# Volume of a tetrahedron with perpendicular opposite edges that are also perpendicular to the segment joining their midpoints

In a tetrahedron (which is not necessarily regular) two opposite edges have the same length $$a$$ and they are perpendicular to each other. Moreover they are each perpendicular to a line of length $$b$$ which joins their midpoints. Express the volume of the tetrahedron in terms of $$a$$ and $$b$$ and prove your answer.

Hint: Can you imagine a more accessible related problem?.

To simplify this problem, what's the unknown here?

To re-examine this problem, we need to find the volume of the tetrahedron and to do this, we need to compute the base and height.

• What are "opposite edges" of a tetrahedron? Does it mean the pairs of edges that don't meet? Can those even be perpendicular? Commented Mar 29, 2019 at 12:49
• @DanielMcLaury Depends on what you mean by "perpendicular". Their directions can certainly be perpendicular. Commented Mar 29, 2019 at 12:51
• "In a tetrahedron (which is not necessarily regular) two opposite edges have the same length a and they are perpendicular to each other. Moreover they are each perpendicular to a line of length b which joins their midpoints." No, that's impossible. If the two edges "have the same length a and they are perpendicular to each other" then the line joining their midpoints has angle $\pi/4$ radians (45 degrees). Commented Mar 29, 2019 at 12:53
• @user247327 Consider the segment from $(-1, 0, 0)$ to $(1, 0, 0)$ and the segment from $(0, -1, 1)$ to $(0, 1, 1)$. The segment joining their midpoints goes from $(0,0,0)$ to $(0,0,1)$, and the three segments are all mutually perpendicular. At least under the definition of "perpendicular" which OP is clearly using (which doesn't require the lines to intersect). Commented Mar 29, 2019 at 13:00
• @blue interesting analogy gives a way to visualise the problem. Commented Mar 30, 2019 at 9:51

Let $$ABCD$$ be the tetrahedron, $$AB=CD=a$$, $$M$$ and $$N$$ midpoints of $$AB$$ and $$CD$$, $$MN=b$$. We know $$AB\perp CD$$ and $$AB,CD\perp MN$$. Consider planes $$\alpha,\beta$$ such that $$\alpha\supseteq AB$$, $$\alpha\parallel CD$$ and $$\beta\supseteq CD$$, $$\beta\parallel AB$$. Then $$\alpha\parallel\beta$$ and $$MN$$ is their common normal.

Project $$C,D$$ on $$\alpha$$ to $$C',D'$$, and $$A,B$$ on $$\beta$$ to $$A',B'$$. Note $$AA'=BB'=CC'=DD'=MN=b$$. Since $$AB\perp CD$$, we see that $$AB\perp C'D'$$. Also $$AB$$ and $$C'D'$$ intersect in $$M$$, and moreover $$M$$ is the midpoint of both $$AB$$ and $$C'D'$$. Thus $$AC'BD'$$ is a square of diagonal $$a$$, so of edge $$a/\sqrt 2$$. Similarly, $$A'CB'D$$ is a square of edge $$a/\sqrt 2$$. $$AB'CD'A'BC'D$$ is a cuboid with edges $$a/\sqrt 2$$, $$a/\sqrt2$$ and $$b$$, hence its volume is $$a^2b/2$$.

Now the volume of $$ABCD$$ is $$a^2b/2$$ minus volumes of $$ABCC'$$, $$ABDD'$$, $$CDAA'$$ and $$CDBB'$$. Each od these tetrahedrons have volume $$a^2b/12$$, since their bases are half of square of side $$a/\sqrt 2$$ and the height is $$b$$. So the volume of $$ABCD$$ is $$a^2b/2-4a^2b/12= a^2b/6$$.

This is the picture:

We may assume that we have a vertical edge connecting the points $$\left(0,0,\pm{1 \over2}a\right)$$ and a horizontal edge connecting the points $$\left(b,\pm{1\over2}a,0\right)$$. The symmetry plane $$z=0$$ cuts the tetrahedron into two pyramids of base area $${1\over2} a b$$ and height $${1\over2}a$$. It follows that the volume of the tetrahedron is $${1\over6} a^2 b$$.

The other answers give appropriate approaches for this simple case, but I thought I'd share a little-known generalization.

We know that, for a tetrahedron, $$\text{volume} = \frac13 \cdot \text{area of base}\cdot\text{altitude}$$ And we know from trig that we can write the area of a triangle in terms of two base sides and their included angle; thus, we have

$$\text{volume} = \frac16\cdot \text{side}\cdot\text{side}\cdot\sin(\text{included angle}) \cdot \text{altitude} \tag{\star}$$

Fun Fact: $$(\star)$$ works even for opposite "$$\text{side}$$"s, provided "$$\text{included angle}$$" is taken to mean the angle determined by the direction vectors of those sides, and "$$\text{altitude}$$" is the extent of the tetrahedron in the direction perpendicular to the two "$$\text{sides}$$"s. (For opposite edges, "altitude" is equivalent to the shortest distance between the lines containing those edges.)

In the problem at hand, the opposites sides both have length $$a$$, the included angle is $$90^\circ$$ (whose sine is $$1$$), and the altitude is $$b$$, so that the volume is $$\frac16a\cdot a\cdot 1\cdot b = \frac16a^2b$$. Easy-peasy.

We can prove $$(\star)$$ using coordinates and a determinant (which would have been overkill for the particular problem, and may be well outside the context in which the problem was posed to OP; nevertheless ...). For instance, we can take our tetrahedron's vertices to be

$$A=(0,0,a) \qquad B = (0,0,b) \qquad C = (h,c\sin\theta,c\cos\theta) \qquad D = (h,d\sin\theta,d\cos\theta)$$

Here, we have opposite side-lengths $$|AB|=|a-b|$$ and $$|CD|=|d-c|$$, and "included angle" $$\theta$$. Moreover, those edges lie in planes $$x=0$$ and $$x=h$$, so that the tetrahedron lies entirely between those planes, giving a corresponding "altitude" of $$h$$. Then we can compute $$\text{volume} = \frac16\,|\Delta|$$ where $$\Delta := \left|\begin{array}{ccc} B_x-A_x & B_y-A_y & B_z-A_z \\ C_x-A_x & C_y-A_y & C_z-A_z \\ D_x-A_x & D_y-A_y & D_z-A_z \end{array}\right| = \left|\begin{array}{ccc} 0 & 0 & b-a \\ h & c\sin\theta & c\cos\theta-a \\ h & d\sin\theta & d\cos\theta-a \end{array}\right| = h (b-a) (d-c)\sin\theta$$ In absolute value, this gives $$(\star)$$. $$\square$$

• why assume adjacent sides of the base triangle are of length a? Commented Mar 31, 2019 at 20:55
• @base10: Um ... I don't assume adjacent sides of the base triangle are of length $a$. I write specifically: "In the problem at hand, the opposites sides both have length $a$, ...". The point of the Fun Fact is that the side-side-sine-altitude formula for volume, which typically applies to any pair of adjacent sides, actually applies to any pair of opposite sides (provided the angle and altitude are interpreted appropriately).
– Blue
Commented Mar 31, 2019 at 22:43