then using your notation $x = x(A)$ and $y = bt+y(A), z = ct+z(A)$. So from $y$ $$t = \frac{y-y(A)}{b}$$ and then plugging this in $z$: $$z = \frac{c}{b}(y-y(A))+z(A) \rightarrow cy-bz-cy(A)+bz(A)=0.$$ The equation of the line is then \begin{align} & cy-bz-cy(A)+bz(A)=0 \\ & x = x(A). \end{align} You can easily generalize to the other cases. Of course $a,b,c$ cannot be zero all together, otherwise you would get just a point $(x,y,z) = (x(A),y(A),z(A))$. So you can always adapt these steps to the other cases.