Given the vector m = 1+2j and n = 3j, find (i) The angle between m and n, (ii) The projection of m on n.

The vectors can be written as $\vec{m} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\vec{n} = \begin{pmatrix} 0 \\ 3 \end{pmatrix}$.

$$\vec{m} \cdot \vec{n} = (1)(0) + (2)(3) = 0 + 6 = 6$$

$$|\vec{m}| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$$ $$|\vec{n}| = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$$

The formula is $\cos \theta = \frac{\vec{m} \cdot \vec{n}}{|\vec{m}| |\vec{n}|}$.

$$\cos \theta = \frac{6}{\sqrt{5} \cdot 3} = \frac{2}{\sqrt{5}}$$ $$\theta = \arccos\left(\frac{2}{\sqrt{5}}\right)$$ $$\theta \approx 26.57^\circ$$

The angle between $\vec{m}$ and $\vec{n}$ is $\boxed{\mathbf{26.57^\circ}}$.

The scalar projection of vector $\vec{A}$ on vector $\vec{B}$ is given by $\text{comp}_{\vec{B}} \vec{A} = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$.
Using the values from previous steps:

$$\text{comp}_{\vec{n}} \vec{m} = \frac{\vec{m} \cdot \vec{n}}{|\vec{n}|} = \frac{6}{3} = 2$$

The projection of $\vec{m}$ on $\vec{n}$ is $\boxed{\mathbf{2}}$.