The problem is as follows.
(i) $P$ be a point not on the line $L$ that passes through the points $Q$ and $R$. Show that the distance $d$ from the point $P$ to the line $L$ is
$$d = \frac{\|\vec{a}\times\vec{b}\|}{\|\vec{b}\|}$$
where $\vec{a}=\vec{QR}$ and $\vec{b}=\vec{QP}$.
(ii) Use this formula to find the distance from the point $(-2,3)$ to the line $3x-4y+5=0$.
My calculations for (i) are as follows.
Let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$.
$$\sin{\theta} = \frac{d}{\|\vec{b}\|}$$
$$d = \|\vec{b}\|\sin{\theta} = \frac{\|\vec{a}\|\|\vec{b}\|\sin{\theta}}{\|\vec{a}\|} = \frac{\|\vec{a}\times\vec{b}\|}{\|\vec{a}\|}$$
This value for $d$ is different from the value given in the problem. Have I made a calculatory mistake or is the problem's assertion incorrect?
Secondly, I don't understand (ii). Is the cross product possible for 2-dimensional vectors? If so, how is it calculated? And if it isn't, does this problem have to be solved using $d = \|\vec{b}\|\sin{\theta}$ or is there another method? Thanks in advance.
About i) You are in fact correct, as can easily be verified using by example.
About ii) In fact the cross product is usually defined on $\mathbb R^3$, but it has an equivalent in $\mathbb R^n$ for $n\geq 2$. For $n=2$ this is simply the map $(x,y)\to(y,-x)$. But you can simply embed the problem into $\mathbb R^3$ by setting a $z$ value to $0$ (obviously this does then not affect the distance).