b
Given

Lets,

\(\vec{A} =\hat{i}+\hat{j}+\sqrt{2} \hat{k}\)

\(\vec{B}(z-\text { axis }) =\hat{k}\)

We know formula

\(\vec{A} \cdot \vec{B}=|\vec{A}||\vec{B}| \cos \theta\)

\(\cos \theta=\frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|}=\frac{(0+0+\sqrt{2})}{\left(\sqrt{1^2+1^2+\sqrt{2}^2}\right) \cdot \sqrt{1^2}}\)

\(=\frac{\sqrt{2}}{\sqrt{4}}=\frac{1}{\sqrt{2}}\)

\(\cos \theta =\frac{1}{\sqrt{2}}\)

\(\theta =\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ}\)