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

<br />
\begin{array}{l}<br />
 \cos ^2 (\alpha ) + \cos ^2 (\beta ) + \cos ^2 (\gamma ) = 1 \\ <br />
  = \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 \\ <br />
 \end{array}<br />

hypotenuse of sides is {\sqrt {18} }

magnitude of hypotenuse for diagonal of one of it sides is given by
getting ||\underline w || = \sqrt {(\sqrt {18} )^2  + 3^2 }  = \sqrt {27}

From this I did this

<br />
\begin{array}{l}<br />
 \cos \theta  = \frac{{\sqrt {18} }}{{\sqrt {27} }} \\ <br />
 \theta  = 35.3\deg  \\ <br />
 \end{array}<br />

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