# Distance from one vertex of a rectangular prism to the plane determined by three other vertices, without using vectors or calculus

The objective is to find the shortest distance from the point $$H$$ to the plane $$BDE$$. The prism $$ABCD.EFGH$$ has $$AB=AD=5\sqrt{2}$$ and $$AE=12$$. I think that these numbers are badly selected by the author.

The following shows my steps to solution but I feel it is too verbose, tedious, and time consuming.

It can be solved easily with either vector (dot and cross products) or calculus (minimizing a distance function of two variables), but because this topic is for students who have not learned those subjects, I insist on only using not more than Pythagorean theorem and basic trigonometry. Analytic approach is not allowed! ### Finding $$DE$$, $$EP$$ and $$HP$$ \begin{align*} DE^2 &=DH^2+EH^2\\ &=12^2 + (5\sqrt2)^2 \\ &=194\\ DE &=\sqrt{194} \end{align*}

\begin{align*} \frac{1}{HP^2} &=\frac{1}{DH^2}+\frac{1}{EH^2}\\ &=\frac{1}{122^2}+\frac{1}{(5\sqrt2)^2}\\ &=\frac{97}{3600}\\ HP &=\frac{60}{\sqrt{97}} \end{align*}

\begin{align*} DE \times EP &=EH^2\\ EP\sqrt{194} &= (5\sqrt2)^2 \\ EP &=\frac{50}{\sqrt{194}} \end{align*}

### Finding $$BD$$, $$\cos E$$, $$EQ$$ and $$PQ$$ \begin{align*} BD &= \sqrt{CD^2+BC^2}\\ &= \sqrt{(5\sqrt2)^2+(5\sqrt2)^2}\\ &= \sqrt{50+50}\\ &= 10 \end{align*}

$$\begin{gather*} BE^2+DE^2-2\times BE\times DE \cos E = BD^2 \\ 194 + 194 - 2\times 194 \cos E = 100\\ \cos E = \frac{72}{97} \end{gather*}$$

\begin{align*} EQ &= EP \sec E\\ &= \frac{50}{\sqrt{194}}\times \frac{97}{72} \\ &= \frac{2425}{36\sqrt{194}} \end{align*}

\begin{align*} PQ &= \sqrt{EQ^2-EP^2}\\ &= \sqrt{\left(\frac{2425}{36\sqrt{194}}\right)^2-\left(\frac{50}{\sqrt{194}}\right)^2}\\ &= \frac{1625}{36\sqrt{194}} \end{align*}

### Finding $$HQ$$ \begin{align*} HQ &= \sqrt{EQ^2+EH^2}\\ &= \sqrt{\left(\frac{2425}{36\sqrt{194}}\right)^2 +\left(5\sqrt{2}\right)^2}\\ &= \frac{5\sqrt{15218}}{72} \end{align*}

### Finding the altitude, of $$\triangle HPQ$$, passing through $$P$$ and finding $$HH'$$ As the badly chosen numbers make the calculation a bit complicated. The process is left behind as your excercise. The altitude is $$t = 1500/\sqrt{738073}$$.

$$HH'$$ then can be found by equating the area from two different bases. \begin{align*} PQ \times HH' &= HQ \times t\\ \frac{1625}{36\sqrt{194}} \times HH' &= \frac{5\sqrt{15218}}{72} \times \frac{1500}{\sqrt{738073}}\\ HH' &= \frac{60}{13} \end{align*}

Thus the shortest distance from the point $$H$$ to the plane $$BDE$$ is $$\tfrac{60}{13}$$. # Question

Is there any shorter way to solve it but with neither using vector nor using calculus?

• I wish I could stop the rotation, I can't get a fix on the geometry while it keeps spinning around. Anyway, if you can find an equation for the plane (say, by locating some of the vertices at the origin and on the axes of a Cartesian coordinate system), then the line from $H$ to the nearest point on the plane has to be parallel to the normal to the plane, and then all you need is a little algebra to find where that nearest point is and what the distance is. – Gerry Myerson Jun 19 '20 at 3:19
• @GerryMyerson: The spinning cube has been changed to the static one. No analytical method allowed as I mentioned in the question. :-) – Artificial Stupidity Jun 19 '20 at 3:22

It's worth noting that $$H$$ and $$A$$ are equidistant from the plane of $$\triangle BDE$$. (Consider the symmetry of the points' positions relative to the midpoint of $$\overline{DE}$$ and then to that plane.) If one can convince oneself of that first, then it might be easier to think-through the problem. But if not, that's okay, too.

