Height of the tetrahedron through Euclidean geometry

Question

The edge lengths of a tetrahedron ABCD are:

AD=BD=CD=1; AB=2sin(a/2), BC=2sin(b/2), CA= 2sin(c/2) where a,b and c are angles ADB,BDC and CDA respectively.

What will be the height of tetrahedron from vertex A?

I know this can be done easily for any triangle in plane 2D geometry by drawing a perpendicular from the desired corner and applying the Pythagoras theorem. Following an analogous method I considered a perpendicular AE of length "h" from A and joined B,D,C to "E" and applied Pythagoras theorem to form three equations but couldn't progress any further. I did try to formulate a fourth equation via summation of volumes of the tetrahedrons ABDE, ADCE and BECA and for that I used Heron's formula at the base BCD but it made the situation very complex and hard to solve. I am looking for a solution via a geometric construction similar to that of finding height in a triangle, if possible.

Answer 1

Allow me first to rename your angles in a logical manner in the following way :

$AB=2 \sin(c/2), BC=2 \sin(a/2), CA= 2 \sin(b/2)$ where a,b and c are angles $BDC,BDC$ and $CDA$ respectively.

(for example I don't find notation $AB=2 \cos(a/2)$ very logical because if we have two letters $A$ and $B$, the third one must be the missing letter $c$).

The volume $V$ of tetrahedron $DABC$ is equal to

$$V=\frac16 \det M \ \ \ \text{where} \ \ \ M:=\pmatrix{|&|&| \\ \vec{DA}&\vec{DB}&\vec{DC}\\|&|&|}$$

Squaring, we get :

$$36 V^2=\det(M)^2=\det(M^T)\det(M)=\det(M^TM)$$

i.e., the determinant of all dot products in this way :

$$36 V^2 = =\begin{vmatrix}\vec{DA}.\vec{DA}& \vec{DA}.\vec{DB}&\vec{DA}.\vec{DC}\\ \vec{DB}.\vec{DA}& \vec{DB}.\vec{DB}&\vec{DB}.\vec{DC}\\ \vec{DC}.\vec{DA}& \vec{DC}.\vec{DB}&\vec{DC}.\vec{DC}\end{vmatrix}$$

$$36 V^2=\begin{vmatrix}1&\cos c&\cos b\\ \cos c & 1 & \cos a\\ \cos b & \cos a & 1\end{vmatrix}$$

$$36 V^2=1+2\cos a \cos b \cos c -(\cos^2a +\cos^2 b + \cos^2 c) \tag{1}$$

Besides, volume $V$ can be computed as :

$$V=\frac13 h \times \text{area(DBC)}=\frac13 h \times \frac12 1 \times 1 \sin a$$

Squaring the previous relationship, we get :

$$36 V^2=h^2 \sin^2 a \tag{2}$$

The value of altitude's length $h$ is obtained from the comparison of (1) and (2) :

$$h^2=\frac{1}{\sin^2 a}(1+2\cos a \cos b \cos c -(\cos^2a +\cos^2 b + \cos^2 c))$$

Answer 2

In this particular case you don't need the general solution. First find the radius of the circumcircle around $\triangle ABC$ as $$ R=\frac{abc}{4[ABC]}, $$ where $a,b,c,$ are side lengths of the triangle $ABC$, and $[ABC]$ is its area, so that the tetrahedron volume can be computed as $$ V=\frac13[ABC]\sqrt{1-R^2} $$ and the height in question is $$ h_A=\frac{3V}{[BCD]}. $$ The triangle areas can be computed by Heron's formula.

Answer 3

Let $D$ be at the origin, then let

$ A = (0, 0, 1) $

$ B = (\sin a, 0, \cos a )$

$ C = ( \cos \phi \sin c , \sin \phi \sin c , \cos c )$

where $ \phi $ is given by

$ \cos b = B \cdot C = \sin a \sin c \cos \phi + \cos a \cos c $

Now we have to find the distance between $A$ and the plane $BCD$ whose equation is determined by its normal vector

$ N = BC \times BD = (C - B) \times (D - B) = B \times C $

$ N = ( - \cos a \sin \phi \sin c , \cos a \cos \phi \sin c, \sin a \sin \phi \sin c ) $

And plane $BCD$ passes through the origin, therefore the distance between $A$ and plane $BCD$ is

$ d = \dfrac{ | \sin a \sin \phi \sin c | }{ \sqrt{ N \cdot N } } = \dfrac{ | \sin a \sin \phi \sin c |}{ \sqrt{\sin^2 c \cos^2 a + \sin^2 a \sin^2 \phi \sin^2 c }} $

which simplifies to

$ d = \dfrac{ | \sin a \sin \phi | }{ \sqrt{ \cos^2 a + \sin^2 a \sin^2 \phi } }$

Answer 4

After reading through others' responses and spending some more time on the problem I have come up with an explanation of my own for this question which I posted and am leaving it as an answer over here for public knowledge and introspection.

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The pictorial representation of the situation is: We will proceed by constructing perpendiculars from point E to BD and CD with the points of intersections being F and G respectively, and joining ED. Also joining AF and AG. With all the constructions done,we can do the labeling for required elements.
Let: $$ \angle EDF=\alpha $$ $$ \angle EDG=c-\alpha $$ $$ \angle ADE= \theta $$ $\therefore \frac{\pi}{2}-\theta$ is also the angle AD make with the plane's normal, i.e. AE. Notice that: $$ cos(a)=cos(\theta)\cdot cos(\alpha) $$ $$ cos(b)=cos(\theta)\cdot cos(c-\alpha) $$ using the above two results we can derive $sin(\theta)$ in terms of angles $a,b,c$, which is essentially the height of the tetrahedron. Now, $$ \frac{cos(a)}{cos(b)}=\frac{cos(\alpha)}{cos(c-\alpha)} $$ $$ \frac{cos(a)}{cos(b)}=\frac{cos(\alpha)}{cos(c)\cdot cos{\alpha}+sin(c)\cdot sin(\alpha)} $$ $$ \frac{cos(b)}{cos(a)}=cos(c)+sin(c)\cdot tan(\alpha) $$ $$ tan(\alpha)=\frac{cos(b)}{sin(c)\cdot cos(a)}-\frac{cos(c)}{sin(c)} $$ we can now apply the identity $tan^2(x)+1=sec^2(x)$ to find out $cos(\alpha)$;

$$ cos(\alpha)=\frac{cos(a)\cdot sin(c)}{\sqrt{cos^2(b)+cos^2(a)-2cos(a)\cdot cos(b)\cdot cos(c)}} $$

but we also know that $cos(\theta)=\frac{cos(a)}{cos(\alpha)}$ $$ \therefore cos(\theta)=\frac{\sqrt{cos^2(a)+cos^2(b)-2cos(a)\cdot cos(b)\cdot cos(c)}}{sin(c)} $$

$\therefore$ using $sin^2(x)=1-cos^2(x)$, we get: $$ (AE)^2=h^2=sin^2(\theta)=\frac{1-cos^2(a)-cos^2(b)-cos^2(c)+2cos(a)\cdot cos(b)\cdot cos(c)}{sin^2(c)} $$