Find angle between diagonal of a cube side & diagonal of one of its faces

The problem states "The angle, in degrees, between the diagonal of a cube size length 3 units and the diagonal of one of its faces is: "

I started off using this

$\displaystyle

\begin{array}{l}

\cos ^2 (\alpha ) + \cos ^2 (\beta ) + \cos ^2 (\gamma ) = 1 \\

= \left( {\frac{{v_1 }}{{||\underline v ||}}} \right)^2 + \left( {\frac{{v_2 }}{{||\underline v ||}}} \right)^2 + \left( {\frac{{v_3 }}{{||\underline v ||}}} \right)^2 = 1 \\

\end{array}

$

hypotenuse of sides is $\displaystyle {\sqrt {18} }$

magnitude of hypotenuse for diagonal of one of it sides is given by

getting $\displaystyle ||\underline w || = \sqrt {(\sqrt {18} )^2 + 3^2 } = \sqrt {27} $

From this I did this

$\displaystyle

\begin{array}{l}

\cos \theta = \frac{{\sqrt {18} }}{{\sqrt {27} }} \\

\theta = 35.3\deg \\

\end{array}

$

It appears to be the correct answer. However is this the correct way to do this via vectors