Find the height of a parallelpiped from its side lengths and angles? I am attempting to find the height of a v-fold for a popup book. 
I've been pondering this problem, and I think the best way to abstract it is by imagining that I'm actually dealing with a full parrallelpiped (instead of the half one that actually exists) and then attempt to find the height of it.
In my research I have found one formula (here) for calculating the height:
h = ∥c∥|cosϕ|
Where c is the vector of the "vertical" side and ϕ is the angle between the vector c and the vertical. 
That all makes sense, but it's not very helpful since I don't know the angle ϕ And I don't have enough experience to know how to calculate it.

As for my level of knowledge: I am an undergraduate student, and I have not yet taken a course in 3D Geometry. I would love to see theorems (since I'm quite interested in math). But I would also really appreciate a formula/algorithm for calculating the height given the side lengths and angles between them, if that is possible.
Thank you for your time!
--Beka
 A: If you know only $c$ and the angles $\alpha = \angle(\vec{a}, \vec{c}), \, \beta = \angle(\vec{b}, \vec{c})$ and $\gamma = \angle(\vec{a}, \vec{b})$, then you can calculate the height $h$ from the tip of $\vec{c}$ to the plane spanned by $\vec{a}$ and $\vec{b}$. You do not need the magnitudes of the latter two vectors. The formula is
$$h = c \, \frac{\sqrt{\,1 + 2\, \cos(\alpha) \cos(\beta) \cos(\gamma) - \cos^2(\alpha) - \cos^2(\beta) - \cos^2(\gamma)\,}}{\sin(\gamma)}$$

A geometric solution. I will follow the notations from the picture. $A$ and $B$ are the orthogonal projections of the segment $OC$ onto the vectors $\vec{a}$ and $\vec{b}$ respectively. Therefore $$\angle \, OAC = \angle \, OBC = 90^{\circ}$$ and thus the two triangles $\Delta \, OAC$ and $\Delta \, OBC$ are right angled, so
$$OA = OC \cos\Big(\angle \, AOC \Big) = c \cos(\alpha) \,\,\, \text{ and } \,\,\, OB = OC \cos\Big(\angle \, BOC \Big) = c \cos(\beta)$$
The point $H$ is the orthogonal projection of $C$ onto the horizontal plane $OAB$. Thus, segment $CH$ is orthogonal to the plane $OAB$.  By construction, $OA \, \perp \, CA$ and $OA \, \perp \, CH$, which means that $OA$ is perpendicular to the whole plane $ACH$ and in particular $OA \, \perp \, HA$. Analogously, $OB \, \perp \, HB$.
So we want to find $h = CH$. Since $CH$ is perpendicular to the plane $OAB$, and in particular $CH \, \perp \, OH$, the triangle $\Delta\, OHC$ is right angled so by Pythagoras' theorem
$$h = CH = \sqrt{\, CH^2 - OH^2 \,} = \sqrt{\, c^2 - OH^2 \,}$$
