If one wants to parametrize a straight line segment in $\mathbb{R}^3$, which goes from $(1,0,0)$ to $(0,1,\pi/2)$, would this approach be correct?
First, we come up with the $xy$-plane equation, which would be $y=1-x$, and then we come up with, say, the $yz$-plane equation of the line, which appears to be $z=\frac{\pi}{2} y$. Next, we simply take $x=t$, $y=1-t$, $z=\frac{\pi}{2}(1-t)$, according to the two plane level curves of the line above.
Conceptually, I think, this should be true, just want to make sure, as sometimes what seems true may contain subtle flaws.
A vector that goes from $\hat{p_1} = (1,0,0)$ to $\hat{p_2}= (0,1,\pi/2)$ is $\hat{v}= \hat{p_2}-\hat{p_1}= (0-1,1-0,\frac{\pi}{2}-0) = (-1,1,\frac{\pi}{2}).$ So if $t$ is a paramenter such that $t\in\mathbb{R},$ then a parametric vector equation in $t$ is $\hat{L}=\hat{p_1}+t\hat{v}$ or $\hat{L} = (1,0,0)+t(-1,1,\frac{\pi}{2}).$
The parametric equations can be obtained from $\hat{L}$ since $\hat{L} = (x(t),y(t),z(t)),$ so we have $x(t) = 1-t, y(t) = t, z(t)=t\frac{\pi}{2}.$
I hope this helps.