We have to use the vector dot theorem
[tex]\begin{gathered} a\circ b=\lbrack a\rbrack\lbrack b\rbrack cos\theta \\ (1,1,2)(\text{ -1, }2,k)=\sqrt{(1^2+1^2+2^2}(\sqrt{(\text{ -}1)^2+2^2+k^2}cos60 \\ \\ \text{ -1 + 2 + 2}k=\sqrt{4}(\sqrt{5+k^2})(\frac{1}{2}) \\ 1+2k=2(\frac{1}{2})(\sqrt{5+k^2}) \\ 1+2k=\sqrt{5+k^2} \\ (1+2k)^2=(\sqrt{5+k^2})^2 \\ 1+4k+4k^2=5+k^2 \\ 0=4\text{ - }4k\text{ - }3k^2 \\ 3k^{^2}+4k\text{ - }4=0 \\ (3k\text{ - }2)(k\text{ + }2)=0 \\ \\ 3k\text{ - }2=0 \\ k=\frac{2}{3} \\ \\ k+2=0 \\ k=\text{ -}2 \end{gathered}[/tex]So the values k can have for the angle to be 60º is -2 and 2/3
