Linear algebra problem with a probable typo

[Portuga]

Homework Statement

Find $m \in \mathbb{R}$ so that the following linear operator in $\mathbb{R}^3$ be an isometry: $$F\left(x,y,z\right)=\left(\frac{1}{\sqrt{3}}x+\frac{1}{\sqrt{3}}y+mz,\frac{-1}{\sqrt{6}}x+\frac{2}{\sqrt{6}}y-\frac{1}{\sqrt{6}}z,-\frac{1}{\sqrt{2}}x+\frac{1}{\sqrt{2}}z\right).$$

Relevant Equations

If $F$ is an isometry, then $\left\Vert F\left(u\right)\right\Vert =\left\Vert u\right\Vert$.

Well, my guess is that there is something wrong with the factors chosen, because $\left\Vert \left(0,1,0\right)\right\Vert =1$ and
\begin{align}
\left\Vert F\left(0,1,0\right)\right\Vert &=\left\Vert \left(\frac{1}{\sqrt{3}}\left(0\right)+\frac{1}{\sqrt{3}}\left(1\right)+m\left(0\right),\frac{-1}{\sqrt{6}}\left(0\right)+\frac{2}{\sqrt{6}}\left(1\right)-\frac{1}{\sqrt{6}}\left(0\right),-\frac{1}{\sqrt{2}}\left(0\right)+\frac{1}{\sqrt{2}}\left(0\right)\right)\right\Vert \\&=\left\Vert \left(\frac{1}{\sqrt{3}},\frac{2}{\sqrt{6}},0\right)\right\Vert \\&=\sqrt{\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{2}{\sqrt{3}}\right)^{2}}\\&=\sqrt{\frac{1}{3}+\frac{4}{3}}\\&=\sqrt{\frac{5}{3}}.
\end{align}
So, am I wright? I mean, it looks like it was some kind of a typing error.

 

[fresh_42]

Have you checked some other vectors if you replace $\dfrac{2}{\sqrt{6}}y$ by $\dfrac{2}{\sqrt{3}}y$ in the second component of $F$?

Another possibility could be a different norm.

 

[FactChecker]

I think that you have a mistake in the third line of your calculation. The y term should be $(2/\sqrt{6})^2$, not $(2/\sqrt{3})^2$

 

[Portuga]

Oh, thanks!

 

[PeroK]

Oh, thanks!

It would have been a good idea to write $F$ as a matrix:
$$F \ \dot = \ \begin{bmatrix}
\frac 1 {\sqrt 3}&\frac 1 {\sqrt 3}&m\\
-\frac 1 {\sqrt 6}&\frac 2 {\sqrt 6}&-\frac 1 {\sqrt 6}\\
-\frac 1 {\sqrt 2}&0&\frac 1 {\sqrt 2}
\end{bmatrix}$$And then things are obvious, hopefully.

 

[Portuga]

Yes, now I got the point.