# solution-verification Determine the position of M

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

enter image description here

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!

• Yes there exists M' which is M mirrored through BD
– Gwen
Commented May 12 at 13:25
• @Gwen But how do i demonstrate that $A'OM'=90$ I tried but could not...
– user1244903
Commented May 12 at 13:28
• $AO',BD,O'M$ are mutually perpendicular to each other. We can imagine them to be as 3 dimensional axes.Take $O'$ as the origin. Say $OM$ is the $x$ axis. Then if $M$ exists on it at position (8,0,0) then $M'$ must exist at (-8,0,0) which lies beyond the prism
– Gwen
Commented May 12 at 13:35
• @Gwen I dont really know much about coordinate geometry...Is there any other way to explain it? Thank you!
– user1244903
Commented May 12 at 13:56
• Okay I'm explaining it in 2d coordinate system. In the answers
– Gwen
Commented May 12 at 14:33

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 $$COM=30°$$

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,

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

• Thank you for your answer! I see you got a different result...Where did I go wrong in my idea? Thanks again!
– user1244903
Commented May 12 at 14:44
• This is wrong. You're not taking into info the height of the prism.
– Gwen
Commented May 12 at 14:50
• @Gwen Yes, you're right. Thanks for pointing this out. I've corrected my solution. Commented May 12 at 15:20
• And I've removed my downvote :)
– Gwen
Commented May 12 at 15:21