so our goal would be to calculate $OH^2$. Since $$\angle \, OAH = 90^{\circ} = \angle \, OBH$$ the quadrilateral $OAHB$ is inscribed in a circle and $OH$ is the diameter of that circumcircle. By the law of sines
$$OH = \frac{AB}{\sin\Big(\angle \, AOB\Big)} = \frac{AB}{\sin(\gamma)}$$
and thus
$$OH^2 = \frac{AB^2}{\sin^2(\gamma)}$$
Next, apply the law of cosines to the triangle $\Delta \, OAB$
$$AB^2 \, = \, OA^2 \, + \, OB^2 \, - \, 2 \cdot OA \cdot OB\cdot \cos(\gamma)$$
$$AB^2 \, = \, c^2 \cos^2(\alpha) \, + \, c^2 \cos^2(\beta) \, - \, 2 c^2 \cos(\alpha) \cos(\beta) \cos(\gamma)$$
$$OH^2 = \frac{c^2 \big( \, \cos^2(\alpha) \, +  \cos^2(\beta) -  2 \cos(\alpha) \cos(\beta) \cos(\gamma) \,)}{\sin^2(\gamma)}$$
$$h = CH = \sqrt{c^2 - OH^3} = \sqrt{\, c^2 \, - \, \frac{c^2 \big( \, \cos^2(\alpha) \, +  \cos^2(\beta) -  2 \cos(\alpha) \cos(\beta) \cos(\gamma) \,\big)}{\sin^2(\gamma)} \,}$$
so after some term rearrangement
$$h = c \, \frac{\sqrt{\,1 + 2\, \cos(\alpha) \cos(\beta) \cos(\gamma) - \cos^2(\alpha) - \cos^2(\beta) - \cos^2(\gamma)\,}}{\sin(\gamma)}$$
A vector solution. As before, you know the angles between the three pairs of vectors, but you know only the length of vector $\vec{c}$. Then, let $\hat{a} = \frac{1}{a} \, \vec{a}$ and let $\hat{b} = \frac{1}{b} \, \vec{b}$ be the two unit vectors aligned with vectors $\vec{a}$ and $\vec{b}$ respectively.
Our next step is to construct a unit vector $\hat{e}$ that is in the plane $OAB$ and is orthogonal to $\hat{a}$. Then, vector $\hat{b}$ should be a linear combination
$$\hat{b} = \lambda \hat{a} + \mu \hat{e}$$
Then,
\begin{align}
&\hat{b}\cdot\hat{b} = \big( \lambda \hat{a} + \mu \hat{e} \big) \cdot \big( \lambda \hat{a} + \mu \hat{e} \big) =  \lambda^2 (\hat{a}\cdot\hat{a}) + 2\lambda \mu (\hat{a}\cdot \hat{e}) + \mu^2(\hat{e}\cdot \hat{e}) \\
&\hat{a} \cdot \hat{b} = \lambda (\hat{a}\cdot \hat{a}) + \mu (\hat{a} \cdot \hat{e})\\
&\hat{e} \cdot \hat{b} = \lambda (\hat{e}\cdot \hat{a}) + \mu (\hat{e} \cdot \hat{e}) 
\end{align}
so when you rewrite the dot products, having in mind that the vectors involved are unit and $\hat{a}$ and $\hat{e}$ are orthogonal, i.e. $\hat{a}\cdot \hat{e} = 0$
\begin{align}
&1 = \lambda^2 + \mu^2\\
&\cos(\gamma) = \lambda\\
&\hat{e} \cdot \hat{b} = \mu 
\end{align}
which means that $\mu = \sin(\gamma)$ and thus
$$\hat{b} = \cos(\gamma) \hat{a} + \sin(\gamma) \hat{e}$$
and consequently
$$\hat{e} = \frac{\hat{b} - \cos(\gamma) \hat{a}}{\sin(\gamma)} $$
Finally, consider the cross product
$$\hat{a} \times\hat{e} = \hat{a} \times \left(\frac{\hat{b} - \cos(\gamma) \hat{a}}{\sin(\gamma)}\right) = \frac{\hat{a} \times \hat{b}}{\sin(\gamma)}$$
which is a third unit vector, orthogonal to $\hat{a},\, \,\hat{b}$ and $\hat{e}$, i.e. to the whole plane $OAB$.
