Equation of line gives equation of plane

Question

Given a 3D line in parametric form $$x = 5 + t$$$$y = 1 +3t$$$$z = 4t$$ I did the following calculation: $$x + y + z = (5 + t) + (1 + 3t) + (4t)$$ Therefore $$x + y + z = 6 + 8t = 6 + 2(4t) = 6 + 2z$$ Finally, $$x + y - z = 6$$

But this is the equation of a plane!!

So where did I go wrong??

Answer 1

A line in $\mathbb{R}^3$ is a subset of $\mathbb{R}^3$. Your set of equations can be written as the set $\{(5+t,1+3t,4t) \in \mathbb{R}^3 : t \in \mathbb{R}\}.$ But when you write $x+y+z$, you've no longer got a set of points in $\mathbb{R}^3$, you've got a set of points in $\mathbb{R}$.

Answer 2

Your work is fine. You found the equation of a plane in which the line lies. However, not every point in the plane will lie on the line. That is, there exist $x,y,z$ such that $x+y-z=6$ but that do not satisfy your original parametric equations for any $t\in\mathbb{R}$.