# 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 $L = \dfrac {\left| \mathbb{a} \times \mathbb{b} \right| } {\left| \mathbb{a}\right| }$.

• @Workaholic +1. Thank you for your input into an old question! – NNOX Apps Mar 27 '16 at 4:53
• You're welcome! :-) Decided to write up an answer after filling out some details. – Workaholic Mar 27 '16 at 14:52

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, $$d=\dfrac{\big|\, \lVert p\lVert \lVert a\lVert\sin(\alpha)\,\big|}{|a|}=\dfrac{\big|\,a\times p\,\big|}{|a|}.\tag*{\small\square}$$