Consider tetrahedron $$HBDE$$. Relative to base $$\triangle HDE$$, it has height $$|AB|$$; relative to base $$\triangle BDE$$, it has height $$|HH'|$$ (our target length). Expressing volume in two ways, we have

$$\frac13 |AB|\cdot|\triangle HDE| = \frac13|HH'|\cdot|\triangle BDE| \tag{1}$$ Therefore, writing $$M$$ for the midpoint of $$\overline{BD}$$, $$|HH'|=\frac{|AB|\cdot|\triangle HDE|}{|\triangle BDE|} = \frac{|AB|\cdot\tfrac12|HD||HE|}{\tfrac12|BD||ME|} = \frac{|AB||AD||AE|}{|BD||ME|} = \frac{|AB|^2|AE|}{|BD||ME|}\tag{2}$$ (Note that you'd get the same formula for $$|AA'|$$, with $$A'$$ the projection of $$A$$ onto the plane, by considering tetrahedron $$ABDE$$, which has the same volume as $$HBDE$$.) By Pythagoras, we have \begin{align} |BD|^2 &= |AB|^2+|AD|^2 \\[4pt] &= 2|AB|^2 \tag{3}\\[4pt] |ME|^2 &= |BE|^2-\left(\tfrac12|BD|\right)^2 \\ &= |AB|^2+|AE|^2-\tfrac12|AB|^2 \\ &= \tfrac12|AB|^2+|AE|^2 \tag{4} \end{align}

Substituting the values $$|AB|=|AD|=5\sqrt2$$ and $$|AE|=12$$, we find $$|BD|^2 = 100 \quad\to\quad |BD|=10 \qquad\qquad |ME|^2 = 169\quad\to\quad |ME|=13 \tag{5}$$ (These make me think that the problem's given lengths are not so "badly selected".) Therefore,

$$|HH'| = \frac{600}{130}=\frac{60}{13}\tag{\star}$$

which agrees with the value derived by OP. $$\square$$

We could have calculated $$|\triangle BDE|$$ via Heron's formula. Alternatively (but equivalently), we could invoke de Gua's Theorem, the dimensionally-enhanced Pythagorean Theorem for right-corner tetrahedra that deserves to be better-known: $$|\triangle BDE|^2 = |\triangle ABD|^2 + |\triangle BAE|^2 + |\triangle BDA|^2 \tag{6}$$ so that \begin{align} |\triangle BDE|^2 &=\left(\tfrac12|AB||AD|\right)+\left(\tfrac12|AB||AE|\right)^2+\left(\tfrac12|AB||AD|\right)^2 \tag{7}\\[4pt] &=\tfrac14\left(|AB|^2|AD|^2+|AB|^2|AE|^2+|AB|^2|AD|^2\right) \tag{8} \\[4pt] &=\tfrac14|AB|^2|AD|^2|AE|^2\left(\frac{1}{|AB|^2}+\frac{1}{|AD|^2}+\frac{1}{|AE|^2}\right) \tag{9} \end{align} and thus, from $$(2)$$,

$$|HH'| = \frac{1}{\sqrt{\dfrac{1}{|AB|^2}+\dfrac{1}{|AD|^2}+\dfrac{1}{|AE|^2}}} \tag{\star\star}$$

which gives the same value of $$60/13$$.

An analytical addendum: With $$A$$ at the origin, and $$B$$, $$D$$, $$E$$ along the $$x$$-, $$y$$-, $$z$$-axes, the plane of $$\triangle BDE$$ has intercept-intercept-intercept form $$\frac{x}{|AB|}+\frac{y}{|AD|}+\frac{z}{|AE|}=1 \tag{10}$$ Thus, the distance from the plane to $$(x,y,z)$$ is given by $$\frac{\left|\dfrac{x}{|AB|}+\dfrac{y}{|AD|}+\dfrac{z}{|AE|}-1\right|}{\sqrt{\dfrac{1}{|AB|^2}+\dfrac{1}{|AD|^2}+\dfrac{1}{|AE|^2}}} \tag{11}$$ Substituting $$(x,y,z)\to A=(0,0,0)$$ or $$(x,y,z)\to H=(0,|AD|,|AE|)$$ gives $$(\star\star)$$.