The three unit pairwise orthogonal vectors $\hat{a}, \, \hat{e} \, \frac{\hat{a} \times \hat{b}}{\sin(\gamma)}$ are linearly independent in 3D, so there unique three numbers $c_a, \, c_e, \, h$ for which the vector $\vec{c}$ can be expressed as
$$\vec{c} = c_a \hat{a} + c_e \hat{e} + h \frac{\hat{a} \times \hat{b}}{\sin(\gamma)}$$
Having in mind that since the three unit vectors $\hat{a}, \, \hat{e} \, \frac{\hat{a} \times \hat{b}}{\sin(\gamma)}$ are pairwise orthogonal, their pairwise dot products are zero, which means that we can
dot product $\vec{c}$ with each of the three vectors $\hat{a}, \, \hat{e} \, \frac{\hat{a} \times \hat{b}}{\sin(\gamma)}$ and find the coefficients:
\begin{align}
&\vec{c} \cdot \hat{a} = \left(c_a \hat{a} + c_e \hat{e} + h \frac{\hat{a} \times \hat{b}}{\sin(\gamma)}\right) \cdot \hat{a} = c_a (\hat{a}\cdot\hat{a}) + c_e (\hat{e}\cdot\hat{a}) + h \left(\frac{\hat{a} \times \hat{b}}{\sin(\gamma)} \cdot\hat{a}\right) = c_a\\ 
&\vec{c} \cdot \hat{e} = \left(c_a \hat{a} + c_e \hat{e} + h \frac{\hat{a} \times \hat{b}}{\sin(\gamma)}\right) \cdot \hat{e} = c_a (\hat{a}\cdot\hat{e}) + c_e (\hat{e}\cdot\hat{e}) + h \left(\frac{\hat{a} \times \hat{b}}{\sin(\gamma)} \cdot\hat{e}\right) = c_e\\
&\vec{c} \cdot \left(\frac{\hat{a} \times \hat{b}}{\sin(\gamma)} \right) = \left(c_a \hat{a} + c_e \hat{e} + h \frac{\hat{a} \times \hat{b}}{\sin(\gamma)}\right) \cdot \left(\frac{\hat{a} \times \hat{b}}{\sin(\gamma)} \right) = h\\
\end{align}
Hence
\begin{align}
&c_a = \vec{c} \cdot \hat{a} = c \, \cos(\alpha)\\ 
&c_e = \vec{c} \cdot \hat{e} = \vec{c} \cdot \left(\frac{\hat{b} - \cos(\gamma) \hat{a}}{\sin(\gamma)}\right) = \frac{(\vec{c} \cdot \hat{b}) - \cos(\gamma) (\hat{c}\cdot \hat{a})}{\sin(\gamma)} = \frac{c\, \cos(\beta) - c\, \cos(\gamma) \cos(\alpha)}{\sin(\gamma)}
\end{align}
Now, by the 3D Pythagoras' theorem
$$c^2 = c_a^2 + c_e^2 + h^2$$
we can express $h^2$ as
$$h^2 = c^2 - c^2 \, \cos^2(\alpha) - c^2 \, \left(\frac{\cos(\beta) -  \cos(\gamma) \cos(\alpha)}{\sin(\gamma)}\right)^2$$
and so
$$h = c \, \sqrt{ \, 1 -  \cos^2(\alpha) - \, \left(\frac{\cos(\beta) -  \cos(\gamma) \cos(\alpha)}{\sin(\gamma)}\right)^2 \, }$$
and after some rearrangement of trigonometric terms you get the same result as before.
A: I will give it a try, but remember that my math is from fifty years ago.
To determine the height of the figure in your drawing, you can ignore all of your drawing except the lines (you may call these vectors) labeled a, c, and a x b, as well as the angle, $\Phi$.  Then, drop a vertical line from the terminus of line c to line a, call it h.  You then have a right triangle.  The angle, from line c to line a is 90 degrees minus $\Phi$, call it $\Theta$.  You know the length of line c.  The length of the vertical line h is given by $h = c (sin(\Theta))$.  
I used h for the height, but you will see it used most often to designate the hypotenuse of a right triangle when studying trigonometry.  The line "labeled" a x b is only retained because it establishes the angle $\Phi$.  Back in the old days, you would either use a slide rule or a trig table for the calculation.  Your angle $\Phi$ has to be either specified or measured.  Of course, a measurement will not be very accurate.
