The y-axis can be represented in vector form by $\hat{j}$ and $-\hat{j}$ Let $\vec{a}$√ = $\sqrt{\widehat{2i}}+\hat{j}+\hat{k}$ and $\vec{b}$ = $\hat{j}$ or $-\hat{j}$ cos θ = $\frac{\hat{a} \cdot b}{|a|^{\hat{}}|b|}$ ∴ cos θ = $\frac{\left(\sqrt{\widehat{2i}}+\hat{j}+\hat{k}\right)\cdot\left(\pm\hat{j}\right)}{\left|\sqrt{\widehat{2i}}+\hat{j}+\hat{k}\right|\left|\hat{j}\right|}$ = ± $\frac{1\times1}{\sqrt{(\sqrt{2})^2+1^2+1^2}\times\sqrt{1^2}}$ = ± $\frac{1}{\sqrt{4}}$ = ± $\frac{1}{2}$ So, the cosine of the angle which the vector $\sqrt{\widehat{2i}}+\hat{j}+\hat{k}$ makes with y axis is ± $\frac{1}{2}$