the normal to the plane $x = z$

Question

How can I find the normal of the plane $x = z$? Since in the standard form of the equation of a plane $(ax+by+cz = d)$ one would simply say that the normal is the vector $(a,b,c)$.

So im thinking that the normal here should be $(1,0,-1)$. But I saw somewhere else that the normal for this plane should be $(1/√2)(1,0,−1)$ so now Im confused.

Answer 1

In some texts the normal is required to have length $1$. The length of $(1,0,-1)$ is $\sqrt{2}$. So, $(1/\sqrt{2},0,-1/\sqrt{2})$ is also perpendicular and has length $1$.

Answer 2

You seem to have done the correct steps in figuring out that $(1, 0, -1)$ is normal to the plane $x=z$, but it's convention to have a normal vector $\vec v$ be such that $|\vec v| = 1$.

Since the length of $(1, 0, -1)$ is $\sqrt2$, the normal vector can be written as $$\displaystyle\frac{1}{\sqrt2}(1, 0, -1) = (\frac1{\sqrt2}, 0, -\frac1{\sqrt2})$$ which has the same direction, but has length $1$.

Answer 3

In this view:

it is easy to see that $$(-1,0,1)$$ is a normal to the plane at stake.

A normal unit vector is

$$\left(-\frac1{\sqrt{2}},0,\frac1{\sqrt{2}}\right).$$