How can we visualize $ax+by+d=0$ in 3D space parallel to z-axis?

Question

In a 3D space, it is a special case of $ax+by+cz=d,$ with $c=0$ and it represents a plane $ax+by=d$ which parallel to the z axis.

But my question is how can I visualise $ax+by=d$ parallel to z-axis in 3D? (Please depicts with 3D cube)

My approach: GBCD, FGED planes are parallel to z-axis.

Answer 1

Let $\vec{N}$ be the normal vector perpendicular to z-axis i.e. the plane $GBDC. $ And the direction cosine of $\vec{N}$ is $(a, b, c)$ i.e. $(\cos \alpha, \cos \beta, \cos \gamma).$

The direction cosine of $\vec{OZ}$ is $(\cos 90°, \cos 90°, \cos 0°).$ i.e $(0,0,1).$

Now $\vec{N}$ and $\vec{OZ}$ perpendicular to each other, therefore $\vec{N}.\vec{OZ}=0\implies(a,b,c).(0,0,1)=0\implies c=0.$ Therefore $ax+by=d$ parallel to z-axis.

The planes $GBCD, GDEF, EOAF, AOBC$ are parallel z-axis which represents $ax+by=d.$

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Special notes to clear the basic concepts.

We could draw 2 normals $\vec{N}$ of z-axis on the above plane $GBCD, GDEF$

On the plane GBDC, $\vec{N}{\perp}z-axis$, $\vec{N}{\parallel}x-axis$ and $\vec{N}{\perp}y-axis.$And in this case the direction cosine of $\vec{N}$ is $(1,0,0).$

On the plane $GDEF$, $\vec{N}{\perp}z-axis$, $\vec{N}{\parallel}y-axis$ and $\vec{N}{\perp}x-axis.$And in this case the direction cosine of $\vec{N}$ is $(0,1,0).$

Conclusion:1 Normal vector $\vec{N}$ parallel to any coordinate axis $\implies$ perpendicular to remaining 2 coordinates axis.

Again plane $GBDC{\perp} x-axis\implies{\parallel} {\to} xz-axis$

Answer 2

Here, maybe I can help you to clear this doubt.

So, here is what I want you to do, go to the following 3D- graphing calculator(geogebra).

https://www.geogebra.org/3d

Use your cursor and move the set of axes, such that you only see the $y$-axis and the $x$-axis(the $z$-axis should be pointing towards you, ie pointing outside the screen)

Now type the equation you stated with fixed numbers instead of variables(only change the coefficient of $z$, what do you notice?

As $c$ approaches $0$, what happens to the plane created, you do even see a plane anymore from your perspective or just a line? How does $c$ affect the plane? (Does it rotate or move it somehow?)

When $c = 0$, does it interact with the $z$-axis when you change the other coefficients?

Answer 3

What @Shooting Stars is trying to explain through his images about how the plane is parallel to the z-axis, not necessarily count that it intersects it. Note that the z-axis is the vertical axis.

Consider the function's derivate, where the function is given by $f(x, y ,z) = 4x + 10y + 0z$, and its derivative with respect to the z-axis is $\frac{d(f(x))}{dz} = 0$, because the plane is not increasing or decreasing with respect to the z-axis. Alternatively, it can be said that it is travelling parallel to the z-axis. Also, it intersects the z-axis for infinite solutions because, note that there are also no translations applied to in the x or y direction. Being parallel to z-axis and intersecting the z-axis are not necessarily the same thing in 3D space. They can both occur or either one can occur. Basically, they can be mutually exclusive.

Secondly, $f(x, y ,z)$ intersects the z-axis on infinite solutions, also because $f(0, 0, a), a \in R$ will always result in an intersection on the z-axis. Consider, at the z-axis, the x, y coordinates are $0$.