Create a tetrahedron with corners $A=(1,2,4)$, $B = (1,0,2)$, $C = (2,1,3)$ and $D = (4,1,1)$. Determine the angle $\alpha$ between the edge $AB$ and the side $BCD$.

The first step I make is displacing the tetrahedron to place $B$ in origin. Thus we get the new coordinates $A' = (0,2,2)$, $B=(0,0,0)$, $C'=(1,1,1)$, $D'=(3,1,-1)$. Let $v_1$, $v_{2}$ and $v_3$ be the vectors from $B'$ to $C'$, $D'$ and $A'$ respectively.The crossproduct $u=v_1 \times v_2$ gives us the normal vector to the side $BCD$. Notice how the angle $\beta$ between $u$ and $v_3$ added with $\alpha$ gives a perpendicular angle (that is, $\pi/2$).

We can easily compute $\beta$: $$cos(\beta)=\frac{u \cdot v_3}{|u| |v_3|}$$

Here's the interesting part. The numerator is the triple product $v_1 \times v_2 \cdot v_3$, also the volume of the parallelepipedum formed by the vectors, is negative! More precisely, it is equal to $-\frac{1}{2\sqrt{3}}$. Notice how this yields a negative angle $\alpha$: $$\alpha + \beta = \pi/2 \leftrightarrow \alpha = \pi/2 - arccos(-\frac{1}{2\sqrt{3}}) = arcsin(-\frac{1}{2\sqrt{3}})$$ The book however says the angle should be $arcsin(\frac{1}{2\sqrt{3}})$. Have I reached the same result or how do I interpret it (geometrically) that the triple product is negative?

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    $\begingroup$ Have you checked that $v_1 \times v_2$ is indeed the normal vector to $(BCD)$ that points inside of the tetrahedron, and not outside ? $\endgroup$ – Dimitri Feb 3 '16 at 13:48
  • $\begingroup$ @Dimitri How can I do this without plotting? $\endgroup$ – Lozansky Feb 3 '16 at 15:37
  • $\begingroup$ Explicitly calculate the vector $u = v_1 \times v_2$ in cartesian coordinates. Then for instance, you can calculate the scalar product $u .BA$. Depending on the sign of this scalar product, you might have to take $v_2 \times v_1$ as a normal vector for your calculation to make sense, and it fixes your sign problem. Draw a little sketch to convince yourself. $\endgroup$ – Dimitri Feb 3 '16 at 15:43
  • $\begingroup$ @Dimitri The scalar product is -4. So I should reverse the order! I still don't understand geometrically why cross-product is anti-commutative but now I know how to fix the error. Thanks! $\endgroup$ – Lozansky Feb 3 '16 at 21:18

Since you do not know in advance if the points of the surface are order in clockwise or counterclockwise order when viewed from $AB$ in general. So it would be better to generalize the way one finds this angle, such that it gives the same result in both cases.

One way to this is by projecting $A$ on to the plane defined by $BCD$. I will define $\vec{v}$, $\vec{u}_1$ and $\vec{u}_2$ as the vector pointing from $B$ to $A$, from $B$ to $C$ and from $B$ to $D$ respectively. Then the projection of $A$ onto the plane can be found with

$$ \vec{n} = \vec{u}_1 \times \vec{u}_2, $$

$$ A_p = B + \vec{v} - \frac{\vec{v} \cdot \vec{n}}{\vec{n} \cdot \vec{n}} \vec{n}. $$

The angle $\angle ABA_p$ should then always be between 0° and 90°.


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