Show that the shortest distance from a point $P$ to the line through $P_0$ with direction vector $\overrightarrow{d}$ is $$\frac{|\overrightarrow{P_0P}\times \overrightarrow{d}|}{|\overrightarrow{d}|}$$
I need help writing the proof for this. So far I have:
Let $u=P_oP$, the $u_1= \text{proj}_ud = \frac{(\overrightarrow{u} \circ \overrightarrow{d})d}{|\overrightarrow{d}|^2}$
So, $|\overrightarrow{u_2}| = |\overrightarrow{u}-\overrightarrow{u_1}| = \frac{u|\overrightarrow{d}|-u\overrightarrow{d}}{|\overrightarrow{d}|}$.
Let $P=(a_1,b_1),P_0=(a_2,b_2),\overrightarrow{d}=(a_3,b_3)$. Now vector $\overrightarrow{P_0P}=(a_2-a_1,b_2-b_1)$. We only need a length, so we can write it as $(a_1-a_2,b_1-b_2)$. Now apply formula for vector cross product $$\frac{|\overrightarrow{P_0P}\times\overrightarrow{d}|}{|\overrightarrow{d}|}=\frac{|(a_1-a_2,b_1-b_2)\times(a_3,b_3)|}{|(a_3,b_3)|}=\frac{\left|\left|\begin{matrix}\overrightarrow{i}&\overrightarrow{j}&\overrightarrow{k}\\a_1-a_2&b_1-b_2&0\\a_3&b_3&0\end{matrix}\right|\right|}{\sqrt{a_3^2+b_3^2}}=\frac{|a_3(b_1-b_2)-b_3(a_1-a_2)|}{\sqrt{a_3^2+b_3^2}}$$ Now let $l$ be the line $PP_0$. Slope of that line will be $k=\frac{b_3}{a_3}$. So, we have $$l:y-b_2=k(x-a_2)\implies l:b_3x-a_3y+a_3b_2-a_2b_3=0$$ Now find distance $$\text{distance}(P,l)=\frac{|b_3a_1-a_3b_1+a_3b_2-a_2b_3|}{\sqrt{a_3^2+b_3^2}}=\frac{|a_3(b_1-b_2)-b_3(a_1-a_2)|}{\sqrt{a_3^2+b_3^2}}$$ This completes the proof.