solution-verification Determine the position of M

Question

The problem

In the regular quadrilateral prism, $ABCDA'B'C'D'$ the edge of the base is equal to $4 \sqrt{6}$, and the volume is $1152$. Determine the position of the point M on the edge CC', so that the planes $(BA'D)$ and $(MBD)$ are perpendicular.

The idea

drawing

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I let O be the center of square ABCD.

If those planes are perpendicular, then the angle they form is 90.

Using the theorem of the 3 perpendicular we get that $MO\perp DB, A'0\perp BD$ and the common side of both planes $(BA'D)$ and $(MBD)$, which make the angle between them be $\angle A'OM=90$

We can calculate using the volume and the edge of the base that the lateral edge is $12$, so AA'=12 and we can also calculate $AO=4\sqrt3$, from here we get that $\angle A0A'=60=> \angle COM=30, => MC=4 $ so we determined the pos of M

The thing is I don't know if this is the only solution... I don't know if M can be outside of CC' Thank you!

Answer 1

What I was trying to say is that, $M'$ can be a point which has equal distance from $O$ as that of $M$ and will keep the triangle perpendicularity relation true. But in your replies, you say that $M'$ needs to be on $CC'$, which isn't possible.

There is only a single point on $CC'$ for which $COM=30°$

Answer 2

The height of the prism is $h = \dfrac{1152}{(4 \sqrt{6})^2} = 12 $

The ratio of the height to the base side length is $ r = \dfrac{12}{4 \sqrt{6}} = \dfrac{3}{\sqrt{6}} = \sqrt{\dfrac{3}{2}} $

So, let $A = (0,0,0), B = (1, 0, 0), C = (1,1,0), D = (0, 1, 0),\\ A' = (0,0,r), B' = (1,0,r), C' = (1,1,r), D' = (0,1,r) $

Then the normal vector to plane $BA'D$ is

$n_1 = A' B \times A' D = ( (1,0,-r) \times (0,1,-r) = (r , r, 1) $

Point $M $ is on $CC'$ , so $M = (1, 1, z) $

The normal vector to plane $MBD$ is

$n_2 = BM \times BD = ( 0,1,z ) \times ( -1,1,0 ) = (-z , -z, 1) $

If these two planes are perpendicular to each other, then the dot product $n_1 \cdot n_2 = 0 $, i.e.

$(r,r,1) \cdot (-z, -z, 1) = 0 $

Therefore,

$ - 2 r z + 1 = 0 $

Hence $z = \dfrac{1}{2 r} = \sqrt{\dfrac{1}{6}}$

The ratio of $z$ to $r$ is

$ \dfrac{z}{r} = \sqrt{ \dfrac{ 2 }{18 } } = \dfrac{1}{3} $

Which means that $M$ is one third of the way from $C$ to $C'$.

Therefore, in the actual dimensions of the prism,

$ Z = \dfrac{h}{3} = 4 $