The y-axis can be represented in vector form by $\hat{\mathit{j}}$ and $-\hat{\mathit{j}}$
Let $\overrightarrow{\mathit{a}}$√ = $\sqrt{2}\hat{\mathit{i}}+\hat{\mathit{j}}+\hat{\mathit{k}}$ and $\overrightarrow{\mathit{b}}$ = $\hat{\mathit{j}}$ or $-\hat{\mathit{j}}$
cos θ = $\frac{\hat{\mathit{a}}.\mathit{b}}{\widehat{\left|\mathit{a}\right|}|\mathit{b}|}$
∴ cos θ = $\frac{\left(\sqrt{2}\hat{\mathit{i}}+\hat{\mathit{j}}+\hat{\mathit{k}}\right).\left(\pm \hat{\mathit{j}}\right)}{\left|\sqrt{2}\hat{\mathit{i}}+\hat{\mathit{j}}+\hat{\mathit{k}}\right||\hat{\mathit{j}}|}$ = ± $\frac{1\times 1}{\sqrt{{\left(\sqrt{2}\right)}^2+{\left(1\right)}^2+{\left(1\right)}^2}\times \sqrt{1^2}}$ = ± $\frac{1}{\sqrt{4}}$ = ± $\frac{1}{2}$
So, the cosine of the angle which the vector
$\sqrt{2}\hat{\mathit{i}}+\hat{\mathit{j}}+\hat{\mathit{k}}$ makes with y axis is ± $\frac{1}{2}$