The line through $(3, 4, -1)$ normal to $x+4y-z=-2$ intersects a coordinate axis, or not?

Question

Does the Line passing through $(3, 4, -1)$ and which is normal to $x+4y-z=-2$, intersect any of the coordinate axes?

So I have found the parametric equations of this line which are:

\begin{align}x&=3+t\\ y&=4+4t\\ z&=-1-t\end{align}

Do I set these equations to zero to find if they intersect the coordinate axes? That's where I'm stuck.

Answer 1

We have

$$\begin{cases}x=3+t\\ y=4+4t\\ z=-1-t\end{cases}$$

Along each of the coordinate axes, at least two of $x,y,z$ must be $0$. So try out the three different cases $(x,0,0),(0,y,0),(0,0,z)$ and see if there is a consistent result(s) to the system in each.

Answer 2

for the $yz$ axis must $$y=z=0$$ and we get $t=-1$ therefore our intersection point is given by $S(2;0;0)$

Answer 3

Note that your parametrization can be written in the form:

\begin{align} x &= (t+1)+2\\ y &= 4(t+1)\\ z &= -(t+1) \end{align}

Note that $y(t) = \lambda z(t)$ for some scalar, while $x(t)$ and $y(t)$ or $x(t)$ and $z(t)$ do not differ only by scalar, but by a non-zero constant as well. This means that $y$ and $z$ can be simultaneously $0$, while $x$ and $y$ or $y$ and $z$ cannot.

Thus, set $y = z = 0$ and solve system

\begin{align}4(t+1)&=0\\ -(t+1) &= 0\end{align}

to get $t = -1$. Substitute it back to get $x = 2$. Thus, your line passes through point $(2,0,0)$.