Math Is Fun Forum

Shelled

Member

Registered: 2014-04-15

Posts: 44

Vector Help

Hi I'm having trouble with a part of this problem.
http://i.imgur.com/usIuMlB.png

I've been wanting to find  the points A and C.
I've calculated:
BD = D -B
     = (24.75, -10.4392, 0)
and the magnitude of BD = 26.87

but I'm completely  lost as where to go on from here.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,635

Re: Vector Help

hi Shelled,

ABCD is a square and you know the diagonal.

So find the midpoint, E, and BE (=ED).

I notice that B and D are in the same z = 11.9312 plane.  Now, A and C may be one above and one below that plane, but it will have to be by equal amounts.

So you can use the dot product to get a constraint on AE.  I will have to try this to see what else is needed so I'll be back on this later. 

EDIT:  AC must be of the form (-10.4391K,-24.7594K, z)

and you can get the x and y components by doing AE.EB = 0 

ENDEDIT

ANOTHER EDIT:

That didn't work.  I just got 0 = 0

Need more input so I looked for AB=BC=CD=DA.

By trial I got A is approximately (77.81,34,069,15.48)

I think I need to know something about how you are expected to do this.  Is it a long project or just a quick (?) exercise in vectors ?

ENDEDIT

Once you have A and C:

Then AA' etc will be parallel to BB' with z = 0.  That should be enough to get A', C' and D'.

Thereafter it should be plain (plane ?) sailing. 

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Shelled

Member

Registered: 2014-04-15

Posts: 44

Re: Vector Help

Thank you Bob.
It's a long project. I think it's a lot clearer now, I'll post here again if I need more help.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,635

Re: Vector Help

hi Shelled,

I've been having a longer think.  This is what you might do:

ABCD must lie in a plane that contains BED, and AC will be perpendicular to BD.  So you can write down the equation of that plane.

A and C must lie on a circle, centre E and radius EB, so you can write down the equation of that circle.

(To get a square, you need AB = BC.  I think you don't need this one.)

So, you could call A (x, y, z) and form equations according to the above.  The given numbers don't make it easy to simplify these.  If I had access (which I don't) to a computer equation solver, then I could probably find those values and hence get  A and C.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Bob

Administrator

Registered: 2010-06-20

Posts: 10,635

Re: Vector Help

hi Shelled,

This is a back of the envelop analysis, done whilst on the way to the Royal Albert Hall to hear Tchaikovsky's  6th.

A and C must lie in a plane that is perpendicular to BED:

As B and D have the same z coordinate, so will E, so the z term above will be zero and the plane is vertical.

If you draw a circle, centre E, and radius BE, it will go through A and C.  This shows the view looking along BED.

But A and C can be any diametrically opposite points on this circle.  So there are an infinite number of such pairs!  AE = BE = CE = DE and AC is at 90 to BD.

But, we also know B'.  The line BB' must be perpendicular to the plane ABCD, so another equation is possible.

So, find the line of intersection of the planes.  AC must lie on it.

Then find where that line intersects the sphere

You'll get a quadratic, hence two solutions.  One will be A and the other C.  You can tell from a rough 3D sketch which is which.

Hope that helps,

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob