Let $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$Now, taking dot product of $\vec{a}$ with $\hat{i}$, we get$\vec{a} \cdot \hat{i} = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \cdot \hat{i} = a_1 \hat{i} \cdot \hat{i} + a_2 \hat{j} \cdot \hat{i} + a_3 \hat{k} \cdot \hat{i}$$\Rightarrow \vec{a} \cdot \hat{i} = a_1 \hat{i} \cdot \hat{i} + a_2 \cdot 0 + a_3 \cdot 0 (\because \hat{j} \cdot \hat{i} = \hat{k} \cdot \hat{i} = 0)$$\Rightarrow \vec{a} \cdot \hat{i} = a_1$Similarly, taking dot product of $\vec{a}$ with $\hat{j}$ and $\hat{k}$, we get$\vec{a} \cdot \hat{j} = a_2 \text{ and } \vec{a} \cdot \hat{k} = a_3$$\Rightarrow (\vec{a} \cdot \hat{i}) \hat{i} + (\vec{a} \cdot \hat{j}) \hat{j} + (\vec{a} \cdot \hat{k}) \hat{k} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} = \vec{a}$If $\vec{a}$ is any non-zero vector, then $(\vec{a} \cdot \hat{i}) \hat{i} + (\vec{a} \cdot \hat{j}) \hat{j} + (\vec{a} \cdot \hat{k}) \hat{k}$ equals $\vec{a}$. OR Let $\vec{a} = \hat{i} - \hat{j}$ and $\vec{b} = \hat{i} + \hat{j}$.Now, projection of vector $\vec{a}$ and $\vec{b}$ is given by,$\frac{1}{|\vec{b}|}(\vec{a} \cdot \vec{b}) = \frac{1}{\sqrt{1+1}} (1.1+(-1)(1)) = \frac{1}{\sqrt{2}} (1-1) = 0$Hence, the projection of vector $\vec{a}$ on $\vec{b}$ is 0 .