Infer distance from a point to a line, from the distance from a point to a plane [Stewart P793 12.4.44] I'm able to prove $44$, but how would one deduce $43$ from it without further industry, forthwith?
 $43$ seems like a reduced, 2D version of $44$? I'm not enquiring about individual proofs.

$44.$ Let $P$ be a point not on the plane that passes through the
  points $Q, R, S$.
  Show that the distance $h$ from $P$ to the plane $ = \dfrac {\left| \left( \mathbb{a} \times \mathbb{b} \right) \cdot \mathbb{p}\right| } {\left| \mathbb{a} \times \mathbb{b}\right| }$. 
  
$43 = 39$ (5th edition). Let $P$ be a point not on the line $L$ that passes through the
  points $Q,R$.
  Show that the distance $d$ from the point P to the line $L = \dfrac {\left| \mathbb{a} \times \mathbb{b} \right| } {\left| \mathbb{a}\right| }$. 
  

 A: It's just the case in which ${\rm T}$ lies in $\rm QR$. Hence $p\perp b$, and thus $|\,p\times b\,|=\lVert p\lVert\,\lVert b\lVert$. Applying the previous result, $$
d=\!\dfrac{\big|\,(a\times b)\cdot p\,\big|}{|\,a\times b\,|}
\!=\!\dfrac{\big|\,(p\times b)\cdot a\,\big|}{\lVert a\lVert\,\lVert b\lVert}
\!=\!\dfrac{\lVert p\lVert \,\lVert b \lVert\,\big|\,u\cdot a\,\big|}{\lVert a\lVert\,\lVert b\lVert}
\!=\!\dfrac{\big|\,(\lVert p\lVert u)\cdot a\,\big|}{|a|}
\!=\!\dfrac{\big|\, \lVert p\lVert \lVert a\lVert \cos(\theta)\,\big|}{|a|},
$$ where $u$ is a unit vector parallel to $b\times p$, and $\theta$ being the angle between $a$ and $u$.

Now, since $u$ and $p$ are perpendicular, and since the angle between $p$ and $a$ is $\alpha$, then it follows that the angle between $u$ and $a$ is $\tfrac\pi2+\alpha$, i.e. $\theta=\tfrac\pi2+\alpha$. Substituting that value in our last expression will result in, $$d=\dfrac{\big|\, \lVert p\lVert \lVert a\lVert\sin(\alpha)\,\big|}{|a|}=\dfrac{\big|\,a\times p\,\big|}{|a|}.\tag*{$\small\square$}$$
