Are these planes always perpendicular?

Question

The picture shows a pyramid (not necessarily a right pyramid).

$V$ is the apex of the pyramid , and $ABCD$ is it's base.

Let $\alpha$ be the plane $AVC$ and $\beta$ the plane $BVD$.

True or false: If $ABCD$ is a square, then $\alpha$ is always perpendicular to $\beta$.

Prove it (using only euclidian geometry; vectors/coordinates are not allowed).

My thought process so far:

I first thought since $ABCD$ is a square, then $AC$ is perpendicular to $BD$. But this doesn't guarantee $\alpha$ is perpendicular to $\beta$. There's no theorem saying that if a line in a plane is perpendicular to a line in another plane, then the planes are perpendicular.

If $V$ were guaranteed to be straight over the center of $ABCD$, then it would be easy to prove the proposition is true. The intersection line between $\alpha$ and $\beta$ would be perpendicular to the base and, since $AC$ is perpendicular to $BD$, the planes would be perpendicular to each other by definition of angle between planes.

I think I'm stuck because the pyramid could be oblique.

Answer 1

Let's just use 3D coordinates and see what happens.

WLOG, let $V = (x_0,y_0,z_0)$ s.t. $z_0 > 0$, and let $A = (-1,0,0), B = (0,-1,0), C = (1,0,0) , D = (0,1,0)$.

The equation of plane $AVC$ is: $\begin{bmatrix} x & y & z & 1 \\ -1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ x_0 & y_0 & z_0 & 1 \end{bmatrix} = 0$

Hence it's normal is along the vector $(0, -2z_0, 2y_0)$.

The equation of plane $BVD$ is: $\begin{bmatrix} x & y & z & 1 \\ 0 & -1 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ x_0 & y_0 & z_0 & 1 \end{bmatrix} = 0$

Hence it's normal is along the vector $(2z_0, 0, -2x_0)$.

Hence the dot product of the two normals is $0$ iff $x_0y_0 = 0$.

So no, in general. Also we can see that if the base is a square, then those planes are perpendicular only when the apex's projection lies on one of the diagonals of the square.

Answer 2

Let point $E$ be the intersection point of the diagonals $AC$ and $BD$ of the square base $ABCD$. Through points $B$ and $C$ draw lines $b$ and $c$ parallel to the line $EV$. Take the point $F$ on line $b$ such that $EF$ is perpendicular to $b$. Furthermore, take the point $G$ on line $c$ such that $EG$ is perpendicular to $c$. Since by construction $EV \, || \, b \, || \, c$, the segments $EF$ and $EG$ are perpendicular to $EV$, which means that the whole plane $(EFG)$ is perpendicular to $EV$ and therefore the angle between the planes $(BVD) = (EVB)$ and $(AVC) = (EVC)$ is the angle $\angle \, FEG$.

Since $EV \, || \, b \, || \, c$ and $EV$ is perpendicular to the plane $(EFG)$, the lines $b$ and $c$ are also perpendicular to $(EFG)$ and in particular $b$ and $c$ are perpendicular to the segment $FG$, i.e. $\angle \, BFG = \angle \, CGF = 90^{\circ}$.

Since the triangle $\Delta\, BCE$ is right-angle isosceles, $EB = EC = a$ and $BC = \sqrt{2}\, a$. By construction, $\angle \, EFB = 90^{\circ}$, so if we denote $\angle \, BEF = \alpha$, for right-angled triangle $\Delta\, BEF$ $$EF = a\,\cos(\alpha) \, \text{ and } \, BF = a\, \sin(\alpha)$$ If we denote $\angle \, CEG = \beta$, absolutely analogous arguments yield $$EG = a\,\cos(\beta) \, \text{ and } \, CG = a\, \sin(\beta)$$ If w look at the quad $BCGF$, it is a trapezoid ($EF = b \, || \, c = CG$) with $FG$ perpendicular to $BF$ and $CG$, by Pythgoras' theorem $$FG^2 = BC^2 - (BF - CG)^2 = 2a^2 - \big(a\sin(\alpha) - a\sin(\beta)\big)^2$$

Finally, if we look at the triangle $\Delta \, EFG$ we know that $$EF = a\cos(\alpha) \,\, \, EG = a\cos(\beta) \,\,\, \text{ and } \,\,\, FG^2 = 2a^2 - \big(a\sin(\alpha) - a\sin(\beta)\big)^2$$ By Pythagoras' theorem (full version) $\Delta EFG = 90^{\circ}$ if and only if $$EF^2 + EG^2 - FG^2 = 0$$ So let's write this down: $$a^2\cos^2(\alpha) + a^2\cos^2(\beta) - 2a^2 + \big(a\sin(\alpha) - a\sin(\beta)\big)^2 = 0$$ Expand and simplify $$a^2\cos^2(\alpha) + a^2\cos^2(\beta) - 2a^2 + a^2\sin^2(\alpha) + a^2\sin^2(\beta) - 2\,a^2 \sin(\alpha)\sin(\beta) = 0$$ which simplifies to $$\sin(\alpha)\sin(\beta) = 0$$ which is possible only when $\alpha = 0$ or $\beta = 0$, when the angles are between $0$ and $90^{\circ}$. So, this shows that the two planes $(AVC)$ and $(BVC)$ are perpendicular if and only if $$\angle \, BEF = 0 \,\, \text{ or } \,\, \angle \, CEG = 0$$ which means that the two planes $(AVC)$ and $(BVC)$ are perpendicular if and only if $EV$ is perpendicular to the diagonal $BD$, or $EV$ is perpendicular to the diagonal $AC$, or $EV$ is perpendicular to both, so to the whole base plane $ABCD$. This can also be reformulated as follows: The two planes $(AVC)$ and $(BVC)$ are perpendicular if and only if $VA = VC$, or $VB = VD$, or $VA = VB = VC = VD$.

Answer 3

The planes are not always perpendicular.

Imagine a point $E$, halfway between $A$ and $B$. The point $V$ can relocate anywhere along the line from $V$ to $E$ and, provided that it is not coincident with $E$, the volume would still be a pyramid with a square base.

Now take the extreme case where $V$ is very, very close to $E$, but not exactly at $E$. To convey the idea graphically, imagine the distance between $V$ and $E$ is the width of an atom... Technically, we still have a square-based pyramid, but for practical purposes it's a flat surface, like a sheet of paper.

In this scenario, the planes you are asking about, $AVC$ and $BVD$, will also lie on this sheet of paper, and would therefore be parallel (or very nearly so).