The vector $\overline{\mathrm{b}}=3\overline{\mathrm{i}}+4\overline{\mathrm{k}}$   is to be written as the sum of a vector $\overline{{\mathrm{b}}_1}$   parallel to  $\overline{\mathrm{a}}=\overline{\mathrm{i}}+\overline{\mathrm{j}}$ and a vector  ${\overline{\mathrm{b}}}_2$ perpendicular to $\overline{\mathrm{a}}$  then ${\overline{\mathrm{b}}}_1$  is equal to

$$\begin{gathered} \overline{\mathrm{b}}=\overline{{\mathrm{b}}_1}+\overline{{\mathrm{b}}_2} \\ \overline{{\mathrm{b}}_1} is parallel to \overline{a} \Rightarrow \overline{b_1}=\lambda \overline{a} and \\ \overline{ b_2} is perpendicular to \overline{a} \Rightarrow \overline{b_2}.\overline{a}=0 \end{gathered}$$

then $\overline{\mathrm{b}}=\lambda \overline{a}+\overline{{\mathrm{b}}_2}$

take dot product with $\overline{a}$

$$\begin{gathered} \overline{b}.\overline{a}=\lambda \left(\overline{a}.\overline{a}\right)+\overline{b_2}.\overline{a} \\ 3=\lambda {\left|\overline{a}\right|}^2 where \left|\overline{a}\right|=\sqrt{2} \\ \lambda =\frac{3}{2} \end{gathered}$$

$$\begin{gathered} \overline{b_1}=\lambda \overline{a} so, \\ \overline{b_1}=\frac{3}{2}\left(\overline{i}+\overline{j}\right) \end{gathered}$$